CBSE Class 10 Quadratic Equations Sure Shot Questions Set 13

ZEROS OF A QUADRATIC POLYNOMIAL

The value of the polynomial \( x^2 – 7x + 10 \) at :
(i) \( x = 1 \) is \( (1)^2 – 7 \times 1 + 10 = 1 – 7 + 10 = 4 \)
(ii) \( x = 2 \) is \( (2)^2 – 7 \times 2 + 10 = 4 – 14 + 10 = 0 \)
(iii) \( x = 3 \) is \( (3)^2 – 7 \times 3 + 10 = 9 – 21 + 10 = – 2 \)
(iv) \( x = 5 \) is \( (5)^2 – 7 \times 5 + 10 = 25 – 35 + 10 = 0 \)

It is observed here that for \( x = 2 \) and \( x = 5 \); the value of polynomial \( x^2 – 7x + 10 \) is zero. These two values of \( x \) are called zeros of the polynomial.

Thus, if for \( x = \alpha \), where \( \alpha \) is a real number, the value of given quadratic polynomial is zero; the real number \( \alpha \) is called zero of the quadratic polynomial.

Question. Show that :
(i) \( x = 3 \) is a zero of quadratic polynomial \( x^2 – 2x – 3 \).
(ii) \( x = – 2 \) is a zero of quadratic polynomial \( 3x^2 + 7x + 2 \).
(iii) \( x = 4 \) is not a zero of quadratic polynomial \( 2x^2 – 7x – 5 \).
Answer: (i) The value of \( x^2 – 2x – 3 \) at \( x = 3 \) is
\( (3)^2 – 2 \times 3 – 3 = 9 – 6 – 3 = 0 \)
\( \Rightarrow x = 3 \) is a zero of quadratic polynomial \( x^2 – 2x – 3 \).
(ii) The value of \( 3x^2 + 7x + 2 \) at \( x = – 2 \) is
\( 3(–2)^2 + 7 (–2) + 2 = 12 – 14 + 2 = 0 \)
\( \Rightarrow x = – 2 \) is a zero of quadratic polynomial \( 3x^2 + 7x + 2 \).
(iii) The value of \( 2x^2 – 7x – 5 \) at \( x = 4 \) is
\( 2(4)^2 – 7(4) – 5 = 32 – 28 – 5 = – 1 \neq 0 \)
\( \Rightarrow x = 4 \) is not a zero of quadratic polynomial \( 2x^2 – 7x – 5 \).

Question. Find the value of \( m \), if \( x = 2 \) is a zero of quadratic polynomial \( 3x^2 – mx + 4 \).
Answer: Since, \( x = 2 \) is a zero of \( 3x^2 – mx + 4 \)
\( \Rightarrow 3(2)^2 – m \times 2 + 4 = 0 \)
\( \Rightarrow 12 – 2m + 4 = 0 \), i.e., \( m = 8 \).

QUADRATIC EQUATION & IT'S ROOTS

Since, \( ax^2 + bx + c, a \neq 0 \) is a quadratic polynomial, \( ax^2 + bx + c = 0, a \neq 0 \) is called a quadratic equation.
(i) \( –x^2 – 7x + 2 = 0 \) is a quadratic equation, as \( –x^2 – 7x + 2 \) is a quadratic polynomial.
(ii) \( 5x^2 – 7x = 0 \) is a quadratic equation.
(iii) \( 5x^2 + 2 = 0 \) is a quadratic equation, but
(iv) \( –7x + 2 = 0 \) is not a quadratic equation.

Question. Which of the following are quadratic equations, give reason :
(i) \( x^2 – 8x + 6 = 0 \)
(ii) \( 3x^2 – 4 = 0 \)
(iii) \( 2x + \frac{5}{x} = x^2 \)
(iv) \( x^2 + \frac{2}{x^2} = 3 \)
Answer: (i) Since, \( x^2 – 8x + 6 \) is a quadratic polynomial \( \Rightarrow x^2 – 8x + 6 = 0 \) is a quadratic equation.
(ii) \( 3x^2 – 4 = 0 \) is a quadratic equation.
(iii) \( 2x + \frac{5}{x} = x^2 \Rightarrow 2x^2 + 5 = x^3 \Rightarrow x^3 – 2x^2 – 5 = 0 \); which is cubic and not a quadratic equation.
(iv) \( x^2 + \frac{2}{x^2} = 3 \Rightarrow x^4 + 2 = 3x^2 \Rightarrow x^4 – 3x^2 + 2 = 0 \); which is biquadratic and not a quadratic equation.

