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

Solve the given two equations and select the correct answer from the given options.
$I.$ $8x^2 + 6x = 5$
$II.$ $12y^2 - 22y + 8 = 0$

If one root of $x^2 - x - k = 0$ is the square of the other,then $k =$

Form a quadratic equation whose one root is $3-\sqrt{5}$ and the sum of roots is $6$.

The value of $a$ for which the equations $x^2 - 3x + a = 0$ and $x^2 + ax - 3 = 0$ have a common root is

Determine $p$ such that the equation $x^{2}+5px+16=0$ has no real roots.

Vedclass Test Series

Exam Paper Generator

Online Exam Module