RBSE Solutions Class 10 Maths Chapter 3 Polynomials Exercise 3.4

Question 1. Solve the following equations by the method of completing square.
(i) \( 3x^2 - 5x + 2 = 0 \)
(ii) \( 5x^2 - 6x - 2 = 0 \)
(iii) \( 4x^2 + 3x + 5 = 0 \)
(iv) \( 4x^2 + 4\sqrt{3}x + 3 = 0 \)
(v) \( 2x^2 + x - 4 = 0 \)
(vi) \( 2x^2 + x + 4 = 0 \)
(vii) \( 4x^2 + 4bx - (a^2 - b^2) = 0 \)
Answer:
(i) Given quadratic equation is \( 3x^2 - 5x + 2 = 0 \).
First, divide the entire equation by 3 to make the coefficient of \( x^2 \) equal to 1.
\( x^2 - \frac{5}{3}x + \frac{2}{3} = 0 \)
Next, move the constant term to the right side of the equation.
\( x^2 - \frac{5}{3}x = - \frac{2}{3} \)
To complete the square, add the square of half the coefficient of x to both sides. The coefficient of x is \( -\frac{5}{3} \), so half of it is \( -\frac{5}{6} \).
\( x^2 - \frac{5}{3}x + \left( \frac{5}{6} \right)^2 = - \frac{2}{3} + \left( \frac{5}{6} \right)^2 \)
The left side now forms a perfect square.
\( \left( x - \frac{5}{6} \right)^2 = - \frac{2}{3} + \frac{25}{36} \)
Now, find a common denominator for the right side.
\( \left( x - \frac{5}{6} \right)^2 = \frac{-2 \times 12}{3 \times 12} + \frac{25}{36} \)
\( \left( x - \frac{5}{6} \right)^2 = \frac{-24 + 25}{36} \)
\( \left( x - \frac{5}{6} \right)^2 = \frac{1}{36} \)
Take the square root of both sides.
\( x - \frac{5}{6} = \pm \sqrt{\frac{1}{36}} \)
\( x - \frac{5}{6} = \pm \frac{1}{6} \)
Now, solve for \( x \) using both the positive and negative values.
Taking the positive sign:
\( x - \frac{5}{6} = \frac{1}{6} \)
\( x = \frac{1}{6} + \frac{5}{6} \)
\( x = \frac{6}{6} \)
\( x = 1 \)
Taking the negative sign:
\( x - \frac{5}{6} = - \frac{1}{6} \)
\( x = - \frac{1}{6} + \frac{5}{6} \)
\( x = \frac{4}{6} \)
\( x = \frac{2}{3} \)
Therefore, the solutions for the given equation are \( x = 1 \) and \( x = \frac{2}{3} \). Completing the square helps turn a regular quadratic equation into a simpler form to solve.
In simple words: We changed the equation so one side was a perfect square, like \( (a-b)^2 \). Then we took the square root of both sides to find the values of \( x \). This method helps find two possible answers for \( x \).

🎯 Exam Tip: When completing the square, remember to add \( (\frac{\text{coefficient of } x}{2})^2 \) to *both* sides of the equation to maintain balance.