Question. In each of the following, determine whether the given values are solutions (roots) of the equation or not :
(i) \( 3x^2 – 2x – 1 = 0; x = 1 \)
(ii) \( x^2 + 6x + 5 = 0; x = – 1, x = – 5 \)
(iii) \( x^2 + \sqrt{2} x – 4 = 0; x = \sqrt{2} , x = – 2 \sqrt{2} \)
Answer: (i) Value of \( 3x^2 – 2x – 1 \) at \( x = 1 \) is \( 3(1)^2 – 2(1) – 1 = 3 – 2 – 1 = 0 = \text{RHS} \). \( \therefore x = 1 \) is a solution of the given equation.
(ii) For \( x = – 1, \text{L.H.S.} = (–1)^2 + 6 (–1) + 5 = 1 – 6 + 5 = 0 = \text{R.H.S.} \Rightarrow x = – 1 \) is a solution of the given equation. For \( x = – 5, \text{L.H.S.} = (–5)^2 + 6(–5) + 5 = 25 – 30 + 5 = 0 = \text{R.H.S.} \Rightarrow x = – 5 \) is a solution of the given equation.
(iii) For \( x = \sqrt{2} , \text{L.H.S.} = x^2 + \sqrt{2} x – 4 = (\sqrt{2})^2 + \sqrt{2} (\sqrt{2}) – 4 = 2 + 2 – 4 = 0 = \text{R.H.S.} \therefore x = \sqrt{2} \) is a solution of the given equation. For \( x = – 2 \sqrt{2} , \text{L.H.S.} = (–2 \sqrt{2})^2 + \sqrt{2} \times (– 2 \sqrt{2}) – 4 = 8 – 4 – 4 = 0 = \text{R.H.S.} \therefore x = – 2 \sqrt{2} \) is a solution of the given equation.

SOLVING A QUADRATIC EQUATION BY FACTORISATION

Since, \( 3x^2 – 5x + 2 \) is a quadratic polynomial; \( 3x^2 – 5x + 2 = 0 \) is a quadratic equation.
Also, \( 3x^2 – 5x + 2 = 3x^2 – 3x – 2x + 2 = 3x(x – 1) – 2(x – 1) = (x – 1) (3x – 2) \).
In the same way: \( 3x^2 – 5x + 2 = 0 \Rightarrow (x – 1) (3x – 2) = 0 \Rightarrow x – 1 = 0 \text{ or } 3x – 2 = 0 \Rightarrow x = 1 \text{ or } x = \frac{2}{3} \); which is the solution of given quadratic equation.

In order to solve the given Quadratic Equation:

• 1. Clear the fractions and brackets, if given.
• 2. By transferring each term to the left hand side; express the given equation as \( ax^2 + bx + c = 0 \) or \( a + bx + cx^2 = 0 \).
• 3. Factorise left hand side of the equation obtained (the right hand side being zero).
• 4. By putting each factor equal to zero; solve it.

Question. Solve :
(i) \( x^2 + 3x – 18 = 0 \)
(ii) \( (x – 4) (5x + 2) = 0 \)
(iii) \( 2x^2 + ax – a^2 = 0 \); where ‘\( a \)’ is a real number.
Answer: (i) \( x^2 + 3x – 18 = 0 \Rightarrow x^2 + 6x – 3x – 18 = 0 \Rightarrow x(x + 6) – 3(x + 6) = 0 \Rightarrow (x + 6) (x – 3) = 0 \Rightarrow x + 6 = 0 \) or \( x – 3 = 0 \Rightarrow x = – 6 \) or \( x = 3 \). \( \therefore \) Roots of the given equation are : \( – 6 \) and \( 3 \).
(ii) \( (x – 4) (5x + 2) = 0 \Rightarrow x – 4 = 0 \) or \( 5x + 2 = 0 \Rightarrow x = 4 \) or \( x = – \frac{2}{5} \).
(iii) \( 2x^2 + ax – a^2 = 0 \Rightarrow 2x^2 + 2ax – ax – a^2 = 0 \Rightarrow 2x(x + a) – a(x + a) = 0 \Rightarrow (x + a) (2x – a) = 0 \Rightarrow x + a = 0 \) or \( 2x – a = 0 \Rightarrow x = – a \) or \( x = \frac{a}{2} \).

