CAT 2025 Lesson : Quadratic Equations - Sum & Product of Roots

For a quadratic function, $f(x) = x^{2} + bx + c$,, it is known that $f(1) = 12$ and $f(4) = 0$. Then, $b^2 + c^2 =$?

Solution

$f(1) = 1^{2} + b + c = 12$
⇒ $b + c = 11 \longrightarrow (1)$

$f(4) = 4^{2} + 4b +c = 0$
⇒ $4b + c = -16 \longrightarrow (1)$

Subtracting equations ⇒ $\text{Eq} (2) - \text{Eq} (1)$

⇒ $3b = -27$
⇒ $b = -9$

Substituting in Eq $(1)$, we get $c = 20$

∴ $b^2 + c^2 = 81 + 400 = 481$

Answer: $481$

Match the following equations with the nature of their roots

Equations	Nature of Roots
(A) $x^{2} - 2x - 7 = 0$	(i) Both roots are positive
(B) $-4x^{2} + 19x - 3 = 0$	(ii) One root is positive while the other is negative
(C) $3x^{2} + 12x + 4 = 0$	(iii) Both roots are negative

Solution

Where $\bm{\alpha}$ and $\bm{\beta}$ are the roots for an equation $ax^{2} + bx + c = 0$,

Sum of Roots (SOR) $= \alpha + \beta = \dfrac{-b}{a}$ and Product of Roots (POR) $= \alpha \beta = \dfrac{c}{a}$

We can find the nature of the roots by examining these values.

(A) $x^{2} - 2x - 7 = 0$     ⇒     $\alpha \beta = -7$

As POR is negative, one root is positive while the other is negative.

(B) $-4x^{2} + 19x - 3 = 0$    ⇒     $\alpha + \beta = \dfrac{19}{4}$ and $\alpha \beta = \dfrac{3}{4}$

∴ As both SOR And POR are positive, both the roots are positive.

(C) $3x^{2} + 12x + 4 = 0$     ⇒     $\alpha + \beta = -4$ and $\alpha \beta = \dfrac{4}{3}$

∴ As SOR is negative while POR is positive, both the roots are negative.

Answer: (A) - (ii); (B) - (i); (C) - (iii)

For what values of $p$ does the equation $x^{2} - (5p - 4) x + 2p - 6 = 0$ have two positive roots?

Solution

Both the roots will be positive if the sum of roots and the product of roots are both positive.

Sum of Roots $= 5p - 4 \gt 0$     ⇒     $p \gt \dfrac{4}{5} \longrightarrow (1)$

Product of Roots $= 2p - 6 \gt 0$     ⇒     $2p - 6 \gt 0$     ⇒     $p \gt 3 \longrightarrow (2)$

Merging the two conditions, when p $\gt$ 3, both conditions are satisfied.

Answer: $p \gt 3$