Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 6 Quadratic Equations

Exercise 6.1

Answer 1.
Answer:
\( (x-8)(x+6) = 0 \)
\( (x-8) = 0 \) or \( (x+6) = 0 \)
\( x = 8 \) or \( x = -6 \)
In simple words: We break the equation into two parts. Each part equals zero. So x can be 8 or -6.

📝 Teacher's Note: Show students that when two things multiply to give zero, one of them must be zero. Like if A × B = 0, then A = 0 or B = 0.

🎯 Exam Tip: Write both solutions clearly: x = 8, x = -6. Check your answer by putting values back in the original equation.

Answer 2.
Answer:
\( (2x + 3)(3x - 7) = 0 \)
\( \implies (2x + 3) = 0, (3x - 7) = 0 \)
\( \implies 2x = -3, 3x = 7 \)
\( \implies x = -\frac{3}{2}, x = \frac{7}{3} \)
In simple words: We solve each bracket separately. The first gives x = -3/2, the second gives x = 7/3.

📝 Teacher's Note: Remind students to solve each bracket step by step. Move numbers to one side, then divide by the coefficient of x.

🎯 Exam Tip: Write fractions in simplest form. Show the steps clearly: 2x = -3, then x = -3/2.

Answer 3.
Answer:
\( 4x^2 + 16x = 0 \)
\( 4x(x + 4) = 0 \)
\( 4x = 0, (x + 4) = 0 \)
\( x = 0, x = -4 \)
In simple words: We take out the common factor 4x first. Then solve each part to get x = 0 or x = -4.

📝 Teacher's Note: Explain that factoring out the common term makes the problem easier. Show students how to spot common factors.

🎯 Exam Tip: Always look for common factors first. Write x = 0 as one solution when you factor out x.

Answer 4.
Answer:
\( 2x^2 - 3x - 9 = 0 \)
\( x^2 - \frac{3}{2}x - \frac{9}{2} = 0 \)
\( x^2 - 3x + \frac{3}{2}x - \frac{9}{2} = 0 \)
\( x(x - 3) + \frac{3}{2}(x - 3) = 0 \)
\( (x - 3)(x + \frac{3}{2}) = 0 \)
\( (x - 3) = 0, (x + \frac{3}{2}) = 0 \)
\( x = 3, x = -\frac{3}{2} \)
In simple words: We split the middle term and group terms to factor. The solutions are x = 3 and x = -3/2.

📝 Teacher's Note: Teach students to look for two numbers that multiply to give ac and add to give b in ax² + bx + c.

🎯 Exam Tip: In factoring, find two numbers that multiply to (-18) and add to (-3). Here it's (-9) and (+6).

Answer 5.
Answer:
\( 2x^2 - x - 6 = 0 \)
\( x^2 - \frac{1}{2}x - 3 = 0 \)
\( x^2 - 2x + \frac{3}{2}x - 3 = 0 \)
\( x(x - 2) + \frac{3}{2}(x - 2) = 0 \)
\( (x - 2)(x + \frac{3}{2}) = 0 \)
\( (x - 2) = 0, (x + \frac{3}{2}) = 0 \)
\( x = 2, x = -\frac{3}{2} \)
In simple words: We factor by splitting the middle term. The answers are x = 2 and x = -3/2.

📝 Teacher's Note: Show students that splitting -x into -2x + 3x/2 helps us group and factor easily.

🎯 Exam Tip: Double-check by substituting both answers back into the original equation to verify they work.

Answer 6.
Answer:
\( 5x^2 - 11x + 2 = 0 \)
\( 5x^2 - 10x - x + 2 = 0 \)
\( 5x(x - 2) - 1(x - 2) = 0 \)
\( (x - 2)(5x - 1) = 0 \)
\( (x - 2) = 0, (5x - 1) = 0 \)
\( x = 2, x = \frac{1}{5} \)
In simple words: We split -11x into -10x - x, then factor by grouping. The solutions are x = 2 and x = 1/5.

📝 Teacher's Note: Help students find that -10 and -1 multiply to give +10 and add to give -11.

🎯 Exam Tip: When factoring 5x² - 11x + 2, look for factors of 5×2 = 10 that add to -11.

Answer 7.
Answer:
\( 4x^2 - 13x - 12 = 0 \)
\( x^2 - \frac{13}{4}x - 3 = 0 \)
\( x^2 - 4x + \frac{3}{4}x - 3 = 0 \)
\( x(x - 4) + \frac{3}{4}(x - 4) = 0 \)
\( (x - 4)(x + \frac{3}{4}) = 0 \)
\( (x - 4) = 0, (x + \frac{3}{4}) = 0 \)
\( x = 4, x = -\frac{3}{4} \)
In simple words: We factor by splitting the middle term. The solutions are x = 4 and x = -3/4.

📝 Teacher's Note: Show students that -16 and +3 multiply to give -48 (which is 4×(-12)) and add to give -13.

🎯 Exam Tip: For 4x² - 13x - 12, find factors of 4×(-12) = -48 that add to -13. These are -16 and +3.

Answer 9.
Answer:
\( 25x(x + 1) = -4 \)
\( x^2 + x = -\frac{4}{25} \)
\( x^2 + x + \frac{4}{25} = 0 \)
\( x^2 + \frac{1}{5}x + \frac{4}{5}x + \frac{4}{25} = 0 \)
\( x(x + \frac{1}{5}) + \frac{4}{5}(x + \frac{1}{5}) = 0 \)
\( (x + \frac{1}{5})(x + \frac{4}{5}) = 0 \)
\( (x + \frac{1}{5}) = 0, (x + \frac{4}{5}) = 0 \)
\( x = -\frac{1}{5}, x = -\frac{4}{5} \)
In simple words: We expand the left side, move everything to one side, then factor. Both solutions are negative fractions.

📝 Teacher's Note: Remind students to expand carefully and rearrange the equation to standard form ax² + bx + c = 0 first.

🎯 Exam Tip: Always move all terms to one side to get zero on the right before factoring.

Answer 10.
Answer:
\( 10x - \frac{1}{x} = 3 \)
\( 10x^2 - 1 = 3x \)
\( 10x^2 - 3x - 1 = 0 \)
\( x^2 - \frac{3}{10}x - \frac{1}{10} = 0 \)
\( x^2 + \frac{1}{5}x - \frac{1}{2}x - \frac{1}{10} = 0 \)
\( x(x + \frac{1}{5}) - \frac{1}{2}(x + \frac{1}{5}) = 0 \)
\( (x + \frac{1}{5})(x - \frac{1}{2}) = 0 \)
\( (x + \frac{1}{5}) = 0, (x - \frac{1}{2}) = 0 \)
\( x = -\frac{1}{5}, x = \frac{1}{2} \)
In simple words: We multiply both sides by x to remove the fraction, then rearrange and factor. The solutions are x = -1/5 and x = 1/2.

📝 Teacher's Note: When equations have fractions with x in denominator, multiply through by x to clear fractions first.

🎯 Exam Tip: After clearing fractions, always check that your solutions don't make the original denominator zero.

Answer 11.
Answer:
\( \frac{2}{x^2} - \frac{5}{x} + 2 = 0 \)
\( 2 - 5x + 2x^2 = 0 \)
\( 2x^2 - 5x + 2 = 0 \)
\( x^2 - \frac{5}{2}x + 1 = 0 \)
\( x^2 - 2x - \frac{1}{2}x + 1 = 0 \)
\( x(x - 2) - \frac{1}{2}(x - 2) = 0 \)
\( (x - 2)(x - \frac{1}{2}) = 0 \)
\( (x - 2) = 0, (x - \frac{1}{2}) = 0 \)
\( x = 2, x = \frac{1}{2} \)
In simple words: We multiply by x² to clear all fractions, then factor the quadratic. The solutions are x = 2 and x = 1/2.

📝 Teacher's Note: When fractions have x² in denominator, multiply the whole equation by x² to clear all fractions at once.

🎯 Exam Tip: After multiplying by x², rearrange to standard form, then factor by splitting the middle term.