(ii) Given quadratic equation is \( 5x^2 - 6x - 2 = 0 \).
First, divide the entire equation by 5 to make the coefficient of \( x^2 \) equal to 1.
\( x^2 - \frac{6}{5}x - \frac{2}{5} = 0 \)
Next, move the constant term to the right side of the equation.
\( x^2 - \frac{6}{5}x = \frac{2}{5} \)
To complete the square, add the square of half the coefficient of x to both sides. The coefficient of x is \( -\frac{6}{5} \), so half of it is \( -\frac{6}{10} \).
\( x^2 - \frac{6}{5}x + \left( \frac{6}{10} \right)^2 = \frac{2}{5} + \left( \frac{6}{10} \right)^2 \)
The left side now forms a perfect square.
\( \left( x - \frac{6}{10} \right)^2 = \frac{2}{5} + \frac{36}{100} \)
Simplify the fraction on the left side and find a common denominator for the right side.
\( \left( x - \frac{3}{5} \right)^2 = \frac{2 \times 20}{5 \times 20} + \frac{36}{100} \)
\( \left( x - \frac{3}{5} \right)^2 = \frac{40 + 36}{100} \)
\( \left( x - \frac{3}{5} \right)^2 = \frac{76}{100} \)
Take the square root of both sides.
\( x - \frac{3}{5} = \pm \sqrt{\frac{76}{100}} \)
\( x - \frac{3}{5} = \pm \frac{\sqrt{76}}{10} \)
Simplify \( \sqrt{76} \) as \( \sqrt{4 \times 19} = 2\sqrt{19} \).
\( x - \frac{3}{5} = \pm \frac{2\sqrt{19}}{10} \)
\( x - \frac{3}{5} = \pm \frac{\sqrt{19}}{5} \)
Now, solve for \( x \) using both the positive and negative values.
Taking the positive sign:
\( x = \frac{3}{5} + \frac{\sqrt{19}}{5} \)
\( x = \frac{3 + \sqrt{19}}{5} \)
Taking the negative sign:
\( x = \frac{3}{5} - \frac{\sqrt{19}}{5} \)
\( x = \frac{3 - \sqrt{19}}{5} \)
Therefore, the solutions for the given equation are \( x = \frac{3 + \sqrt{19}}{5} \) and \( x = \frac{3 - \sqrt{19}}{5} \). The quadratic formula is a general way to solve such equations if completing the square seems too long.
In simple words: We followed the same steps: divide, move the constant, add the square of half the middle term, and then take the square root. The answers here have square roots because the numbers didn't divide perfectly.

🎯 Exam Tip: When simplifying square roots like \( \sqrt{76} \), always look for perfect square factors (like 4, 9, 16, etc.) to extract them and simplify the radical.

(iii) Given quadratic equation is \( 4x^2 + 3x + 5 = 0 \).
First, divide the entire equation by 4 to make the coefficient of \( x^2 \) equal to 1.
\( x^2 + \frac{3}{4}x + \frac{5}{4} = 0 \)
Next, move the constant term to the right side of the equation.
\( x^2 + \frac{3}{4}x = - \frac{5}{4} \)
To complete the square, add the square of half the coefficient of x to both sides. The coefficient of x is \( \frac{3}{4} \), so half of it is \( \frac{3}{8} \).
\( x^2 + \frac{3}{4}x + \left( \frac{3}{8} \right)^2 = - \frac{5}{4} + \left( \frac{3}{8} \right)^2 \)
The left side now forms a perfect square.
\( \left( x + \frac{3}{8} \right)^2 = - \frac{5}{4} + \frac{9}{64} \)
Now, find a common denominator for the right side.
\( \left( x + \frac{3}{8} \right)^2 = \frac{-5 \times 16}{4 \times 16} + \frac{9}{64} \)
\( \left( x + \frac{3}{8} \right)^2 = \frac{-80 + 9}{64} \)
\( \left( x + \frac{3}{8} \right)^2 = \frac{-71}{64} \)
Since the square of any real number cannot be negative, there are no real roots for this equation. The right-hand side is a negative number, so taking its square root would involve imaginary numbers.
In simple words: After doing all the steps, we ended up with a square number on one side equal to a negative number. This means there are no real answers for \( x \), because you can't get a negative number by squaring a real number.

🎯 Exam Tip: If the right-hand side of the equation becomes negative after completing the square, it means the quadratic equation has no real solutions.

(iv) Given quadratic equation is \( 4x^2 + 4\sqrt{3}x + 3 = 0 \).
First, divide the entire equation by 4 to make the coefficient of \( x^2 \) equal to 1.
\( x^2 + \sqrt{3}x + \frac{3}{4} = 0 \)
Next, move the constant term to the right side of the equation.
\( x^2 + \sqrt{3}x = - \frac{3}{4} \)
To complete the square, add the square of half the coefficient of x to both sides. The coefficient of x is \( \sqrt{3} \), so half of it is \( \frac{\sqrt{3}}{2} \).
\( x^2 + \sqrt{3}x + \left( \frac{\sqrt{3}}{2} \right)^2 = - \frac{3}{4} + \left( \frac{\sqrt{3}}{2} \right)^2 \)
The left side now forms a perfect square.
\( \left( x + \frac{\sqrt{3}}{2} \right)^2 = - \frac{3}{4} + \frac{3}{4} \)
\( \left( x + \frac{\sqrt{3}}{2} \right)^2 = 0 \)
Take the square root of both sides.
\( x + \frac{\sqrt{3}}{2} = 0 \)
\( x = - \frac{\sqrt{3}}{2} \)
In this case, since the right side was 0, there is only one distinct real root, which means the quadratic equation has two equal roots. This often happens when the quadratic is a perfect square to begin with.
In simple words: We completed the square and found that the equation equals zero. This means \( x \) has only one value that makes the equation true. Both roots are the same.

