OP Malhotra Class 11 Maths Solutions Chapter 10 Quadratic Equations Exercise 10 (A)

Find the roots of the following equations.

Question 1. \( 2x^2 + x - 3 = 0 \)
Answer: The given equation is \( 2x^2 + x - 3 = 0 \).
By comparing this with the standard quadratic equation \( ax^2 + bx + c = 0 \), we find that \( a = 2 \), \( b = 1 \), and \( c = -3 \).
Then, we use the quadratic formula to find the roots:
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Now, we substitute the values of \( a, b, \) and \( c \):
\[ x = \frac{-(1) \pm \sqrt{(1)^2-4(2)(-3)}}{2(2)} \]
\[ x = \frac{-1 \pm \sqrt{1 + 24}}{4} \]
\[ x = \frac{-1 \pm \sqrt{25}}{4} \]
\[ x = \frac{-1 \pm 5}{4} \]
This gives us two possible values for \( x \):
\( x = \frac{-1 + 5}{4} \) or \( x = \frac{-1 - 5}{4} \)
\( \implies x = \frac{4}{4} \) or \( x = \frac{-6}{4} \)
\( \implies x = 1 \) or \( x = -\frac{3}{2} \)
Quadratic equations can have up to two roots, which are the values of x that make the equation true.
Therefore, the roots of the equation are \( 1 \) and \( -\frac{3}{2} \).
In simple words: We found the values of a, b, and c from the equation and put them into the quadratic formula. This helped us get two answers for x that make the equation correct.

🎯 Exam Tip: Always remember the quadratic formula accurately and pay close attention to the signs when substituting values, especially for \( -b \) and \( -4ac \).

Question 2. \( 6x^2 + 7x - 20 = 0 \)
Answer: The given equation is \( 6x^2 + 7x - 20 = 0 \).
Comparing this with the standard quadratic form \( ax^2 + bx + c = 0 \), we identify \( a = 6 \), \( b = 7 \), and \( c = -20 \).
We apply the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Substitute the values into the formula:
\[ x = \frac{-(7) \pm \sqrt{(7)^2-4(6)(-20)}}{2(6)} \]
\[ x = \frac{-7 \pm \sqrt{49 + 480}}{12} \]
\[ x = \frac{-7 \pm \sqrt{529}}{12} \]
\[ x = \frac{-7 \pm 23}{12} \]
Now, we calculate the two possible roots:
\( x = \frac{-7 + 23}{12} \) or \( x = \frac{-7 - 23}{12} \)
\( \implies x = \frac{16}{12} \) or \( x = \frac{-30}{12} \)
\( \implies x = \frac{4}{3} \) or \( x = -\frac{5}{2} \)
The discriminant, which is \( b^2 - 4ac \), tells us the nature of the roots; here, \( \sqrt{529} \) is a real number, so the roots are real.
Therefore, the roots of the equation are \( \frac{4}{3} \) and \( -\frac{5}{2} \).
In simple words: We used the quadratic formula after finding a, b, and c from the equation. This gave us two solutions for x.

🎯 Exam Tip: When dealing with negative signs in the quadratic formula, be extra careful with calculations, especially under the square root, to avoid common errors.