Answer 12.
Answer:
\( \sqrt{2}x^2 - 3x - 2\sqrt{2} = 0 \)
\( x^2 - \frac{3}{\sqrt{2}}x - 2 = 0 \)
\( x^2 + \frac{1}{\sqrt{2}}x - 2\sqrt{2}x - 2 = 0 \)
\( x(x + \frac{1}{\sqrt{2}}) - 2\sqrt{2}(x + \frac{1}{\sqrt{2}}) = 0 \)
\( (x + \frac{1}{\sqrt{2}})(x - 2\sqrt{2}) = 0 \)
\( (x + \frac{1}{\sqrt{2}}) = 0, (x - 2\sqrt{2}) = 0 \)
\( x = -\frac{1}{\sqrt{2}}, x = 2\sqrt{2} \)
In simple words: Even with square roots, we can factor by splitting the middle term. The solutions have square roots in them.

📝 Teacher's Note: Don't be scared of square roots in quadratic equations. The same factoring method works.

🎯 Exam Tip: Keep square roots in your final answer unless asked to rationalize. Write answers clearly with proper radical signs.

Answer 13.
Answer:
\( a^2x^2b - 3abx + 2b^2 = 0 \)
\( x^2 - \frac{3b}{a}x + 2(\frac{b}{a})^2 = 0 \)
\( x^2 - \frac{b}{a}x - 2\frac{b}{a}x + 2(\frac{b}{a})^2 = 0 \)
\( x(x - \frac{b}{a}) - 2\frac{b}{a}(x - \frac{b}{a}) = 0 \)
\( (x - \frac{b}{a})(x - 2\frac{b}{a}) = 0 \)
\( (x - \frac{b}{a}) = 0, (x - 2\frac{b}{a}) = 0 \)
\( x = \frac{b}{a}, x = 2\frac{b}{a} \)
In simple words: Even when we have letters instead of numbers, we can still factor. The solutions are x = b/a and x = 2b/a.

📝 Teacher's Note: Treat a and b as constants (fixed numbers). The same factoring rules apply with letters.

🎯 Exam Tip: Divide the whole equation by the leading coefficient first to make factoring easier.

Answer 14.
Answer:
\( x^2 - (\sqrt{2} + 1)x + \sqrt{2} = 0 \)
\( x^2 - x - \sqrt{2}x + \sqrt{2} = 0 \)
\( x(x - 1) - \sqrt{2}(x - 1) = 0 \)
\( (x - 1)(x - \sqrt{2}) = 0 \)
\( (x - 1) = 0, (x - \sqrt{2}) = 0 \)
\( x = 1, x = \sqrt{2} \)
In simple words: We split the middle term and factor by grouping. One solution is 1, the other is √2.

📝 Teacher's Note: Show students how -√2 - 1 can be split into -1 - √2 to make grouping possible.

🎯 Exam Tip: Look for ways to split the middle term so that common factors appear in the grouped terms.

Answer 15.
Answer:
\( x^2 - (\sqrt{3} + 1)x + \sqrt{3} = 0 \)
\( x^2 - x - \sqrt{3}x + \sqrt{3} = 0 \)
\( x(x - 1) - \sqrt{3}(x - 1) = 0 \)
\( (x - 1)(x - \sqrt{3}) = 0 \)
\( (x - 1) = 0, (x - \sqrt{3}) = 0 \)
\( x = 1, x = \sqrt{3} \)
In simple words: We factor by grouping terms. The solutions are x = 1 and x = √3.

📝 Teacher's Note: Notice the pattern: when we have -(√n + 1)x + √n, it often factors as (x - 1)(x - √n).

🎯 Exam Tip: Practice recognizing patterns like this. It makes factoring much faster in exams.

Answer 16.
\( 4x^2 + 4bx - (a^2 - b^2) = 0 \)
\( x^2 + bx - \frac{(a^2 - b^2)}{4} = 0 \)
\( x^2 + \frac{(a+b)}{2}x - \frac{(a-b)}{2}x - \frac{(a^2 - b^2)}{4} = 0 \)
\( x\{x + \frac{(a+b)}{2}\} - \frac{(a-b)}{2}\{x + \frac{(a+b)}{2}\} = 0 \)
\( \{x + \frac{(a+b)}{2}\}\{x - \frac{(a-b)}{2}\} = 0 \)
\( \{x + \frac{(a+b)}{2}\} = 0, \{x - \frac{(a-b)}{2}\} = 0 \)
\( x = -\frac{(a+b)}{2}, x = \frac{(a-b)}{2} \)

📝 Teacher's Note: This is factoring by grouping. First divide by 4, then split the middle term. Students often forget to simplify fractions at the end.

🎯 Exam Tip: Write each step clearly. Show the factoring by grouping method. Final answer must have both values of x.

Answer 17.
\( ax^2 + (4a^2 - 3b)x - 12ab = 0 \)
\( x^2 + 4ax - 3\frac{b}{a}x - 12b = 0 \)
\( x(x + 4a) - 3\frac{b}{a}(x + 4a) = 0 \)
\( (x + 4a)(x - 3\frac{b}{a}) = 0 \)
\( x = -4a, x = 3\frac{b}{a} \)

📝 Teacher's Note: When coefficient of x² is not 1, divide everything by that coefficient first. This makes factoring easier to see.

🎯 Exam Tip: Always divide by the coefficient of x² first. Keep fractions in simplest form. Show both solutions clearly.

Answer 18.
\( \left(x - \frac{1}{2}\right)^2 = 4 \)
\( x^2 - x + \frac{1}{4} = 4 \)
\( x^2 - x - \frac{15}{4} = 0 \)
\( x^2 + \frac{3}{2}x - \frac{5}{2}x - \frac{15}{4} = 0 \)
\( x(x + \frac{3}{2}) - \frac{5}{2}(x + \frac{3}{2}) = 0 \)
\( (x + \frac{3}{2})(x - \frac{5}{2}) = 0 \)
\( x = -\frac{3}{2}, x = \frac{5}{2} \)

📝 Teacher's Note: Perfect square equations can be solved by taking square root of both sides, but factoring method is also correct. Both methods give same answer.

🎯 Exam Tip: Expand the perfect square carefully. Make sure all fraction arithmetic is correct. Check your answer by substituting back.

Answer 19.
\( x^2 - 4\sqrt{2}x + 6 = 0 \)
\( x^2 - \sqrt{2}x - 3\sqrt{2}x + 6 = 0 \)
\( x(x - \sqrt{2}) - 3\sqrt{2}(x - \sqrt{2}) = 0 \)
\( (x - \sqrt{2})(x - 3\sqrt{2}) = 0 \)
\( (x - \sqrt{2}) = 0, (x - 3\sqrt{2}) = 0 \)
\( x = \sqrt{2}, x = 3\sqrt{2} \)

📝 Teacher's Note: When dealing with surds (square roots), factor them just like regular numbers. Split the middle term so that the factors contain the same surd terms.

🎯 Exam Tip: Keep surds in exact form. Don't convert to decimals unless asked. Write \( \sqrt{2} \) and \( 3\sqrt{2} \) as final answers.

Answer 20.
\( \frac{x + 3}{x + 2} = \frac{3x - 7}{2x - 3} \)
\( (x + 3)(2x - 3) = (3x - 7)(x + 2) \)
\( 2x^2 + 6x - 3x - 9 = 3x^2 - 7x + 6x - 14 \)
\( 2x^2 + 3x - 9 = 3x^2 - x - 14 \)
\( (3 - 2)x^2 + (-1 - 3)x + (-14 + 9) = 0 \)
\( x^2 - 4x - 5 = 0 \)
\( x^2 + x - 5x - 5 = 0 \)
\( x(x + 1) - 5(x + 1) = 0 \)
\( (x + 1)(x - 5) = 0 \)
\( (x + 1) = 0, (x - 5) = 0 \)
\( x = -1, x = 5 \)

📝 Teacher's Note: When solving rational equations, cross multiply first. Then expand both sides carefully. Common mistake is wrong signs when moving terms.

🎯 Exam Tip: Always cross multiply when fractions are equal. Check that your solutions don't make any denominator zero. Show all steps clearly.