Question. Solve the following quadratic equations :
(i) \( x^2 + 5x = 0 \)
(ii) \( x^2 = 3x \)
(iii) \( x^2 = 4 \)
Answer: (i) \( x^2 + 5x = 0 \Rightarrow x(x + 5) = 0 \Rightarrow x = 0 \) or \( x + 5 = 0 \Rightarrow x = 0 \) or \( x = – 5 \).
(ii) \( x^2 = 3x \Rightarrow x^2 – 3x = 0 \Rightarrow x(x – 3) = 0 \Rightarrow x = 0 \) or \( x = 3 \).
(iii) \( x^2 = 4 \Rightarrow x = \pm 2 \).

Question. Solve the following quadratic equations :
(i) \( 7x^2 = 8 – 10x \)
(ii) \( 3(x^2 – 4) = 5x \)
(iii) \( x(x + 1) + (x + 2) (x + 3) = 42 \)
Answer: (i) \( 7x^2 = 8 – 10x \Rightarrow 7x^2 + 10x – 8 = 0 \Rightarrow 7x^2 + 14x – 4x – 8 = 0 \Rightarrow 7x(x + 2) – 4(x + 2) = 0 \Rightarrow (x + 2) (7x – 4) = 0 \Rightarrow x = – 2 \) or \( x = \frac{4}{7} \).
(ii) \( 3(x^2 – 4) = 5x \Rightarrow 3x^2 – 5x – 12 = 0 \Rightarrow 3x^2 – 9x + 4x – 12 = 0 \Rightarrow 3x(x – 3) + 4(x – 3) = 0 \Rightarrow (x – 3) (3x + 4) = 0 \Rightarrow x = 3 \) or \( x = – \frac{4}{3} \).
(iii) \( x(x + 1) + (x + 2) (x + 3) = 42 \Rightarrow x^2 + x + x^2 + 3x + 2x + 6 – 42 = 0 \Rightarrow 2x^2 + 6x – 36 = 0 \Rightarrow x^2 + 3x – 18 = 0 \Rightarrow x^2 + 6x – 3x – 18 = 0 \Rightarrow x(x + 6) – 3(x + 6) = 0 \Rightarrow (x + 6) (x – 3) = 0 \Rightarrow x = – 6 \) or \( x = 3 \).

Question. Solve for \( x : 12abx^2 – (9a^2 – 8b^2) x – 6ab = 0 \)
Answer: Given equation is: \( 12abx^2 – 9a^2x + 8b^2x – 6ab = 0 \Rightarrow 3ax(4bx – 3a) + 2b(4bx – 3a) = 0 \Rightarrow (4bx – 3a) (3ax + 2b) = 0 \Rightarrow 4bx – 3a = 0 \) or \( 3ax + 2b = 0 \Rightarrow x = \frac{3a}{4b} \) or \( x = – \frac{2b}{3a} \).

SOLVING A QUADRATIC EQUATION BY COMPLETING THE SQUARE

Every quadratic equation can be converted in the form : \( (x + a)^2 – b^2 = 0 \) or \( (x – a)^2 – b^2 = 0 \).
Steps :
1. Bring, if required, all the term of the quadratic equation to the left hand side.
2. Express the terms containing \( x \) as \( x^2 + 2xy \) or \( x^2 – 2xy \).
3. Add and subtract \( y^2 \) to get \( x^2 + 2xy + y^2 – y^2 \) or \( x^2 – 2xy + y^2 – y^2 \); which gives \( (x + y)^2 – y^2 \) or \( (x – y)^2 – y^2 \).