Question 3. \( 36x^2 + 23 = 60x \)
Answer: First, we rewrite the equation \( 36x^2 + 23 = 60x \) into the standard form \( ax^2 + bx + c = 0 \).
Subtract \( 60x \) from both sides to get:
\( 36x^2 - 60x + 23 = 0 \)
Comparing this with the standard form, we identify \( a = 36 \), \( b = -60 \), and \( c = 23 \).
Now, we apply the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Substitute the identified values:
\[ x = \frac{-(-60) \pm \sqrt{(-60)^2-4(36)(23)}}{2(36)} \]
\[ x = \frac{60 \pm \sqrt{3600 - 3312}}{72} \]
\[ x = \frac{60 \pm \sqrt{288}}{72} \]
To simplify \( \sqrt{288} \), we find its prime factors: \( 288 = 144 \times 2 = 12^2 \times 2 \). So, \( \sqrt{288} = 12\sqrt{2} \).
\[ x = \frac{60 \pm 12\sqrt{2}}{72} \]
Divide all terms by 12:
\[ x = \frac{12(5 \pm \sqrt{2})}{12(6)} \]
\[ x = \frac{5 \pm \sqrt{2}}{6} \]
It's important to arrange the equation in standard form before identifying the coefficients for the quadratic formula.
Therefore, the roots of the equation are \( \frac{5 + \sqrt{2}}{6} \) and \( \frac{5 - \sqrt{2}}{6} \).
In simple words: We first put the equation in the right order. Then we found a, b, and c and used the quadratic formula to find the two answers for x.

🎯 Exam Tip: Always rearrange the quadratic equation to the standard form \( ax^2 + bx + c = 0 \) before identifying coefficients to avoid errors in signs.

Question 4. \( x^2 - 2x + 5 = 0 \)
Answer: The given quadratic equation is \( x^2 - 2x + 5 = 0 \).
By comparing it with the standard form \( ax^2 + bx + c = 0 \), we find \( a = 1 \), \( b = -2 \), and \( c = 5 \).
Now, we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Substitute the values:
\[ x = \frac{-(-2) \pm \sqrt{(-2)^2-4(1)(5)}}{2(1)} \]
\[ x = \frac{2 \pm \sqrt{4 - 20}}{2} \]
\[ x = \frac{2 \pm \sqrt{-16}}{2} \]
Since we have a negative number under the square root, we use the imaginary unit \( i \), where \( \sqrt{-1} = i \). So, \( \sqrt{-16} = \sqrt{16 \times -1} = 4i \).
\[ x = \frac{2 \pm 4i}{2} \]
Divide by 2:
\[ x = \frac{2}{2} \pm \frac{4i}{2} \]
\[ x = 1 \pm 2i \]
When the discriminant \( b^2 - 4ac \) is negative, the quadratic equation has complex roots involving the imaginary unit \( i \).
Therefore, the roots of the equation are \( 1 + 2i \) and \( 1 - 2i \).
In simple words: We used the formula, but because we got a negative number under the square root, our answers for x have an "i" in them, which means they are complex numbers.

🎯 Exam Tip: Recognize that a negative discriminant (\( b^2-4ac < 0 \)) indicates complex conjugate roots, which should be expressed using the imaginary unit \( i \).

Question 5. \( 3x^2 - 17x + 25 = 0 \)
Answer: The given quadratic equation is \( 3x^2 - 17x + 25 = 0 \).
Comparing this to the standard form \( ax^2 + bx + c = 0 \), we have \( a = 3 \), \( b = -17 \), and \( c = 25 \).
Now, we use the quadratic formula to find the roots:
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Substitute the values into the formula:
\[ x = \frac{-(-17) \pm \sqrt{(-17)^2-4(3)(25)}}{2(3)} \]
\[ x = \frac{17 \pm \sqrt{289 - 300}}{6} \]
\[ x = \frac{17 \pm \sqrt{-11}}{6} \]
Again, because we have a negative number under the square root, we use the imaginary unit \( i \). So, \( \sqrt{-11} = \sqrt{11}i \).
\[ x = \frac{17 \pm \sqrt{11}i}{6} \]
Complex roots always appear in conjugate pairs, meaning if \( A + Bi \) is a root, then \( A - Bi \) must also be a root.
Therefore, the roots of the equation are \( \frac{17 + \sqrt{11}i}{6} \) and \( \frac{17 - \sqrt{11}i}{6} \).
In simple words: We put the numbers from the equation into the formula. The answer had a square root of a negative number, so we used 'i' to write the complex number solutions.

