JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

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Application +4 | -1

\alpha, \beta, \gamma

are the roots of the equation,

x^3+a x^2+b x+c=0

, then

\left(1-\alpha^2\right)\left(1-\beta^2\right)\left(1-\gamma^2\right)

is equal to:

View answer and explanation

Correct option: A

(1+b)^2-(a+c)^2

Option A (

(1+b)^2-(a+c)^2

) is correct because by substituting

y = 1-x^2

and simplifying, we can find the product of the roots of the transformed equation.

More Questions from QUADRATIC EQUATIONS

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\alpha ,\beta

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\alpha ,\;\beta

{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

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