Answer 21.
\( \frac{2x}{x-4} + \frac{2x-5}{x-3} = \frac{25}{3} \)
\( \frac{6x}{x-4} + \frac{6x-15}{x-3} = 25 \)
\( 6x(x-3) + (6x-15)(x-4) = 25(x-4)(x-3) \)
\( 6x^2 - 18x + 6x^2 - 15x - 24x + 60 = 25(x^2 - 4x - 3x + 12) \)
\( 12x^2 - 57x + 60 = 25x^2 - 175x + 300 \)
\( 13x^2 - 118x + 240 = 0 \)
\( x^2 - \frac{118}{13}x + \frac{240}{13} = 0 \)
\( x^2 - 6x - \frac{40}{13}x + \frac{240}{13} = 0 \)
\( x(x - 6) - \frac{40}{13}(x - 6) = 0 \)
\( (x - 6)(x - \frac{40}{13}) = 0 \)
\( x = 6, x = \frac{40}{13} \)

📝 Teacher's Note: For equations with multiple fractions, find common denominator or cross multiply. This creates a quadratic equation. Be very careful with arithmetic.

🎯 Exam Tip: Multiply through by LCM to clear fractions first. Check your solutions in the original equation to make sure denominators don't become zero.

Answer 22.
\( \frac{x+3}{x-2} - \frac{1-x}{x} = \frac{17}{4} \)
\( \frac{4x+12}{x-2} - \frac{4-4x}{x} = 17 \)
\( x(4x+12) - (4-4x)(x-2) = 17x(x-2) \)
\( 4x^2 + 12x - (4x - 4x^2 - 8 + 8x) = 17x^2 - 34x \)
\( 4x^2 + 12x - 4x + 4x^2 + 8 - 8x = 17x^2 - 34x \)
\( 8x^2 + 8 = 17x^2 - 34x \)
\( 9x^2 - 34x - 8 = 0 \)
\( x^2 - \frac{34}{9}x - \frac{8}{9} = 0 \)
\( x^2 - 4x + \frac{2}{9}x - \frac{8}{9} = 0 \)
\( x(x - 4) + \frac{2}{9}(x - 4) = 0 \)
\( (x - 4)(x + \frac{2}{9}) = 0 \)
\( x = 4, x = -\frac{2}{9} \)

📝 Teacher's Note: When you have fractions with different denominators, find the LCM. Multiply through to clear all fractions before solving.

🎯 Exam Tip: Always check that your solutions don't make any denominator zero in the original equation. Show the factoring step clearly.

Answer 23.
\( \frac{1}{x-2} + \frac{2}{x-1} = \frac{6}{x} \)
\( \frac{(x-1) + 2(x-2)}{(x-2)(x-1)} = \frac{6}{x} \)
\( x(x-1) + 2x(x-2) = 6(x-2)(x-1) \)
\( x^2 - x + 2x^2 - 4x = 6(x^2 - 2x - x + 2) \)
\( 3x^2 - 5x = 6x^2 - 18x + 12 \)
\( 3x^2 - 13x + 12 = 0 \)
\( x^2 - \frac{13}{3}x + 4 = 0 \)
\( x^2 - 3x - \frac{4}{3}x + 4 = 0 \)
\( x(x - 3) - \frac{4}{3}(x - 3) = 0 \)
\( (x - 3)(x - \frac{4}{3}) = 0 \)
\( x = 3, x = \frac{4}{3} \)

📝 Teacher's Note: For three fractions, combine the left side first, then cross multiply. Check that none of your solutions make any denominator zero.

🎯 Exam Tip: Combine fractions on one side before cross multiplying. Always verify your answers in the original equation. Missing this check loses marks.

Answer 24.
\( \frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{5}{6} \)
\( \frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)} = \frac{5}{6} \)
\( \frac{x^2 + 2x + 1 - (x^2 - 2x + 1)}{x^2 - x + x - 1} = \frac{5}{6} \)
\( \frac{x^2 + 2x + 1 - x^2 + 2x - 1}{x^2 - 1} = \frac{5}{6} \)
\( \frac{4x}{x^2 - 1} = \frac{5}{6} \)
\( 6(4x) = 5(x^2 - 1) \)
\( 24x = 5x^2 - 5 \)
\( 5x^2 - 24x - 5 = 0 \)
\( x^2 - \frac{24}{5}x - 1 = 0 \)
\( x^2 + \frac{1}{5}x - 5x - 1 = 0 \)
\( x(x + \frac{1}{5}) - 5(x + \frac{1}{5}) = 0 \)
\( (x + \frac{1}{5})(x - 5) = 0 \)
\( x = 5, x = -\frac{1}{5} \)

📝 Teacher's Note: Use the difference of squares formula: \( (a+b)^2 - (a-b)^2 = 4ab \). This makes the calculation much easier.

🎯 Exam Tip: Look for patterns like difference of squares to simplify. Show the cross multiplication step clearly. Keep fractions exact.

Answer 25.
\( \frac{x-1}{2x+1} + \frac{2x+1}{x-1} = \frac{5}{2} \)
\( \frac{(x-1)^2 + (2x+1)^2}{(2x+1)(x-1)} = \frac{5}{2} \)
\( \frac{(x^2 - 2x + 1) + (4x^2 + 4x + 1)}{2x^2 + x - 2x - 1} = \frac{5}{2} \)
\( \frac{5x^2 + 2x + 2}{2x^2 - x - 1} = \frac{5}{2} \)
\( 10x^2 + 4x + 4 = 10x^2 - 5x - 5 \)
\( -9x - 9 = 0 \)
\( x + 1 = 0 \)
\( x = -1 \)

📝 Teacher's Note: When you have \( \frac{a}{b} + \frac{b}{a} \), combine over common denominator \( ab \). The numerator becomes \( a^2 + b^2 \).

🎯 Exam Tip: Expand the squares carefully. Many terms will cancel when you cross multiply. Always check your solution doesn't make denominators zero.

Answer 26.
\( \frac{m}{n}x^2 + \frac{n}{m} = 1 - 2x \)
Multiply by mn
\( m^2x^2 + n^2 = mn - 2mnx \)
\( m^2x^2 + 2mnx + n^2 = mn \)
\( (mx + n)^2 = mn \)
\( mx + n = \pm\sqrt{mn} \)
\( mx = -n \pm \sqrt{mn} \)
\( x = \frac{-n \pm \sqrt{mn}}{m} \)

📝 Teacher's Note: Multiply through by mn to clear fractions. Look for perfect square patterns. The left side becomes a perfect square.

🎯 Exam Tip: Clear fractions first by multiplying by LCM. Recognize \( (mx + n)^2 \) pattern. Final answer uses ± symbol for both solutions.

Answer 27.
\( \frac{1}{(x-1)(x-2)} + \frac{1}{(x-2)(x-3)} + \frac{1}{(x-3)(x-4)} = \frac{1}{6} \)
\( \frac{(x-3)(x-4) + (x-1)(x-4) + (x-1)(x-2)}{(x-1)(x-2)(x-3)(x-4)} = \frac{1}{6} \)
\( \frac{x^2 - 3x - 4x + 12 + x^2 - x - 4x + 4 + x^2 - x - 2x + 2}{(x-1)(x-2)(x-3)(x-4)} = \frac{1}{6} \)
\( \frac{3x^2 - 15x + 18}{(x-1)(x-2)(x-3)(x-4)} = \frac{1}{6} \)
\( \frac{3(x^2 - 5x + 6)}{(x-1)(x-2)(x-3)(x-4)} = \frac{1}{6} \)
\( \frac{3(x-3)(x-2)}{(x-1)(x-2)(x-3)(x-4)} = \frac{1}{6} \)
\( \frac{3}{(x-1)(x-4)} = \frac{1}{6} \)
\( x^2 - 5x + 4 = 18 \)
\( x^2 - 5x - 14 = 0 \)
\( x^2 + 2x - 7x - 14 = 0 \)
\( x(x + 2) - 7(x + 2) = 0 \)
\( (x + 2)(x - 7) = 0 \)
\( x = -2, x = 7 \)

📝 Teacher's Note: For partial fractions addition, find common denominator first. Look for factoring patterns in the numerator. The \( (x-2) \) and \( (x-3) \) cancel out.

🎯 Exam Tip: Factor the numerator after combining fractions. Cancel common factors before cross multiplying. This makes the calculation much simpler.