🎯 Exam Tip: When the discriminant is negative, express the roots in the form \( \frac{-b \pm i\sqrt{|b^2-4ac|}}{2a} \), remembering that the \( \pm \) produces two conjugate roots.

Question 6. \( x^2 + 3x - 3 = 0 \), giving your answer correct to two decimal places.
Answer: The given quadratic equation is \( x^2 + 3x - 3 = 0 \).
Comparing it to the standard form \( ax^2 + bx + c = 0 \), we get \( a = 1 \), \( b = 3 \), and \( c = -3 \).
Now, we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Substitute the values:
\[ x = \frac{-(3) \pm \sqrt{(3)^2-4(1)(-3)}}{2(1)} \]
\[ x = \frac{-3 \pm \sqrt{9 + 12}}{2} \]
\[ x = \frac{-3 \pm \sqrt{21}}{2} \]
We know that \( \sqrt{21} \approx 4.5826 \). So, we substitute this value:
\[ x = \frac{-3 \pm 4.5826}{2} \]
This gives us two roots:
\( x_1 = \frac{-3 + 4.5826}{2} \) or \( x_2 = \frac{-3 - 4.5826}{2} \)
\( \implies x_1 = \frac{1.5826}{2} \) or \( x_2 = \frac{-7.5826}{2} \)
\( \implies x_1 = 0.7913 \) or \( x_2 = -3.7913 \)
Rounding to two decimal places, as required by the question:
\( x = 0.79 \) or \( x = -3.79 \)
When giving answers to a specified number of decimal places, always calculate one more decimal place than required and then round the final answer.
Therefore, the roots of the equation, correct to two decimal places, are \( 0.79 \) and \( -3.79 \).
In simple words: We used the formula to find x. Since we needed the answer to two decimal places, we found the square root of 21 and then rounded our final answers.

🎯 Exam Tip: When rounding to a specific number of decimal places, calculate at least one extra decimal place before rounding to ensure accuracy.

Question 7. \( 5x^2 - x + 4 = 0 \)
Answer: The given quadratic equation is \( 5x^2 - x + 4 = 0 \).
Comparing this to the standard form \( ax^2 + bx + c = 0 \), we have \( a = 5 \), \( b = -1 \), and \( c = 4 \).
Now, we apply the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Substitute the values:
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2-4(5)(4)}}{2(5)} \]
\[ x = \frac{1 \pm \sqrt{1 - 80}}{10} \]
\[ x = \frac{1 \pm \sqrt{-79}}{10} \]
Since we have a negative number under the square root, we use the imaginary unit \( i \). So, \( \sqrt{-79} = \sqrt{79}i \).
\[ x = \frac{1 \pm \sqrt{79}i}{10} \]
A negative discriminant indicates that the roots are not real numbers but are complex numbers.
Therefore, the roots of the equation are \( \frac{1 + \sqrt{79}i}{10} \) and \( \frac{1 - \sqrt{79}i}{10} \).
In simple words: We put the numbers into the formula. The answer had a square root of a negative number, so the solutions for x are complex numbers, written with 'i'.

🎯 Exam Tip: Always remember that \( \sqrt{-k} = i\sqrt{k} \) for any positive number \( k \), which is key for simplifying complex roots.