🎯 Exam Tip: When a quadratic equation's discriminant (b²-4ac) is zero, it indicates that the equation has exactly one real root, also known as two equal roots.

(v) Given quadratic equation is \( 2x^2 + x - 4 = 0 \).
First, divide the entire equation by 2 to make the coefficient of \( x^2 \) equal to 1.
\( x^2 + \frac{1}{2}x - 2 = 0 \)
Next, move the constant term to the right side of the equation.
\( x^2 + \frac{1}{2}x = 2 \)
To complete the square, add the square of half the coefficient of x to both sides. The coefficient of x is \( \frac{1}{2} \), so half of it is \( \frac{1}{4} \).
\( x^2 + \frac{1}{2}x + \left( \frac{1}{4} \right)^2 = 2 + \left( \frac{1}{4} \right)^2 \)
The left side now forms a perfect square.
\( \left( x + \frac{1}{4} \right)^2 = 2 + \frac{1}{16} \)
Now, find a common denominator for the right side.
\( \left( x + \frac{1}{4} \right)^2 = \frac{2 \times 16}{1 \times 16} + \frac{1}{16} \)
\( \left( x + \frac{1}{4} \right)^2 = \frac{32 + 1}{16} \)
\( \left( x + \frac{1}{4} \right)^2 = \frac{33}{16} \)
Take the square root of both sides.
\( x + \frac{1}{4} = \pm \sqrt{\frac{33}{16}} \)
\( x + \frac{1}{4} = \pm \frac{\sqrt{33}}{4} \)
Now, solve for \( x \) using both the positive and negative values.
Taking the positive sign:
\( x = - \frac{1}{4} + \frac{\sqrt{33}}{4} \)
\( x = \frac{\sqrt{33} - 1}{4} \)
Taking the negative sign:
\( x = - \frac{1}{4} - \frac{\sqrt{33}}{4} \)
\( x = \frac{- \sqrt{33} - 1}{4} \)
Therefore, the solutions for the given equation are \( x = \frac{\sqrt{33} - 1}{4} \) and \( x = \frac{- \sqrt{33} - 1}{4} \). This method is useful for transforming quadratic equations into a solvable form.
In simple words: We used the method of completing the square to change the equation. This allowed us to find two different answers for \( x \), one with \( +\sqrt{33} \) and one with \( -\sqrt{33} \).

🎯 Exam Tip: Always make sure to consider both the positive and negative square roots when solving for \( x \) to get all possible solutions.

(vi) Given quadratic equation is \( 2x^2 + x + 4 = 0 \).
First, divide the entire equation by 2 to make the coefficient of \( x^2 \) equal to 1.
\( x^2 + \frac{1}{2}x + 2 = 0 \)
Next, move the constant term to the right side of the equation.
\( x^2 + \frac{1}{2}x = -2 \)
To complete the square, add the square of half the coefficient of x to both sides. The coefficient of x is \( \frac{1}{2} \), so half of it is \( \frac{1}{4} \).
\( x^2 + \frac{1}{2}x + \left( \frac{1}{4} \right)^2 = -2 + \left( \frac{1}{4} \right)^2 \)
The left side now forms a perfect square.
\( \left( x + \frac{1}{4} \right)^2 = -2 + \frac{1}{16} \)
Now, find a common denominator for the right side.
\( \left( x + \frac{1}{4} \right)^2 = \frac{-2 \times 16}{1 \times 16} + \frac{1}{16} \)
\( \left( x + \frac{1}{4} \right)^2 = \frac{-32 + 1}{16} \)
\( \left( x + \frac{1}{4} \right)^2 = \frac{-31}{16} \)
Since the square of any real number cannot be negative, there are no real roots for this equation. The right-hand side is a negative number, so taking its square root would result in an imaginary number.
In simple words: When we tried to solve this equation by making a perfect square, we got a negative number on one side. This means there are no real numbers for \( x \) that can make this equation true.