Answer 28.
Answer:
\( 7x + \frac{3}{x} = 35 \cdot \frac{3}{5} \)

\( 7x^2 + 3 = \frac{178}{5}x \)

\( 7x^2 - \frac{178}{5}x + 3 = 0 \)

\( x^2 - \frac{178}{35}x + \frac{3}{7} = 0 \)

\( x^2 - 5x - \frac{3}{35}x + \frac{3}{7} = 0 \)

\( x(x - 5) - \frac{3}{35}(x - 5) = 0 \)

\( (x - 5)(x - \frac{3}{35}) = 0 \)

\( x = 5, x = \frac{3}{35} \)
In simple words: We solved the equation by making it into a standard quadratic form. Then we factored it to find two answers for x.

📝 Teacher's Note: Show students how to convert mixed numbers to improper fractions first. This makes the algebra easier. Common mistake is forgetting to multiply through by x at the start.

🎯 Exam Tip: Always write both values of x clearly. Show all steps when clearing fractions. You get marks for method even if final answer has small errors.

Answer 29.
Answer:
\( \frac{a}{x-a} + \frac{b}{x-b} = \frac{2c}{x-c} \)

\( \frac{a(x-b) + b(x-a)}{(x-a)(x-b)} = \frac{2c}{x-c} \)

\( \frac{ax - ab + bx - ab}{x^2 - ax - bx + ab} = \frac{2c}{x-c} \)

\( \frac{(a+b)x - 2ab}{x^2 - (a+b)x + ab} = \frac{2c}{x-c} \)

\( \{(a+b)x - 2ab\}(x-c) = 2c\{x^2 - (a+b)x + ab\} \)

\( (a+b)x^2 - 2abx - c(a+b)x + 2abc = 2cx^2 - 2c(a+b)x + 2abc \)

\( (a+b)x^2 - [2ab + c(a+b)]x + 2abc = 2cx^2 - 2c(a+b)x + 2abc \)

\( (a+b-2c)x^2 = (2ab + ac + bc - 2ca - 2bc)x \)

\( (a+b-2c)x^2 = (2ab - ac - bc)x \)

\( x = 0, x = \frac{(2ab - ac - bc)}{(a+b-2c)} \)
In simple words: We combined fractions on the left side, then cross-multiplied. After expanding and simplifying, we got a quadratic that factors to give two solutions.

📝 Teacher's Note: This is a complex rational equation. Teach students to find common denominator first, then cross-multiply. Go slowly through the algebra expansion.

🎯 Exam Tip: Write "x = 0" as one solution clearly. Show all algebraic steps. Check your final answer by substituting back into original equation if time allows.

Answer 30.
Answer:
\( x^2 + 2ab = (2a + b)x \)

\( x^2 + 2ab = 2ax + bx \)

\( x^2 - 2ax - bx + 2ab = 0 \)

\( x(x - 2a) - b(x - 2a) = 0 \)

\( (x - 2a)(x - b) = 0 \)

\( x = 2a, x = b \)
In simple words: We rearranged the equation into standard quadratic form. Then we factored by grouping to find the two values of x.

📝 Teacher's Note: Show students the factoring by grouping method clearly. Group the first two terms and last two terms separately, then factor out the common binomial.

🎯 Exam Tip: Write both answers clearly: x = 2a and x = b. Show the factoring steps - examiners give marks for correct method.

Answer 31.
Answer:
\( (a+b)^2x^2 - 4abx - (a-b)^2 = 0 \)

\( (a+b)^2x^2 - [(a+b)^2 - (a-b)^2]x - (a-b)^2 = 0 \)

\( (a+b)^2x^2 - (a+b)^2x + (a-b)^2x - (a-b)^2 = 0 \)

\( \{(a+b)^2x\}(x-1) + \{(a-b)^2\}(x-1) = 0 \)

\( (x-1)[(a+b)^2x + (a-b)^2] = 0 \)

\( x = 1, x = -\frac{(a-b)^2}{(a+b)^2} = -\frac{(a-b)}{(a+b)} \)
In simple words: We factored the quadratic equation by recognizing patterns in the coefficients. One solution is x = 1, and the other involves the ratio of (a-b) to (a+b).

📝 Teacher's Note: This uses the identity (a+b)² - (a-b)² = 4ab. Teach students to recognize these algebraic identities to simplify complex expressions.

🎯 Exam Tip: Write x = 1 clearly as the first solution. For the second solution, simplify the fraction properly. Show how you used algebraic identities.

Answer 32.
Answer:
\( a(x^2 + 1) - x(a^2 + 1) = 0 \)

\( ax^2 + a - a^2x - x = 0 \)

\( x^2 + 1 - ax - \frac{1}{a}x = 0 \)

\( x^2 - ax - \frac{1}{a}x + 1 = 0 \)

\( x(x - a) - \frac{1}{a}(x - a) = 0 \)

\( (x - a)(x - \frac{1}{a}) = 0 \)

\( x = a, x = \frac{1}{a} \)
In simple words: We rearranged the equation and factored by grouping. The two solutions are x = a and x = 1/a, which are reciprocals of each other.

📝 Teacher's Note: Point out to students that the solutions a and 1/a are reciprocals. This is a nice pattern that helps students remember the answer form.

🎯 Exam Tip: Write both solutions clearly: x = a and x = 1/a. Show the factoring steps. Remember that when a appears in the answer, it should be treated as a constant.

Answer 33.
Answer:
\( x^2 - x - a(a + 1) = 0 \)

\( x^2 + ax - (a + 1)x - a(a + 1) = 0 \)

\( x(x + a) - (a + 1)\{(x + a)\} = 0 \)

\( (x + a)\{x - (a + 1)\} = 0 \)

\( x = -a, x = (a + 1) \)
In simple words: We factored the quadratic equation by grouping terms. The two solutions are x = -a and x = (a + 1).

📝 Teacher's Note: Show students how to expand a(a+1) = a² + a first. Then demonstrate the factoring by grouping technique step by step.

🎯 Exam Tip: Write the solutions clearly: x = -a and x = (a+1). Show how you grouped the terms for factoring. Check by substituting back if you have time.

Answer 34.
Answer:
\( x^2 + \left(a + \frac{1}{a}\right)x + 1 = 0 \)

\( x^2 + ax + \frac{1}{a}x + 1 = 0 \)

\( x(x + a) + \frac{1}{a}(x + a) = 0 \)

\( (x + a)(x + \frac{1}{a}) = 0 \)

\( x = -a, x = -\frac{1}{a} \)
In simple words: We factored the quadratic equation by grouping. The solutions are x = -a and x = -1/a, which are negative reciprocals.

📝 Teacher's Note: Point out that the solutions -a and -1/a are negative reciprocals. This pattern helps students remember and check their work.

🎯 Exam Tip: Write both solutions clearly with proper negative signs: x = -a and x = -1/a. Show the factoring by grouping method clearly.

Answer 35.
Answer:
\( abx^2 + (b^2 - ac)x - bc = 0 \)

\( abx^2 + b^2x - acx - bc = 0 \)

\( bx(ax + b) - c(ax + b) = 0 \)

\( (ax + b)(bx - c) = 0 \)

\( x = -\frac{b}{a}, x = \frac{c}{b} \)
In simple words: We factored the quadratic by grouping terms. The two solutions are x = -b/a and x = c/b.

📝 Teacher's Note: Show students how to group terms with common factors. The key is recognizing that both groups should have the same binomial factor (ax + b).

🎯 Exam Tip: Write the fractions clearly: x = -b/a and x = c/b. Show the factoring steps. Make sure the signs are correct in your final answer.

Answer 36.
Answer:
\( a^2b^2x^2 + b^2x - a^2x - 1 = 0 \)

\( b^2x(a^2x + 1) - 1(a^2x + 1) = 0 \)

\( (a^2x + 1)(b^2x - 1) = 0 \)

\( x = -\frac{1}{a^2}, x = \frac{1}{b^2} \)
In simple words: We factored by grouping. The solutions are x = -1/a² and x = 1/b², involving the squares of a and b.

📝 Teacher's Note: Emphasize that the answers involve squares (a² and b²), not just a and b. Students often forget to include the squares in their final answer.

🎯 Exam Tip: Write the solutions with proper squares: x = -1/a² and x = 1/b². Double-check that you have factored correctly by expanding back.