Question. Find the roots of the following quadratic equations (if they exist) by the method of completing the square.
(i) \( 2x^2 – 7x + 3 = 0 \)
(ii) \( 4x^2 + 4\sqrt{3} x + 3 = 0 \)
(iii) \( 2x^2 + x + 4 = 0 \)
Answer: (i) \( 2x^2 – 7x + 3 = 0 \Rightarrow x^2 – \frac{7}{2}x + \frac{3}{2} = 0 \Rightarrow x^2 – 2 \times x \times \frac{7}{4} + (\frac{7}{4})^2 – (\frac{7}{4})^2 + \frac{3}{2} = 0 \Rightarrow (x – \frac{7}{4})^2 – \frac{49}{16} + \frac{3}{2} = 0 \Rightarrow (x – \frac{7}{4})^2 – \frac{25}{16} = 0 \Rightarrow x – \frac{7}{4} = \pm \frac{5}{4} \Rightarrow x = 3 \) or \( x = \frac{1}{2} \).
(ii) \( 4x^2 + 4\sqrt{3}x + 3 = 0 \Rightarrow x^2 + \sqrt{3}x + \frac{3}{4} = 0 \Rightarrow x^2 + 2 \times x \times \frac{\sqrt{3}}{2} + (\frac{\sqrt{3}}{2})^2 – (\frac{\sqrt{3}}{2})^2 + \frac{3}{4} = 0 \Rightarrow (x + \frac{\sqrt{3}}{2})^2 = 0 \Rightarrow x = – \frac{\sqrt{3}}{2} \).
(iii) \( 2x^2 + x + 4 = 0 \Rightarrow x^2 + \frac{x}{2} + 2 = 0 \Rightarrow x^2 + 2 \times x \times \frac{1}{4} + (\frac{1}{4})^2 – (\frac{1}{4})^2 + 2 = 0 \Rightarrow (x + \frac{1}{4})^2 + \frac{31}{16} = 0 \Rightarrow (x + \frac{1}{4})^2 = – \frac{31}{16} \). This is not possible as the square of a real number cannot be negative.

SOLVING A QUADRATIC EQUATION BY USING QUADRATIC FORMULA

Hindu Method (Sri Dharacharya Method) :
By completing the perfect square as \( ax^2 + bx + c = 0 \Rightarrow x^2 + \frac{b}{a}x + \frac{c}{a} = 0 \). Adding and subtracting \( (\frac{b}{2a})^2 \), we get \( (x + \frac{b}{2a})^2 – \frac{b^2 – 4ac}{4a^2} = 0 \). Which gives, \( x = \frac{–b \pm \sqrt{b^2 – 4ac}}{2a} \). Hence the Quadratic equation \( ax^2 + bx + c = 0 (a \neq 0) \) has two roots, given by \( \alpha = \frac{–b + \sqrt{b^2 – 4ac}}{2a} \) and \( \beta = \frac{–b – \sqrt{b^2 – 4ac}}{2a} \). Every quadratic equation has at most two and only two real roots.

Question. Solve the following quadratic equations by using quadratic formula :
(i) \( x^2 – 7x + 12 = 0 \)
(ii) \( 3x^2 – x – 10 = 0 \)
Answer: (i) Comparing \( x^2 – 7x + 12 = 0 \) with \( ax^2 + bx + c = 0 \); we get : \( a = 1, b = – 7 \) and \( c = 12 \). \( b^2 – 4ac = (–7)^2 – 4 \times 1 \times 12 = 1 \). \( x = \frac{–b \pm \sqrt{b^2 – 4ac}}{2a} = \frac{7 \pm 1}{2} = 4 \) or \( 3 \).
(ii) For \( 3x^2 – x – 10 = 0 \), \( a = 3, b = – 1, c = – 10 \). \( b^2 – 4ac = (–1)^2 – 4 \times 3 \times (– 10) = 121 \). \( x = \frac{1 \pm 11}{6} = 2 \) or \( – \frac{5}{3} \).

Question. For a quadratic equation \( ax^2 + bx + c = 0, a \neq 0 \), prove that : \( x = \frac{–b \pm \sqrt{b^2 – 4ac}}{2a} \).
Answer: \( ax^2 + bx + c = 0 \Rightarrow 4a^2x^2 + 4abx + 4ac = 0 \) [Multiplying by ‘\( 4a \)’]
\( \Rightarrow (2ax)^2 + 2 \times 2ax \times b + b^2 – b^2 + 4ac = 0 \)
\( \Rightarrow (2ax + b)^2 = b^2 – 4ac \Rightarrow 2ax + b = \pm \sqrt{b^2 – 4ac} \Rightarrow x = \frac{–b \pm \sqrt{b^2 – 4ac}}{2a} \).