Question 8. \( \sqrt{3}x^2 - \sqrt{2}x + 3\sqrt{3} = 0 \)
Answer: The given quadratic equation is \( \sqrt{3}x^2 - \sqrt{2}x + 3\sqrt{3} = 0 \).
Comparing this to the standard form \( ax^2 + bx + c = 0 \), we identify \( a = \sqrt{3} \), \( b = -\sqrt{2} \), and \( c = 3\sqrt{3} \).
Now, we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Substitute the values:
\[ x = \frac{-(-\sqrt{2}) \pm \sqrt{(-\sqrt{2})^2-4(\sqrt{3})(3\sqrt{3})}}{2(\sqrt{3})} \]
\[ x = \frac{\sqrt{2} \pm \sqrt{2 - 4 \times 3 \times 3}}{2\sqrt{3}} \]
\[ x = \frac{\sqrt{2} \pm \sqrt{2 - 36}}{2\sqrt{3}} \]
\[ x = \frac{\sqrt{2} \pm \sqrt{-34}}{2\sqrt{3}} \]
Since we have a negative number under the square root, we use the imaginary unit \( i \). So, \( \sqrt{-34} = \sqrt{34}i \).
\[ x = \frac{\sqrt{2} \pm \sqrt{34}i}{2\sqrt{3}} \]
Even when coefficients are irrational, the quadratic formula can still be used to find the roots, which might also be irrational or complex.
Therefore, the roots of the equation are \( \frac{\sqrt{2} + \sqrt{34}i}{2\sqrt{3}} \) and \( \frac{\sqrt{2} - \sqrt{34}i}{2\sqrt{3}} \).
In simple words: This equation had square roots in it, but we still used the same formula. We found a, b, and c with the square roots, and the answer for x also came out with square roots and 'i' because of a negative number inside the square root.

🎯 Exam Tip: Be careful with the arithmetic of square roots and multiplication when substituting coefficients into the quadratic formula, especially when they are irrational.

Question 9. \( \frac{x^2+8}{11} = 5x - x^2 - 5 \)
Answer: First, we need to simplify the given equation \( \frac{x^2+8}{11} = 5x - x^2 - 5 \).
Multiply both sides by 11 to remove the fraction:
\( x^2 + 8 = 11(5x - x^2 - 5) \)
Distribute 11 on the right side:
\( x^2 + 8 = 55x - 11x^2 - 55 \)
Now, rearrange all terms to one side to get the standard quadratic form \( ax^2 + bx + c = 0 \):
\( x^2 + 11x^2 - 55x + 8 + 55 = 0 \)
\( 12x^2 - 55x + 63 = 0 \)
Comparing this with the standard quadratic equation, we have \( a = 12 \), \( b = -55 \), and \( c = 63 \).
Now, we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Substitute the values:
\[ x = \frac{-(-55) \pm \sqrt{(-55)^2-4(12)(63)}}{2(12)} \]
\[ x = \frac{55 \pm \sqrt{3025 - 3024}}{24} \]
\[ x = \frac{55 \pm \sqrt{1}}{24} \]
\[ x = \frac{55 \pm 1}{24} \]
This gives two roots:
\( x_1 = \frac{55 + 1}{24} \) or \( x_2 = \frac{55 - 1}{24} \)
\( \implies x_1 = \frac{56}{24} \) or \( x_2 = \frac{54}{24} \)
Simplify the fractions:
\( \implies x_1 = \frac{7}{3} \) or \( x_2 = \frac{9}{4} \)
Clearing fractions and rearranging terms are often the first steps in solving equations that don't immediately look like standard quadratic forms.
Therefore, the roots of the equation are \( \frac{7}{3} \) and \( \frac{9}{4} \).
In simple words: First, we cleaned up the equation by getting rid of the fraction and putting all terms on one side. Then, we used the quadratic formula with the new numbers to find the two answers for x.

🎯 Exam Tip: Always simplify and rearrange complex equations into the standard \( ax^2 + bx + c = 0 \) form before applying the quadratic formula.