Answer 37.
Answer:
\( \frac{x-1}{x-2} + \frac{x-3}{x-4} = 3\frac{1}{3} \)

\( \frac{(x-1)(x-4) + (x-3)(x-2)}{(x-2)(x-4)} = \frac{10}{3} \)

\( \frac{x^2 - 5x + 4 + x^2 - 5x + 6}{x^2 - 6x + 8} = \frac{10}{3} \)

\( \frac{2x^2 - 10x + 10}{x^2 - 6x + 8} = \frac{10}{3} \)

\( 6x^2 - 30x + 30 = 10x^2 - 60x + 80 \)

\( 4x^2 - 30x + 50 = 0 \)

\( 2x^2 - 15x + 25 = 0 \)

\( x^2 - \frac{15}{2}x + \frac{25}{2} = 0 \)

\( x^2 - 5x - \frac{5}{2}x + \frac{25}{2} = 0 \)

\( x(x - 5) - \frac{5}{2}(x - 5) = 0 \)

\( (x - 5)(x - \frac{5}{2}) = 0 \)

\( x = 5, x = \frac{5}{2} \)
In simple words: We combined the fractions on the left side, cross-multiplied, and solved the resulting quadratic equation. The solutions are x = 5 and x = 5/2.

📝 Teacher's Note: This is a complex rational equation. Teach students to find common denominator first, then cross-multiply carefully. Watch out for arithmetic errors in the expansion.

🎯 Exam Tip: Convert mixed numbers to improper fractions first. Show all steps clearly when combining fractions and cross-multiplying. Write final answers as x = 5 and x = 5/2.

Exercise 6.2

Answer 1.
Answer:
(i) \( 2x^2 - 5x + 3 = 0 \)
\( 2x^2 - 5x + 3 = 0 \)
Discriminant = \( b^2 - 4ac \)
\( = (-5)^2 - 4(2)(3) \)
\( = 25 - 24 \)
\( = 1 \)

(ii) \( x^2 + 2x + 4 = 0 \)
\( x^2 + 2x + 4 = 0 \)
Discriminant = \( b^2 - 4ac \)
\( = (2)^2 - 4(1)(4) \)
\( = 4 - 16 \)
\( = -12 \)

(iii) \( 2x^2 - 3x + 1 = 0 \)
\( 2x^2 - 3x + 1 = 0 \)
Discriminant = \( b^2 - 4ac \)
\( = (-3)^2 - 4(2)(1) \)
\( = 9 - 8 \)
\( = 1 \)

(iv) \( 10x - \frac{1}{x} = 3 \)
\( 10x - \frac{1}{x} = 3 \)
\( 10x^2 - 3x - 1 = 0 \)
Discriminant = \( b^2 - 4ac \)
\( = (-3)^2 - 4(10)(-1) \)
\( = 9 + 40 \)
\( = 49 \)

(v) \( x^2 + 2x - 2 = 0 \)
\( x^2 + 2x - 2 = 0 \)
Discriminant = \( b^2 - 4ac \)
\( = (2)^2 - 4(1)(-2) \)
\( = 4 + 8 \)
\( = 12 \)

(vi) \( 4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0 \)
\( 4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0 \)
Discriminant = \( b^2 - 4ac \)
\( = (5)^2 - 4(4\sqrt{3})(-2\sqrt{3}) \)
\( = 25 + 96 \)
\( = 121 \)
In simple words: We found the discriminant for each quadratic equation using the formula b² - 4ac. The discriminant tells us about the nature of the roots.

📝 Teacher's Note: Teach students that discriminant > 0 means real and different roots, discriminant = 0 means real and equal roots, discriminant < 0 means no real roots.

🎯 Exam Tip: Always write the standard form ax² + bx + c = 0 first. Identify a, b, c clearly. Use the formula b² - 4ac and show the calculation step by step.

Answer 2.
Answer:
(i) \( 2x^2 + x - 1 = 0 \)
\( 2x^2 + x - 1 = 0 \)
\( b^2 - 4ac \)
\( = (1)^2 - 4(2)(-1) \)
\( = 1 + 8 \)
\( = 9 \)
Since 9 is a perfect square and greater than 0, hence the roots are real and rational.

(ii) \( x^2 - 4x + 4 = 0 \)
\( x^2 - 4x + 4 = 0 \)
\( b^2 - 4ac \)
\( = (-4)^2 - 4(1)(4) \)
\( = 16 - 16 \)
\( = 0 \)
Since discriminant is 0, hence the roots are real and equal.

(iii) \( x^2 + 3x + 1 = 0 \)
\( x^2 + 3x + 1 = 0 \)
\( b^2 - 4ac \)
\( = (3)^2 - 4(1)(1) \)
\( = 9 - 4 \)
\( = 5 \)
Since discriminant is positive, hence the roots are real and irrational.

(iv) \( 4x^2 - 8x + 5 = 0 \)
\( 4x^2 - 8x + 5 = 0 \)
\( b^2 - 4ac \)
\( = (-8)^2 - 4(4)(5) \)
\( = 64 - 100 \)
\( = -36 \)
Since discriminant is negative, hence the roots are imaginary.

(v) \( 2x^2 + 5x - 6 = 0 \)
\( 2x^2 + 5x - 6 = 0 \)
\( b^2 - 4ac \)
\( = (5)^2 - 4(2)(-6) \)
\( = 25 + 48 \)
\( = 73 \)
Since discriminant is positive, hence the roots are real and irrational.

(vi) \( 2x^2 - 3x + 4 = 0 \)
\( 2x^2 - 3x + 4 = 0 \)
\( b^2 - 4ac \)
\( = (-3)^2 - 4(2)(4) \)
\( = 9 - 32 \)
\( = -23 \)
Since discriminant is negative, hence the roots are imaginary.

(vii) \( (x - 1)(2x - 7) = 0 \)
\( (x - 1)(2x - 7) = 0 \)
\( 2x^2 - 2x - 7x + 7 = 0 \)
\( 2x^2 - 9x + 7 = 0 \)
\( b^2 - 4ac \)
\( = (-9)^2 - 4(2)(7) \)
\( = 81 - 56 \)
\( = 25 \)
Since discriminant is a perfect square, hence the roots are real and rational.

(viii) \( x^2 - 5x + 7 = 0 \)
\( x^2 - 5x + 7 = 0 \)
\( b^2 - 4ac \)
\( = (-5)^2 - 4(1)(7) \)
\( = 25 - 28 \)
\( = -3 \)
Since discriminant is negative, hence the roots are imaginary.

In simple words: We use the discriminant formula \( b^2 - 4ac \). If it is positive, roots are real. If it is zero, roots are equal. If it is negative, roots are imaginary.

📝 Teacher's Note: Teach students the simple discriminant rule. Positive means real roots. Zero means equal roots. Negative means imaginary roots. Make them practice with simple examples first.

🎯 Exam Tip: Always write the discriminant formula first. Then substitute values. Write the conclusion about nature of roots clearly. These steps get you full marks.

