Let f(x) be a quadratic polynomial such that
f(–1) + f(2) = 0. If one of the roots of f(x) = 0
is 3, then its other root lies in :
Solution
Let the other root is $\alpha$.
<br><br>$\therefore$ f(x)
=
a(x
–
3)
(x
–
$\alpha$)
<br><br>f(2) = a($\alpha$– 2)
<br><br>f(–1) = 4a(1 + $\alpha$)
<br><br>Given f(–1) + f(2) = 0
<br><br>$\Rightarrow$a($\alpha$ – 2 + 4 + 4$\alpha$) = 0
<br><br>$\Rightarrow$ 5$\alpha$ = -2 As a $\ne$ 0
<br><br>$\Rightarrow$ $\alpha$ = $-\frac{2}{5}$ = - 0.4
<br><br>$\therefore$ $\alpha$ $\in$ (–1, 0)
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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