Question. Solve, by using quadratic formula, each of the following equations :
(i) \( 2x^2 + 5\sqrt{3}x + 6 = 0 \)
(ii) \( 3x^2 + 2\sqrt{5}x – 5 = 0 \)
Answer: (i) \( a = 2, b = 5\sqrt{3}, c = 6 \). \( b^2 – 4ac = 27 \). \( \sqrt{b^2 – 4ac} = 3\sqrt{3} \). \( x = \frac{–5\sqrt{3} \pm 3\sqrt{3}}{4} = – \frac{\sqrt{3}}{2} \) or \( – 2\sqrt{3} \).
(ii) \( a = 3, b = 2\sqrt{5}, c = – 5 \). \( b^2 – 4ac = 80 \). \( \sqrt{b^2 – 4ac} = 4\sqrt{5} \). \( x = \frac{–2\sqrt{5} \pm 4\sqrt{5}}{6} = \frac{\sqrt{5}}{3} \) or \( – \sqrt{5} \).

Question. Using the quadratic formula, solve the equation: \( a^2b^2x^2 – (4b^4 – 3a^4)x – 12a^2b^2 = 0 \).
Answer: \( A = a^2b^2, B = – (4b^4 – 3a^4), C = –12a^2b^2 \). \( B^2 – 4AC = (4b^4 – 3a^4)^2 + 48a^4b^4 = (4b^4 + 3a^4)^2 \). \( \sqrt{B^2 – 4AC} = 4b^4 + 3a^4 \). \( x = \frac{(4b^4 – 3a^4) \pm (4b^4 + 3a^4)}{2a^2b^2} = \frac{4b^2}{a^2} \) or \( – \frac{3a^2}{b^2} \).

NATURE OR CHARACTER OF THE ROOTS OF A QUADRATIC EQUATION

The nature of the roots depends on the value of \( b^2 – 4ac \). \( b^2 – 4ac \) is called the discriminant \( D \).
\( \bullet \) If \( D > 0 \), roots are real and unequal. If \( D \) is a perfect square, roots are rational. If not, roots are irrational.
\( \bullet \) If \( D = 0 \), roots are real and equal.
\( \bullet \) If \( D < 0 \), roots are imaginary (not real).

Question. Without solving, examine the nature of roots of the equations :
(i) \( 2x^2 + 2x + 3 = 0 \)
(ii) \( 2x^2 – 7x + 3 = 0 \)
(iii) \( x^2 – 5x – 2 = 0 \)
(iv) \( 4x^2 – 4x + 1 = 0 \)
Answer: (i) \( D = 2^2 – 4 \times 2 \times 3 = – 20 \). Roots are imaginary.
(ii) \( D = (–7)^2 – 4 \times 2 \times 3 = 25 \). Roots are rational and unequal.
(iii) \( D = (–5)^2 – 4 \times 1 \times (– 2) = 33 \). Roots are irrational and unequal.
(iv) \( D = (–4)^2 – 4 \times 4 \times 1 = 0 \). Roots are real and equal.

Question. For what value of \( m \), are the roots of the equation \( (3m + 1)x^2 + (11 + m)x + 9 = 0 \) equal?
Answer: \( D = (11 + m)^2 – 4(3m + 1) \times 9 = m^2 – 86m + 85 \). For equal roots, \( D = 0 \Rightarrow (m – 85) (m – 1) = 0 \Rightarrow m = 85 \) or \( m = 1 \).

SUM AND PRODUCT OF THE ROOTS

Let \( \alpha \) and \( \beta \) be the two roots of \( ax^2 + bx + c = 0 \).
Sum of roots: \( \alpha + \beta = – \frac{b}{a} = – \frac{\text{coefficient of } x}{\text{coefficient of } x^2} \).
Product of roots: \( \alpha\beta = \frac{c}{a} = \frac{\text{constant term}}{\text{coefficient of } x^2} \).

TO CONSTRUCT A QUADRATIC EQUATION WHOSE ROOTS ARE GIVEN

The equation is: \( x^2 – (\text{sum of roots})x + \text{product of roots} = 0 \).

Question. For each quadratic equation given below, find the sum of the roots and the product of the roots :
(i) \( x^2 + 3x – 6 = 0 \)
(ii) \( 2x^2 + 5\sqrt{3}x + 6 = 0 \)
(iii) \( 3x^2 + 2\sqrt{5}x – 5 = 0 \)
Answer: (i) Sum \( = –3 \), Product \( = –6 \).
(ii) Sum \( = – \frac{5\sqrt{3}}{2} \), Product \( = 3 \).
(iii) Sum \( = – \frac{2\sqrt{5}}{3} \), Product \( = – \frac{5}{3} \).