🎯 Exam Tip: Always check the sign of the constant term after moving it and adding the squared term; if it's negative, real solutions don't exist.

(vii) Given quadratic equation is \( 4x^2 + 4bx - (a^2 - b^2) = 0 \).
First, divide the entire equation by 4 to make the coefficient of \( x^2 \) equal to 1.
\( x^2 + bx - \frac{(a^2 - b^2)}{4} = 0 \)
Next, move the constant term to the right side of the equation.
\( x^2 + bx = \frac{(a^2 - b^2)}{4} \)
To complete the square, add the square of half the coefficient of x to both sides. The coefficient of x is \( b \), so half of it is \( \frac{b}{2} \).
\( x^2 + bx + \left( \frac{b}{2} \right)^2 = \frac{(a^2 - b^2)}{4} + \left( \frac{b}{2} \right)^2 \)
The left side now forms a perfect square.
\( \left( x + \frac{b}{2} \right)^2 = \frac{a^2 - b^2}{4} + \frac{b^2}{4} \)
Combine the terms on the right side since they have a common denominator.
\( \left( x + \frac{b}{2} \right)^2 = \frac{a^2 - b^2 + b^2}{4} \)
\( \left( x + \frac{b}{2} \right)^2 = \frac{a^2}{4} \)
Take the square root of both sides.
\( x + \frac{b}{2} = \pm \sqrt{\frac{a^2}{4}} \)
\( x + \frac{b}{2} = \pm \frac{a}{2} \)
Now, solve for \( x \) using both the positive and negative values.
Taking the positive sign:
\( x = - \frac{b}{2} + \frac{a}{2} \)
\( x = \frac{a - b}{2} \)
Taking the negative sign:
\( x = - \frac{b}{2} - \frac{a}{2} \)
\( x = \frac{-(a + b)}{2} \)
Therefore, the solutions for the given equation are \( x = \frac{a - b}{2} \) and \( x = - \frac{a + b}{2} \). Algebraic expressions can also be solved using the completing the square method.
In simple words: We used the same method even with letters instead of numbers. We made a perfect square on one side and then took the square root to find two possible answers for \( x \), expressed using \( a \) and \( b \).

🎯 Exam Tip: Be careful with signs when dealing with algebraic expressions, especially when moving terms or simplifying square roots of variables.

Question 2. Find the roots of the following equations (if exists), using Shridhar Acharya quadratic formula.
(i) \( 2x^2 - 2\sqrt{2}x + 1 = 0 \)
(ii) \( 9x^2 + 7x - 2 = 0 \)
(iii) \( x + \frac{1}{x} = 3, x \neq 0 \)
(iv) \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
Answer:
(i) Given quadratic equation is \( 2x^2 - 2\sqrt{2}x + 1 = 0 \).
Compare this equation with the general quadratic equation \( ax^2 + bx + c = 0 \).
We have \( a = 2 \), \( b = -2\sqrt{2} \), and \( c = 1 \).
First, calculate the discriminant \( D = b^2 - 4ac \). This helps us know if real roots exist.
\( D = (-2\sqrt{2})^2 - 4 \times 2 \times 1 \)
\( D = (4 \times 2) - 8 \)
\( D = 8 - 8 \)
\( D = 0 \)
Since the discriminant is 0, the equation has real and equal roots.
Now, apply the Shridhar Acharya (quadratic) formula to find the roots:
\( x = \frac{-b \pm \sqrt{D}}{2a} \)
Substitute the values of \( a \), \( b \), and \( D \).
\( x = \frac{-(-2\sqrt{2}) \pm \sqrt{0}}{2 \times 2} \)
\( x = \frac{2\sqrt{2}}{4} \)
\( x = \frac{\sqrt{2}}{2} \)
We can simplify this by multiplying the numerator and denominator by \( \sqrt{2} \).
\( x = \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{\sqrt{2}} \)
\( x = \frac{2}{2\sqrt{2}} \)
\( x = \frac{1}{\sqrt{2}} \)
Since \( D = 0 \), both roots are the same. So the roots are \( x = \frac{1}{\sqrt{2}} \) and \( x = \frac{1}{\sqrt{2}} \). This formula provides a direct way to find roots without factoring or completing the square.
In simple words: We used a special formula to find the answers for \( x \). First, we checked a value called the discriminant. Since it was zero, we knew there was only one answer for \( x \), which we then found using the formula.

