If a(p + q)^2 + 2bpq + c = 0 and a(p + r)^2 + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given equations are $a(p + q)^2 + 2bpq + c = 0$ and $a(p + r)^2 + 2bpr + c = 0$.These equations imply that $q$ and $r$ are the roots of the quadra

Vedclass · Accepted Answer

$p^2 + \frac{c}{a}$

Vedclass · Answer

(A) Given equations are $a(p + q)^2 + 2bpq + c = 0$ and $a(p + r)^2 + 2bpr + c = 0$.These equations imply that $q$ and $r$ are the roots of the quadratic equation in $x$:$a(p + x)^2 + 2bpx + c = 0$.Expanding the equation: $a(p^2 + 2px + x^2) + 2bpx + c = 0$.$ax^2 + 2apx + ap^2 + 2bpx + c = 0$.Rearranging in terms of $x$: $ax^2 + 2x(ap + bp) + (ap^2 + c) = 0$.$ax^2 + 2px(a + b) + (ap^2 + c) = 0$.For a quadratic equation $Ax^2 + Bx + C = 0$,the product of roots is given by $\frac{C}{A}$.Here,$A = a$,$B = 2p(a + b)$,and $C = ap^2 + c$.Therefore,the product of roots $qr = \frac{ap^2 + c}{a} = p^2 + \frac{c}{a}$.

If $a(p + q)^2 + 2bpq + c = 0$ and $a(p + r)^2 + 2bpr + c = 0$,then $qr$ =

Similar Questions

$\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$ is equal to

If $\alpha, \beta$ are the roots of $x^2 - px + q = 0$ and $\alpha', \beta'$ are the roots of $x^2 - p'x + q' = 0$,then the value of $(\alpha - \alpha')^2 + (\beta - \alpha')^2 + (\alpha - \beta')^2 + (\beta - \beta')^2$ is

Solve the given two equations and select the correct option.
$I.$ $x = \sqrt[3]{2744}$
$II.$ $y = \sqrt{487}$

Solve the given two equations and select the correct answer from the given options.
$I.$ $\frac{12}{\sqrt{x}} - \frac{23}{\sqrt{x}} = 5\sqrt{x}$
$II.$ $\frac{\sqrt{y}}{12} - \frac{5\sqrt{y}}{12} = \frac{1}{\sqrt{y}}$

The roots of the equation $3^{2x} - 10 \cdot 3^x + 9 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module