Question:
Let $f (x) = px^2 + qx + r$ , where $p, q, r$ are constants and $p \neq 0$ . If $f (5) = -3 f (2)$
and $f (-4) = 0$ , then the other root of $f$ is
Let $f (x) = px^2 + qx + r$ , where $p, q, r$ are constants and $p \neq 0$ . If $f (5) = -3 f (2)$
and $f (-4) = 0$ , then the other root of $f$ is
Updated On: May 22, 2024
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The Correct Option is A
Solution and Explanation
$f(x)=p x^{2}+q x +r$
$f(-4)=0$
$\Rightarrow \, 16 p-4 q +r=0$ ...(i)
One root is $x=-4$
and $f(5)=-3 f(2)$
$25 p+5 q +r=-3(4 p+2 q +r)$
$\Rightarrow 37 p+11 q+4 r=0$ ...(ii)
E (ii) - E (i), we get
$\Rightarrow -27 p+27 q =0$
$\Rightarrow p =0$
Then, equation is $p x^{2}+q x +r=0$
Roots $=-4, \alpha$
Sum of roots $=-4 x+\alpha=-\frac{p}{q}=-1$
So, another root $\alpha=3$.
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
