Let f(x) be a quadratic polynomial such that f($-$2) + f(3) = 0. If one of the roots of f(x) = 0 is $-$1, then the sum of the roots of f(x) = 0 is equal to :
Solution
<p>$\because$ x = $-$1 be the roots of f(x) = 0</p>
<p>$\therefore$ Let $f(x) = A(x + 1)(x - 1)$ ...... (i)</p>
<p>Now, $f( - 2) + f(3) = 0$</p>
<p>$\Rightarrow A[ - 1( - 2 - b) + 4(3 - b)] = 0$</p>
<p>$b = {{14} \over 3}$</p>
<p>$\therefore$ Second root of f(x) = 0 will be ${{14} \over 3}$</p>
<p>$\therefore$ Sum of roots $= {{14} \over 3} - 1 = {{11} \over 3}$</p>
About this question
Subject: Mathematics · Chapter: Complex Numbers and Quadratic Equations · Topic: Complex Numbers and Argand Plane
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