Question 10. \( \frac{2x}{x-4}+\frac{2x-5}{x-3}=8 \frac{1}{3} \)
Answer: First, we need to simplify the given equation \( \frac{2x}{x-4} + \frac{2x-5}{x-3} = 8\frac{1}{3} \).
Combine the fractions on the left side by finding a common denominator, which is \( (x-4)(x-3) \):
\[ \frac{2x(x-3) + (2x-5)(x-4)}{(x-4)(x-3)} = \frac{25}{3} \]
Expand the numerator and denominator:
\[ \frac{(2x^2 - 6x) + (2x^2 - 8x - 5x + 20)}{x^2 - 3x - 4x + 12} = \frac{25}{3} \]
\[ \frac{2x^2 - 6x + 2x^2 - 13x + 20}{x^2 - 7x + 12} = \frac{25}{3} \]
\[ \frac{4x^2 - 19x + 20}{x^2 - 7x + 12} = \frac{25}{3} \]
Now, cross-multiply to eliminate the denominators:
\( 3(4x^2 - 19x + 20) = 25(x^2 - 7x + 12) \)
\( 12x^2 - 57x + 60 = 25x^2 - 175x + 300 \)
Rearrange all terms to one side to form a standard quadratic equation:
\( 0 = 25x^2 - 12x^2 - 175x + 57x + 300 - 60 \)
\( 0 = 13x^2 - 118x + 240 \)
So, the quadratic equation is \( 13x^2 - 118x + 240 = 0 \).
Comparing this to \( ax^2 + bx + c = 0 \), we identify \( a = 13 \), \( b = -118 \), and \( c = 240 \).
Now, we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Substitute the values:
\[ x = \frac{-(-118) \pm \sqrt{(-118)^2-4(13)(240)}}{2(13)} \]
\[ x = \frac{118 \pm \sqrt{13924 - 12480}}{26} \]
\[ x = \frac{118 \pm \sqrt{1444}}{26} \]
\[ x = \frac{118 \pm 38}{26} \]
This gives two roots:
\( x_1 = \frac{118 + 38}{26} \) or \( x_2 = \frac{118 - 38}{26} \)
\( \implies x_1 = \frac{156}{26} \) or \( x_2 = \frac{80}{26} \)
Simplify the fractions:
\( \implies x_1 = 6 \) or \( x_2 = \frac{40}{13} \)
Always check if the denominators are zero at the calculated roots, as these values would be extraneous solutions for the original rational equation. For this equation, \( x \neq 3 \) and \( x \neq 4 \), so our roots are valid.
Therefore, the roots of the equation are \( 6 \) and \( \frac{40}{13} \).
In simple words: We first combined the fractions and then cross-multiplied to get a normal quadratic equation. After finding a, b, and c, we used the formula to get the two answers for x.

🎯 Exam Tip: For rational equations that lead to quadratics, always check if any calculated root would make the original denominators zero, as those roots must be excluded.

Question 11. The number of real solutions of the equation \( x^2 - 3|x| + 2 = 0 \) is
(a) 3
(b) 4
(c) 1
(d) 3
Answer: (b) 4
The given equation is \( x^2 - 3|x| + 2 = 0 \).
We know that \( x^2 = |x|^2 \) for any real number \( x \). So, we can rewrite the equation in terms of \( |x| \):
\( |x|^2 - 3|x| + 2 = 0 \)
This is a quadratic equation where the variable is \( |x| \). We can factor this equation:
\( (|x| - 1)(|x| - 2) = 0 \)
This means either \( |x| - 1 = 0 \) or \( |x| - 2 = 0 \).
Solving for \( |x| \):
\( |x| = 1 \) or \( |x| = 2 \)
Now, we find the values of \( x \) for each case:
If \( |x| = 1 \), then \( x = 1 \) or \( x = -1 \).
If \( |x| = 2 \), then \( x = 2 \) or \( x = -2 \).
Thus, the real solutions are \( x = 1, -1, 2, -2 \).
There are a total of 4 distinct real solutions for the equation. When solving equations with absolute values, remember that \( |x|=k \) has two solutions for \( x \) if \( k>0 \).
In simple words: We changed the equation so it uses absolute values, making it like a simple quadratic. Then we found that absolute x could be 1 or 2. This means x can be 1, -1, 2, or -2, so there are 4 answers.

🎯 Exam Tip: When an equation contains both \( x^2 \) and \( |x| \), remember that \( x^2 = |x|^2 \) can simplify the problem, often leading to multiple solutions for \( x \).