JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Medium

1 of 1

Analyzing +4 | -1

\alpha ,\beta

are the roots of the equation

\lambda \left( {{x^2} - x} \right) + x + 5 = 0

and if

{\lambda _1}\;\mathrm{and}\;{\lambda _2}

are two values of

\lambda

obtained from

\frac{\alpha }{\beta } + \frac{\beta }{\alpha } = \frac{4}{5}

then

\frac{{{\lambda _1}}}{{\lambda _2^2}} + \frac{{{\lambda _2}}}{{\lambda _1^2}}

is equal to

View answer and explanation

Correct option: D

4048

Option D (4048) is correct because it is the result of substituting the values of

\lambda_1 + \lambda_2

and

\lambda_1 \lambda_2

into the expression after simplifying the given equation.

More Questions from QUADRATIC EQUATIONS

Practice All Chapter MCQs →

\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0

\alpha ,\beta

{x^2} - 3x + a = 0,a \in R

\alpha < 1 < \beta ,

\alpha ,\;\beta

{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)