JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

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Application +4 | -1

Let

f(x)=x^2+2 x+5

and

\alpha, \beta

be roots of

f\left(\frac{1}{t}\right)=0

, then

\alpha+\beta=

View answer and explanation

Correct option: A

\frac{-2}{5}

Option A (-2/5) is correct because it is the sum of the roots of the transformed quadratic equation.

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{x^2} - px + q = 0

{\alpha _1},{\beta _1}

{x^2} - qx + p = 0,

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{x^2} + 4ax + 2 > 0

x

a

f\left( x \right) = a{x^2} + bx + c\;

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

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9{x^2} - 18\left| x \right| + 5 = 0

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)