Question. Construct the quadratic equation whose roots are given below -
(i) \( 3, – 3 \)
(ii) \( 3 + \sqrt{3}, 3 – \sqrt{3} \)
(iii) \( \frac{2 + \sqrt{5}}{2}, \frac{2 – \sqrt{5}}{2} \)
Answer: (i) Sum \( = 0 \), Product \( = –9 \). Equation: \( x^2 – 9 = 0 \).
(ii) Sum \( = 6 \), Product \( = 6 \). Equation: \( x^2 – 6x + 6 = 0 \).
(iii) Sum \( = 2 \), Product \( = – \frac{1}{4} \). Equation: \( 4x^2 – 8x – 1 = 0 \).

Question. If \( a \) and \( c \) are such that the quadratic equation \( ax^2 – 5x + 3 = 0 \) has 10 as the sum of the roots and also as the product of the roots, find \( a \) and \( c \).
Answer: Sum \( = \frac{5}{a} = 10 \Rightarrow a = \frac{1}{2} \). Product \( = \frac{c}{a} = 10 \Rightarrow c = 10 \times \frac{1}{2} = 5 \).

Question. If one of the roots of the quadratic equation \( 2x^2 + px + 4 = 0 \) is 2, find the value of \( p \). Also find the value of the other root.
Answer: Substituting \( x = 2: 2(2)^2 + p(2) + 4 = 0 \Rightarrow 12 + 2p = 0 \Rightarrow p = – 6 \). Equation becomes \( 2x^2 – 6x + 4 = 0 \Rightarrow x^2 – 3x + 2 = 0 \Rightarrow (x – 2)(x – 1) = 0 \). The other root is 1.

Question. find the value(s) of \( p \) so that the given equation has equal roots.
(i) \( 3x^2 – 5x + p = 0 \)
(ii) \( 2px^2 – 8x + p = 0 \)
Answer: (i) \( 25 – 12p = 0 \Rightarrow p = \frac{25}{12} \).
(ii) \( 64 – 8p^2 = 0 \Rightarrow p^2 = 8 \Rightarrow p = \pm 2\sqrt{2} \).

EQUATIONS REDUCIBLE TO QUADRATIC EQUATIONS

Type 1 : Equations of the form \( ax^4 + bx^2 + c = 0 \). Substitute \( x^2 = y \) and solve.
Type 2 : Equations of the form \( px + \frac{q}{x} = r \). Multiply by \( x \) and solve.
Type 3 : Equations involving one radical: \( \sqrt{a – x^2} = bx + c \). Square both sides and solve.

Question. Solve : \( x + \frac{5}{x} = 6 \)
Answer: \( x^2 + 5 = 6x \Rightarrow x^2 – 6x + 5 = 0 \Rightarrow (x – 5) (x – 1) = 0 \Rightarrow x = 5, 1 \).

PROBLEMS ON QUADRATIC EQUATIONS

Question. Find two consecutive natural numbers, whose product is equal to 20.
Answer: Let numbers be \( x, x + 1 \). \( x(x + 1) = 20 \Rightarrow x^2 + x – 20 = 0 \Rightarrow (x + 5)(x – 4) = 0 \). As \( x \) is natural, \( x = 4 \). Numbers are 4 and 5.

Question. The sum of the squares of two consecutive whole numbers is 61. Find the numbers.
Answer: \( x^2 + (x + 1)^2 = 61 \Rightarrow 2x^2 + 2x – 60 = 0 \Rightarrow x^2 + x – 30 = 0 \Rightarrow (x + 6)(x – 5) = 0 \). Numbers are 5 and 6.

Question. The sum of two natural numbers is 8. If the sum of their reciprocals is \( \frac{8}{15} \), find the two numbers.
Answer: Let numbers be \( x \) and \( 8 – x \). \( \frac{1}{x} + \frac{1}{8 – x} = \frac{8}{15} \Rightarrow \frac{8}{8x – x^2} = \frac{8}{15} \Rightarrow x^2 – 8x + 15 = 0 \Rightarrow (x – 5)(x – 3) = 0 \). Numbers are 5 and 3.

Question. Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
Answer: Let parts be \( x \) (larger) and \( 16 – x \) (smaller). \( 2x^2 – (16 – x)^2 = 164 \Rightarrow x^2 + 32x – 420 = 0 \Rightarrow (x + 42)(x – 10) = 0 \). Larger part \( = 10 \), smaller part \( = 6 \).