Answer 3.
Answer:
(i) \( 16x^2 = 24x + 1 \)
\( 16x^2 = 24x + 1 \)
\( 16x^2 - 24x - 1 = 0 \)
\( x^2 - \frac{3}{2}x - \frac{1}{16} = 0 \)
\( a = 1; b = -\frac{3}{2}; c = -\frac{1}{16} \)
\( D = b^2 - 4ac \)
\( = (-\frac{3}{2})^2 - 4(1)(-\frac{1}{16}) \)
\( = \frac{9}{4} + \frac{1}{4} \)
\( = \frac{10}{4} \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-(-\frac{3}{2}) \pm \sqrt{\frac{10}{4}}}{2 \times 1} \)
\( x = \frac{\frac{3}{2} \pm \sqrt{\frac{10}{4}}}{2} \)
\( x = \frac{3 + \sqrt{10}}{4}, x = \frac{3 - \sqrt{10}}{4} \)

(ii) \( x^2 + 10x - 8 = 0 \)
\( x^2 + 10x - 8 = 0 \)
\( a = 1; b = 10; c = -8 \)
\( D = b^2 - 4ac \)
\( = (10)^2 - 4(1)(-8) \)
\( = 100 + 32 \)
\( = 132 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-10 \pm \sqrt{132}}{2} \)
\( x = \frac{-10 \pm \sqrt{\frac{132}{4}}}{2} \)
\( x = -5 + \sqrt{33}, x = -5 - \sqrt{33} \)

(iii) \( 2x^2 - 2\sqrt{6}x + 3 = 0 \)
\( 2x^2 - 2\sqrt{6}x + 3 = 0 \)
\( a = 2; b = -2\sqrt{6}; c = 3 \)
\( D = b^2 - 4ac \)
\( = (-2\sqrt{6})^2 - 4(2)(3) \)
\( = 24 - 24 \)
\( = 0 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-(-2\sqrt{6}) \pm 0}{2 \times 2} \)
\( x = \frac{\sqrt{6}}{2} \)

(iv) \( 3x^2 + 2\sqrt{5}x - 5 = 0 \)
\( 3x^2 + 2\sqrt{5}x - 5 = 0 \)
\( a = 3; b = 2\sqrt{5}; c = -5 \)
\( D = b^2 - 4ac \)
\( = (2\sqrt{5})^2 - 4(3)(-5) \)
\( = 20 + 60 \)
\( = 80 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-(2\sqrt{5}) \pm \sqrt{80}}{6} \)
\( x = \frac{-(2\sqrt{5}) \pm 4\sqrt{5}}{6} \)
\( x = \frac{-2\sqrt{5} + 4\sqrt{5}}{6}, x = \frac{-2\sqrt{5} - 4\sqrt{5}}{6} \)
\( x = \frac{\sqrt{5}}{3}, x = -\sqrt{5} \)

(v) \( 2x^2 + 5\sqrt{3}x + 6 = 0 \)
\( 2x^2 + 5\sqrt{3}x + 6 = 0 \)
\( a = 2; b = 5\sqrt{3}; c = 6 \)
\( D = b^2 - 4ac \)
\( D = (5\sqrt{3})^2 - 4(2)(6) \)
\( = 75 - 48 \)
\( = 27 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-(5\sqrt{3}) \pm 3\sqrt{3}}{4} \)
\( x = \frac{-5\sqrt{3} + 3\sqrt{3}}{4}, x = \frac{-5\sqrt{3} - 3\sqrt{3}}{4} \)
\( x = \frac{-2\sqrt{3}}{4}, x = \frac{-8\sqrt{3}}{4} \)
\( x = -\frac{\sqrt{3}}{2}, x = -2\sqrt{3} \)

(vi) \( \frac{5}{4}x^2 - 2\sqrt{5}x + 4 = 0 \)
\( \frac{5}{4}x^2 - 2\sqrt{5}x + 4 = 0 \)
\( 5x^2 - 8\sqrt{5}x + 16 = 0 \)
\( a = 5; b = -8\sqrt{5}; c = 16 \)
\( D = b^2 - 4ac \)
\( = (-8\sqrt{5})^2 - 4(5)(15) \)
\( = 40 - 300 \)
\( = -260 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-(-8\sqrt{5}) \pm \sqrt{-260}}{2 \times 5} \)
\( x = \frac{8\sqrt{5} \pm \sqrt{-260}}{2 \times 5} \)
\( x = \frac{4\sqrt{5}}{5} \) (Since \( \sqrt{-260} \) is not possible)

(vii) \( 3x^2 - 5x + \frac{25}{12} = 0 \)
\( 3x^2 - 5x + \frac{25}{12} = 0 \)
\( a = 3; b = -5; c = \frac{25}{12} \)
\( D = b^2 - 4ac \)
\( = (-5)^2 - 4(3)(\frac{25}{12}) \)
\( = 25 - 25 \)
\( = 0 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-(-5) \pm 0}{6} \)
\( x = \frac{5}{6} \)

(viii) \( 4x^2 + 12x + 9 = 0 \)
\( 4x^2 + 12x + 9 = 0 \)
\( a = 4; b = 12; c = 9 \)
\( D = b^2 - 4ac \)
\( = (12)^2 - 4(4)(9) \)
\( = 144 - 144 \)
\( = 0 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-12 \pm 0}{8} \)
\( x = -\frac{3}{2} \)

(ix) \( x^2 - 7x - 5 = 0 \)
\( x^2 - 7x - 5 = 0 \)
\( a = 1; b = -7; c = -5 \)
\( D = b^2 - 4ac \)
\( = (-7)^2 - 4(1)(-5) \)
\( = 49 + 20 \)
\( = 69 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{7 \pm \sqrt{69}}{2} \)

In simple words: We use the quadratic formula to solve these equations. Substitute the values of a, b, and c into the formula and calculate the roots step by step.

📝 Teacher's Note: Show students how to identify a, b, and c from the equation first. Then use the quadratic formula. Practice with simple numbers before using surds and fractions.

🎯 Exam Tip: Always write the quadratic formula first. Show all steps clearly. When discriminant is negative, write "no real roots" or "imaginary roots". Check your final answer by substituting back.

Question (x). \( x^2 - 4x - 1 = 0 \)
Answer:
Given: \( x^2 - 4x - 1 = 0 \)
Where \( a = 1; b = -4; c = -1 \)

Step 1: Calculate the discriminant.
\( D = b^2 - 4ac \)
\( = (-4)^2 - 4(1)(-1) \)
\( = 16 + 4 \)
\( = 20 \)

Step 2: Apply the quadratic formula.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{4 \pm \sqrt{20}}{2} \)
\( x = \frac{4 \pm 2\sqrt{5}}{2} \)

Step 3: Simplify to get final answer.
\( x = 2 + \sqrt{5}, x = 2 - \sqrt{5} \)
In simple words: We used the quadratic formula to solve this equation. The discriminant is 20, which is positive, so we get two real solutions.

📝 Teacher's Note: Make sure students write the quadratic formula first. Show them that \( \sqrt{20} = 2\sqrt{5} \). A common mistake is forgetting to simplify the square root.

🎯 Exam Tip: Always show all steps: find a, b, c values, calculate D, then apply formula. Write final answer clearly with both values of x.

Question (xi). \( 6x^2 + 7x - 10 = 0 \)
Answer:
Given: \( 6x^2 + 7x - 10 = 0 \)
Where \( a = 6; b = 7; c = -10 \)

Step 1: Calculate the discriminant.
\( D = b^2 - 4ac \)
\( = (7)^2 - 4(6)(-10) \)
\( = 49 + 240 \)
\( = 289 \)

Step 2: Apply the quadratic formula.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-7 \pm \sqrt{289}}{12} \)
In simple words: We found the discriminant is 289, which is a perfect square (17²). This makes our calculation easier.

📝 Teacher's Note: Point out that 289 = 17². When discriminant is a perfect square, the roots will be rational numbers (simple fractions).

🎯 Exam Tip: Check if the discriminant is a perfect square. If yes, your final answer will be simple fractions, not surds.

Question (xii). \( x^2 - 6x + 4 = 0 \)
Answer:
Given: \( x^2 - 6x + 4 = 0 \)
Where \( a = 1; b = -6; c = 4 \)

Step 1: Calculate the discriminant.
\( D = b^2 - 4ac \)
\( = (-6)^2 - 4(1)(4) \)
\( = 36 - 16 \)
\( = 20 \)

Step 2: Apply the quadratic formula.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{6 \pm \sqrt{20}}{2} \)
\( x = \frac{6 \pm 2\sqrt{5}}{2} \)

Step 3: Simplify to get final answer.
\( x = 3 + \sqrt{5}, x = 3 - \sqrt{5} \)
In simple words: The discriminant is 20, so we get two real solutions with square roots in them.

📝 Teacher's Note: Show students how to simplify \( \frac{6 \pm 2\sqrt{5}}{2} \) by dividing both parts by 2. This is where many students make errors.

🎯 Exam Tip: Always simplify your final answer. Don't leave it as \( \frac{6 \pm 2\sqrt{5}}{2} \). Reduce to \( 3 \pm \sqrt{5} \).

