JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

\alpha \;\mathrm{and}\;\beta

be the roots of the equation

2{x^2} + 2\left( {a + b} \right)x + {a^2} + {b^2} = 0,

then the equation whose roots are

{\left( {\alpha + \beta } \right)^2}\;\mathrm{and}\;{\left( {\alpha - \beta } \right)^2}

, is

View answer and explanation

Correct option: B

{x^2} - 4abx - {\left( {{a^2} - {b^2}} \right)^2} = 0

Option B (

{x^2} - 4abx - {\left( {{a^2} - {b^2}} \right)^2} = 0

) is correct because it is the equation formed with roots

(\alpha + \beta)^2

and

(\alpha - \beta)^2

after using the relationships between roots and coefficients of the given quadratic equation.

More Questions from QUADRATIC EQUATIONS

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{x^2} - px + q = 0

{\alpha _1},{\beta _1}

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

\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}},

{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\;

\;a \ne 0,

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

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