🎯 Exam Tip: If the discriminant \( D \) is 0, the quadratic equation has two identical real roots. If \( D > 0 \), there are two distinct real roots. If \( D < 0 \), there are no real roots.

(ii) Given quadratic equation is \( 9x^2 + 7x - 2 = 0 \).
Compare this equation with the general quadratic equation \( ax^2 + bx + c = 0 \).
We have \( a = 9 \), \( b = 7 \), and \( c = -2 \).
First, calculate the discriminant \( D = b^2 - 4ac \).
\( D = (7)^2 - 4 \times 9 \times (-2) \)
\( D = 49 - (-72) \)
\( D = 49 + 72 \)
\( D = 121 \)
Since the discriminant is greater than 0, the equation has two distinct real roots.
Now, apply the Shridhar Acharya (quadratic) formula to find the roots:
\( x = \frac{-b \pm \sqrt{D}}{2a} \)
Substitute the values of \( a \), \( b \), and \( D \).
\( x = \frac{-7 \pm \sqrt{121}}{2 \times 9} \)
\( x = \frac{-7 \pm 11}{18} \)
Now, solve for \( x \) using both the positive and negative values.
Taking the positive sign:
\( x = \frac{-7 + 11}{18} \)
\( x = \frac{4}{18} \)
\( x = \frac{2}{9} \)
Taking the negative sign:
\( x = \frac{-7 - 11}{18} \)
\( x = \frac{-18}{18} \)
\( x = -1 \)
Therefore, the solutions for the given equation are \( x = \frac{2}{9} \) and \( x = -1 \). The quadratic formula quickly helps find roots for equations that might be hard to factor directly.
In simple words: We used the quadratic formula. After finding the discriminant was a positive number, we knew there would be two different answers. We then put the numbers into the formula to find these two answers for \( x \).

🎯 Exam Tip: Simplify fractions like \( \frac{4}{18} \) to their lowest terms \( \frac{2}{9} \) for full marks.

(iii) Given equation is \( x + \frac{1}{x} = 3, x \neq 0 \).
First, convert this into a standard quadratic equation by multiplying the entire equation by \( x \).
\( x \cdot x + x \cdot \frac{1}{x} = 3 \cdot x \)
\( x^2 + 1 = 3x \)
Now, rearrange the terms to get the standard form \( ax^2 + bx + c = 0 \).
\( x^2 - 3x + 1 = 0 \)
Compare this equation with \( ax^2 + bx + c = 0 \).
We have \( a = 1 \), \( b = -3 \), and \( c = 1 \).
First, calculate the discriminant \( D = b^2 - 4ac \).
\( D = (-3)^2 - 4 \times 1 \times 1 \)
\( D = 9 - 4 \)
\( D = 5 \)
Since the discriminant is greater than 0, the equation has two distinct real roots.
Now, apply the Shridhar Acharya (quadratic) formula to find the roots:
\( x = \frac{-b \pm \sqrt{D}}{2a} \)
Substitute the values of \( a \), \( b \), and \( D \).
\( x = \frac{-(-3) \pm \sqrt{5}}{2 \times 1} \)
\( x = \frac{3 \pm \sqrt{5}}{2} \)
Therefore, the solutions for the given equation are \( x = \frac{3 + \sqrt{5}}{2} \) and \( x = \frac{3 - \sqrt{5}}{2} \). The quadratic formula is useful even when the equation doesn't look like a standard quadratic at first.
In simple words: We first changed the equation into a normal quadratic form. Then, we used the quadratic formula. Because the discriminant was positive, we found two different answers for \( x \), which include a square root.

🎯 Exam Tip: Always convert equations like \( x + \frac{1}{x} = 3 \) into the standard quadratic form \( ax^2 + bx + c = 0 \) before applying the quadratic formula.