Question (xiii). \( 5x^2 - 19x + 17 = 0 \)
Answer:
Given: \( 5x^2 - 19x + 17 = 0 \)
Where \( a = 5; b = -19; c = 17 \)

Step 1: Calculate the discriminant.
\( D = b^2 - 4ac \)
\( = (-19)^2 - 4(5)(17) \)
\( = 361 - 340 \)
\( = 21 \)

Step 2: Apply the quadratic formula.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{19 \pm \sqrt{21}}{10} \)

Final answer: \( x = \frac{19 + \sqrt{21}}{10}, x = \frac{19 - \sqrt{21}}{10} \)
In simple words: The discriminant is 21, which is not a perfect square. So our answer has square roots that cannot be simplified further.

📝 Teacher's Note: Students often confuse the sign of b in the formula. Here b = -19, so -b = +19. Practice this carefully.

🎯 Exam Tip: When discriminant is not a perfect square, leave the square root as it is. Don't try to calculate decimal values unless asked.

Question (xiv). \( 15x^2 - 28 = x \)
Answer:
Step 1: Rearrange in standard form.
\( 15x^2 - 28 = x \)
\( 15x^2 - x - 28 = 0 \)
Where \( a = 15; b = -1; c = -28 \)

Step 2: Calculate the discriminant.
\( D = b^2 - 4ac \)
\( = (-1)^2 - 4(15)(-28) \)
\( = 1 + 1680 \)
\( = 1681 \)

Step 3: Apply the quadratic formula.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{1 \pm \sqrt{1681}}{30} \)
\( x = \frac{1 \pm 41}{30} \)

Step 4: Calculate both solutions.
\( x = \frac{1 + 41}{30} = \frac{42}{30} = \frac{7}{5} \)
\( x = \frac{1 - 41}{30} = \frac{-40}{30} = -\frac{4}{3} \)

Final answer: \( x = \frac{7}{5}, x = -\frac{4}{3} \)
In simple words: First we moved all terms to one side. Then we found that 1681 = 41², so we got nice fraction answers.

📝 Teacher's Note: Always rearrange the equation to standard form first. Show students that 1681 is a perfect square (41²).

🎯 Exam Tip: When the equation is not in standard form, rearrange it first. Check if your discriminant is a perfect square for easier calculation.

Question (xv). \( 4 - 11x = 3x^2 \)
Answer:
Step 1: Rearrange in standard form.
\( 4 - 11x = 3x^2 \)
\( 3x^2 + 11x - 4 = 0 \)
Where \( a = 3; b = 11; c = -4 \)

Step 2: Calculate the discriminant.
\( D = b^2 - 4ac \)
\( = (11)^2 - 4(3)(-4) \)
\( = 121 + 48 \)
\( = 169 \)

Step 3: Apply the quadratic formula.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-11 \pm \sqrt{169}}{6} \)
\( x = \frac{-11 \pm 13}{6} \)

Step 4: Calculate both solutions.
\( x = \frac{-11 + 13}{6} = \frac{2}{6} = \frac{1}{3} \)
\( x = \frac{-11 - 13}{6} = \frac{-24}{6} = -4 \)

Final answer: \( x = \frac{1}{3}, x = -4 \)
In simple words: We rearranged the equation and found that 169 = 13². This gave us simple fraction answers.

📝 Teacher's Note: Make sure students move all terms to get the equation in ax² + bx + c = 0 form. The discriminant 169 = 13².

🎯 Exam Tip: Always check your final answers by substituting back into the original equation. This catches calculation errors.

Question (xvi). \( 25x^2 + 30x + 7 = 0 \)
Answer:
Given: \( 25x^2 + 30x + 7 = 0 \)
Where \( a = 25; b = 30; c = 7 \)

Step 1: Calculate the discriminant.
\( D = b^2 - 4ac \)
\( = (30)^2 - 4(25)(7) \)
\( = 900 - 700 \)
\( = 200 \)

Step 2: Apply the quadratic formula.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-30 \pm \sqrt{200}}{50} \)
\( x = \frac{-30 \pm 10\sqrt{2}}{50} \)
\( x = \frac{-3 \pm \sqrt{2}}{5} \)

Final answer: \( x = \frac{-3 + \sqrt{2}}{5}, x = \frac{-3 - \sqrt{2}}{5} \)
In simple words: We simplified \( \sqrt{200} = 10\sqrt{2} \) and then reduced the fraction by dividing by 10.

📝 Teacher's Note: Show students how \( \sqrt{200} = \sqrt{100 \times 2} = 10\sqrt{2} \). This simplification step is important.

🎯 Exam Tip: Always simplify square roots completely. \( \sqrt{200} \) should be written as \( 10\sqrt{2} \). Then reduce the fraction.

Question (xvii). \( 16x^2 - 24x = 1 \)
Answer:
Step 1: Rearrange in standard form.
\( 16x^2 - 24x = 1 \)
\( 16x^2 - 24x - 1 = 0 \)
Where \( a = 16; b = -24; c = -1 \)

Step 2: Calculate the discriminant.
\( D = b^2 - 4ac \)
\( = (-24)^2 - 4(16)(-1) \)
\( = 576 + 64 \)
\( = 640 \)

Step 3: Apply the quadratic formula.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{24 \pm 8\sqrt{10}}{32} \)
\( x = \frac{3 \pm \sqrt{10}}{4} \)

Final answer: \( x = \frac{3 + \sqrt{10}}{4}, x = \frac{3 - \sqrt{10}}{4} \)
In simple words: We simplified \( \sqrt{640} = 8\sqrt{10} \) and then reduced the fraction by dividing by 8.

📝 Teacher's Note: Students need to practice simplifying \( \sqrt{640} = \sqrt{64 \times 10} = 8\sqrt{10} \). This is a common type of calculation.

🎯 Exam Tip: Don't leave the answer as \( \frac{24 \pm 8\sqrt{10}}{32} \). Always simplify by dividing by the common factor (8 here).

Question (xviii). \( 3x^2 + 2\sqrt{5}x - 5 = 0 \)
Answer:
Given: \( 3x^2 + 2\sqrt{5}x - 5 = 0 \)
Where \( a = 3; b = 2\sqrt{5}; c = -5 \)

Step 1: Calculate the discriminant.
\( D = b^2 - 4ac \)
\( = (2\sqrt{5})^2 - 4(3)(-5) \)
\( = 20 + 60 \)
\( = 80 \)

Step 2: Apply the quadratic formula.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-(2\sqrt{5}) \pm \sqrt{80}}{6} \)
\( x = \frac{-2\sqrt{5} \pm 4\sqrt{5}}{6} \)
\( x = \frac{-2\sqrt{5} + 4\sqrt{5}}{6}, x = \frac{-2\sqrt{5} - 4\sqrt{5}}{6} \)
\( x = \frac{\sqrt{5}}{3}, x = -\sqrt{5} \)

Final answer: \( x = \frac{\sqrt{5}}{3}, x = -\sqrt{5} \)
In simple words: This equation has surds (square roots) in the coefficient. We simplified \( \sqrt{80} = 4\sqrt{5} \) and combined like terms.

📝 Teacher's Note: This is a special type where the coefficient has a surd. Show students that \( (2\sqrt{5})^2 = 4 \times 5 = 20 \) and \( \sqrt{80} = 4\sqrt{5} \).

🎯 Exam Tip: When coefficients have surds, be very careful with squaring. \( (2\sqrt{5})^2 = 20 \), not \( 4\sqrt{5} \).

Question (xix). \( 3x^2 + 12 = 32x \)
Answer:
Step 1: Rearrange in standard form.
\( 3x^2 + 12 = 32x \)
\( 3x^2 - 32x + 12 = 0 \)
Where \( a = 3; b = -32; c = 12 \)

Step 2: Calculate the discriminant.
\( D = b^2 - 4ac \)
\( = (-32)^2 - 4(3)(12) \)
\( = 1024 - 144 \)
\( = 880 \)

Step 3: Apply the quadratic formula.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{32 \pm \sqrt{880}}{6} \)
\( x = \frac{32 \pm 4\sqrt{55}}{6} \)
\( x = \frac{16 \pm 2\sqrt{55}}{3} \)

Final answer: \( x = \frac{16 + 2\sqrt{55}}{3}, x = \frac{16 - 2\sqrt{55}}{3} \)
In simple words: We simplified \( \sqrt{880} = 4\sqrt{55} \) and then reduced the fraction by dividing by 2.