(iv) Given quadratic equation is \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \).
Compare this equation with the general quadratic equation \( ax^2 + bx + c = 0 \).
We have \( a = \sqrt{2} \), \( b = 7 \), and \( c = 5\sqrt{2} \).
First, calculate the discriminant \( D = b^2 - 4ac \).
\( D = (7)^2 - 4 \times \sqrt{2} \times 5\sqrt{2} \)
\( D = 49 - (4 \times 5 \times \sqrt{2} \times \sqrt{2}) \)
\( D = 49 - (20 \times 2) \)
\( D = 49 - 40 \)
\( D = 9 \)
Since the discriminant is greater than 0, the equation has two distinct real roots.
Now, apply the Shridhar Acharya (quadratic) formula to find the roots:
\( x = \frac{-b \pm \sqrt{D}}{2a} \)
Substitute the values of \( a \), \( b \), and \( D \).
\( x = \frac{-7 \pm \sqrt{9}}{2 \times \sqrt{2}} \)
\( x = \frac{-7 \pm 3}{2\sqrt{2}} \)
Now, solve for \( x \) using both the positive and negative values.
Taking the positive sign:
\( x = \frac{-7 + 3}{2\sqrt{2}} \)
\( x = \frac{-4}{2\sqrt{2}} \)
\( x = \frac{-2}{\sqrt{2}} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \).
\( x = \frac{-2\sqrt{2}}{2} \)
\( x = -\sqrt{2} \)
Taking the negative sign:
\( x = \frac{-7 - 3}{2\sqrt{2}} \)
\( x = \frac{-10}{2\sqrt{2}} \)
\( x = \frac{-5}{\sqrt{2}} \)
To rationalize the denominator, multiply by \( \frac{\sqrt{2}}{\sqrt{2}} \).
\( x = \frac{-5\sqrt{2}}{2} \)
Therefore, the solutions for the given equation are \( x = -\sqrt{2} \) and \( x = \frac{-5\sqrt{2}}{2} \). Even with square roots as coefficients, the quadratic formula provides a direct solution.
In simple words: We used the quadratic formula, even though the equation had square roots in it. The discriminant was positive, so we got two different answers for \( x \). We then simplified these answers to remove square roots from the bottom part.

🎯 Exam Tip: Remember to rationalize the denominator if your final answer has a square root in the bottom part, for example, change \( \frac{2}{\sqrt{2}} \) to \( \sqrt{2} \).

Question 2. (v) Find the roots of the following equations (if exists), using Shridhar Acharya quadratic formula.
(v) \( x^2 + 4x + 5 = 0 \)
Answer: The given quadratic equation is \( x^2 + 4x + 5 = 0 \).
Comparing this to the general quadratic equation \( ax^2 + bx + c = 0 \), we get \( a=1, b=4, \) and \( c=5 \).
Now, we calculate the discriminant \( D = b^2 - 4ac \).
\( D = (4)^2 - 4(1)(5) \)
\( D = 16 - 20 \)
\( D = -4 \)
Since the discriminant is negative \( (D < 0) \), the given equation does not have real roots. This means there are no real numbers that can solve the equation.
In simple words: When we check the numbers in the equation, we find that the discriminant (a special value) is negative. A negative discriminant means the equation has no real number solutions.

🎯 Exam Tip: Always calculate the discriminant (\( b^2 - 4ac \)) first when using the quadratic formula, as a negative value immediately tells you that real roots do not exist, saving further calculation.

Question 3. Find two consecutive positive integers, sum of whose square is 290.
Answer: Let the two consecutive odd positive integers be \( x \) and \( x+2 \).
According to the problem, the sum of their squares is 290.
So, \( x^2 + (x+2)^2 = 290 \)
\( x^2 + x^2 + 4x + 4 = 290 \)
\( 2x^2 + 4x + 4 = 290 \)
\( 2x^2 + 4x - 286 = 0 \)
Now, divide the entire equation by 2:
\( x^2 + 2x - 143 = 0 \)
To solve this quadratic equation, we factor it:
\( x^2 + 13x - 11x - 143 = 0 \)
\( x(x + 13) - 11(x + 13) = 0 \)
\( (x + 13)(x - 11) = 0 \)
This gives us two possible values for \( x \):
\( x - 11 = 0 \implies x = 11 \)
\( x + 13 = 0 \implies x = -13 \)
Since we are looking for positive integers, we choose \( x = 11 \).
The first integer is \( x = 11 \).
The second consecutive odd positive integer is \( x+2 = 11+2 = 13 \).
Therefore, the two consecutive odd positive integers are 11 and 13.
In simple words: We set up an equation where the squares of two numbers that are two apart add up to 290. Solving this equation gave us 11 and 13 as the numbers.