📝 Teacher's Note: Show students how to find \( \sqrt{880} = \sqrt{16 \times 55} = 4\sqrt{55} \). Practice finding the largest perfect square factor.

🎯 Exam Tip: Always look for common factors to simplify. Here we divided numerator and denominator by 2 to get the final answer.

Question (xx). \( x^2 + \frac{1}{2}x = 3 \)
Answer:
Step 1: Rearrange in standard form.
\( x^2 + \frac{1}{2}x = 3 \)
\( 2x^2 + x = 6 \) (multiply by 2 to clear fractions)
\( 2x^2 + x - 6 = 0 \)
Where \( a = 2; b = 1; c = -6 \)

Step 2: Calculate the discriminant.
\( D = b^2 - 4ac \)
\( = (1)^2 - 4(2)(-6) \)
\( = 1 + 48 \)
\( = 49 \)

Step 3: Apply the quadratic formula.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-1 \pm \sqrt{49}}{4} \)
\( x = \frac{-1 \pm 7}{4} \)

Step 4: Calculate both solutions.
\( x = \frac{-1 + 7}{4} = \frac{6}{4} = \frac{3}{2} \)
\( x = \frac{-1 - 7}{4} = \frac{-8}{4} = -2 \)

Final answer: \( x = \frac{3}{2}, x = -2 \)
In simple words: We cleared the fraction by multiplying the whole equation by 2. Then we got a perfect square discriminant (49 = 7²).

📝 Teacher's Note: When there are fractions, multiply the whole equation by the LCM to clear them. This makes calculations easier.

🎯 Exam Tip: Always clear fractions first by multiplying the entire equation. This prevents calculation errors with fractional coefficients.

Question (xxi). \( \sqrt{3}x^2 + 10x - 8\sqrt{3} = 0 \)
Answer:
Given: \( \sqrt{3}x^2 + 10x - 8\sqrt{3} = 0 \)
Where \( a = \sqrt{3}; b = 10; c = -8\sqrt{3} \)

Step 1: Calculate the discriminant.
\( D = b^2 - 4ac \)
\( = (10)^2 - 4(\sqrt{3})(-8\sqrt{3}) \)
\( = 100 + 96 \)
\( = 196 \)

Step 2: Apply the quadratic formula.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-10 \pm \sqrt{196}}{2\sqrt{3}} \)
\( x = \frac{-10 \pm 14}{2\sqrt{3}} \)

Step 3: Calculate both solutions.
\( x = \frac{-10 + 14}{2\sqrt{3}} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}} \)
\( x = \frac{-10 - 14}{2\sqrt{3}} = \frac{-24}{2\sqrt{3}} = \frac{-12}{\sqrt{3}} \)

Final answer: \( x = \frac{2}{\sqrt{3}}, x = \frac{-12}{\sqrt{3}} \)
In simple words: This equation has \( \sqrt{3} \) in the coefficients. We found that 196 = 14² which made our calculation easier.

📝 Teacher's Note: Students should know that \( \sqrt{3} \times (-8\sqrt{3}) = -8 \times 3 = -24 \). Practice multiplying surds.

🎯 Exam Tip: When coefficients have surds, be careful with multiplication. \( 4 \times \sqrt{3} \times (-8\sqrt{3}) = -96 \).

Question (xxii). \( 2x^2 - 2\sqrt{6}x + 3 = 0 \)
Answer:
Given: \( 2x^2 - 2\sqrt{6}x + 3 = 0 \)
Where \( a = 2; b = -2\sqrt{6}; c = 3 \)

Step 1: Calculate the discriminant.
\( D = b^2 - 4ac \)
\( = (-2\sqrt{6})^2 - 4(2)(3) \)
\( = 4 \times 6 - 24 \)
\( = 24 - 24 \)
\( = 0 \)

Step 2: Since D = 0, we have equal roots.
\( x = \frac{-b}{2a} = \frac{2\sqrt{6}}{4} = \frac{\sqrt{6}}{2} \)

Final answer: \( x = \frac{\sqrt{6}}{2} \) (repeated root)
In simple words: The discriminant is zero, which means we get only one answer (or two equal answers). This is called a repeated root.

📝 Teacher's Note: When D = 0, the quadratic has equal roots. Use the formula \( x = \frac{-b}{2a} \) directly.

🎯 Exam Tip: If discriminant = 0, write "equal roots" and use \( x = \frac{-b}{2a} \). Don't forget to mention it's a repeated root.

Question xxiii. \(3a^2x^2 + 8abx + 4b^2 = 0, a \neq 0\)
Answer:
\(3a^2x^2 + 8abx + 4b^2 = 0\)

\(x^2 + \frac{8b}{3a}x + \frac{4b^2}{3a^2} = 0\)

Here, \(a = 1; b = \frac{8b}{3a}; c = \frac{4b^2}{3a^2}\)

\(D = b^2 - 4ac\)

\(= (\frac{8b}{3a})^2 - 4(1)(\frac{4b^2}{3a^2})\)

\(= \frac{64b^2}{9a^2} - \frac{16b^2}{3a^2}\)

\(= \frac{64b^2 - 48b^2}{9a^2} = \frac{16b^2}{9a^2}\)

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(x = \frac{-\frac{8b}{3a} \pm \sqrt{\frac{16b^2}{9a^2}}}{2}\)

\(x = \frac{-\frac{8b}{3a} + \frac{4b}{3a}}{2}, x = \frac{-\frac{8b}{3a} - \frac{4b}{3a}}{2}\)

\(x = \frac{-4b}{6a}, x = \frac{-12b}{6a}\)

\(x = -\frac{2b}{3a}, x = -\frac{2b}{a}\)

In simple words: We used the quadratic formula to solve this equation. First we found the discriminant, then used the formula to get two answers for x.

📝 Teacher's Note: Show students how to divide the whole equation by \(3a^2\) first. This makes the coefficient of \(x^2\) equal to 1, which makes calculations easier.

🎯 Exam Tip: Always simplify the equation first by dividing by the coefficient of \(x^2\). Write each step clearly and don't skip the discriminant calculation.

Question xxiv. \(x^2 + \frac{1}{2}x - 1 = 0\)
Answer:
\(x^2 + \frac{1}{2}x - 1 = 0\)

\(2x^2 + x - 2 = 0\)

Here, \(a = 2; b = 1; c = -2\)

\(D = b^2 - 4ac\)

\(= (1)^2 - 4(2)(-2)\)

\(= 1 + 16\)

\(= 17\)

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(x = \frac{-1 \pm \sqrt{17}}{4}\)

\(x = \frac{-1 + \sqrt{17}}{4}, x = \frac{-1 - \sqrt{17}}{4}\)

In simple words: We first cleared the fraction by multiplying the whole equation by 2. Then we used the quadratic formula to find the two roots.

📝 Teacher's Note: When there are fractions, multiply the whole equation by the denominator first. This removes fractions and makes calculations simpler.

🎯 Exam Tip: Always clear fractions at the beginning. Write the discriminant value clearly. Leave answers in surd form unless asked to find decimal values.

Question xxv. \(x^2 - 4\sqrt{15}x - 4 = 0\)
Answer:
\(x^2 - 4\sqrt{15}x - 4 = 0\)

Here, \(a = 1; b = -4\sqrt{15}; c = -4\)

\(D = b^2 - 4ac\)

\(= (-4\sqrt{15})^2 - 4(1)(-4)\)

\(= 240 + 16\)

\(= 256\)

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(x = \frac{4\sqrt{15} \pm \sqrt{256}}{2}\)

\(x = \frac{4\sqrt{15} + 16}{2}, x = \frac{4\sqrt{15} - 16}{2}\)

\(x = 2\sqrt{15} + 8, x = 2\sqrt{15} - 8\)

In simple words: This equation has a square root in the coefficient. We calculated the discriminant carefully and got 256, which is a perfect square (16). So the final answers are simple.

📝 Teacher's Note: When squaring \(4\sqrt{15}\), remind students that \((4\sqrt{15})^2 = 16 \times 15 = 240\). Show this step clearly to avoid confusion.

🎯 Exam Tip: When the discriminant is a perfect square, the final answers will be simpler. Check if \(\sqrt{256} = 16\) to avoid calculation errors.