🎯 Exam Tip: Always check if the question specifies "consecutive integers" or "consecutive odd/even integers" as this changes the setup (x, x+1 vs. x, x+2).

Question 4. The difference of square of two numbers is 45 and square of the smaller number is 4 times the larger number. Find two numbers.
Answer: Let the two numbers be \( x \) (the larger number) and \( y \) (the smaller number).
From the first condition, the difference of their squares is 45:
\( x^2 - y^2 = 45 \) (Equation 1)
From the second condition, the square of the smaller number is 4 times the larger number:
\( y^2 = 4x \) (Equation 2)
Now, substitute Equation 2 into Equation 1:
\( x^2 - 4x = 45 \)
Rearrange into a standard quadratic equation:
\( x^2 - 4x - 45 = 0 \)
To solve this, we can factor the quadratic equation:
\( x^2 - 9x + 5x - 45 = 0 \)
\( x(x - 9) + 5(x - 9) = 0 \)
\( (x - 9)(x + 5) = 0 \)
This gives two possible values for \( x \):
\( x - 9 = 0 \implies x = 9 \)
\( x + 5 = 0 \implies x = -5 \)
If \( x = -5 \), then \( y^2 = 4(-5) = -20 \). A square of a real number cannot be negative, so \( x = -5 \) is not a valid solution for a real number \( y \).
Therefore, the larger number must be \( x = 9 \).
Now, substitute \( x = 9 \) back into Equation 2 to find \( y \):
\( y^2 = 4(9) \)
\( y^2 = 36 \)
\( y = \pm \sqrt{36} \)
\( y = \pm 6 \)
So, the two numbers can be 9 and 6, or 9 and -6.
In simple words: We made two equations from the given information about the two numbers. By solving these equations, we found that the larger number is 9, and the smaller number can be either 6 or -6.

🎯 Exam Tip: When dealing with square roots and finding numbers, always consider both positive and negative solutions, and then check them against the problem's context (e.g., "positive numbers").

Question 5. Divide 16 into two part such that twice the square of larger part is more, then 164 from the square of the smaller part.
Answer: Let the two parts of 16 be \( x \) and \( 16-x \).
Assume \( x \) is the larger part.
According to the problem, twice the square of the larger part is 164 more than the square of the smaller part.
So, we can write the equation as:
\( 2x^2 = (16-x)^2 + 164 \)
First, expand the term \( (16-x)^2 \):
\( (16-x)^2 = 16^2 - 2(16)(x) + x^2 = 256 - 32x + x^2 \)
Substitute this back into the equation:
\( 2x^2 = (256 - 32x + x^2) + 164 \)
Combine the constant terms on the right side:
\( 2x^2 = 256 + 164 - 32x + x^2 \)
\( 2x^2 = 420 - 32x + x^2 \)
Move all terms to one side to form a quadratic equation:
\( 2x^2 - x^2 + 32x - 420 = 0 \)
\( x^2 + 32x - 420 = 0 \)
Now, we solve this quadratic equation by factoring:
\( x^2 + 42x - 10x - 420 = 0 \)
\( x(x + 42) - 10(x + 42) = 0 \)
\( (x + 42)(x - 10) = 0 \)
This gives two possible values for \( x \):
\( x - 10 = 0 \implies x = 10 \)
\( x + 42 = 0 \implies x = -42 \)
Since \( x \) represents a part of 16, it must be a positive value. Thus, \( x = -42 \) is not a valid solution.
Therefore, the larger part is \( x = 10 \).
The smaller part is \( 16 - x = 16 - 10 = 6 \).
So, the two parts are 10 and 6.
In simple words: We split 16 into two parts and wrote an equation based on the given conditions. Solving this equation showed that the two parts are 10 and 6.

🎯 Exam Tip: Clearly define the variables for each part (e.g., x and 16-x) and carefully set up the equation based on the word problem to avoid errors in calculation.