JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

\alpha \ne \beta

and

{\alpha ^2} = 5\alpha - 3,\;{\beta ^2} = 5\beta - 3

, then the equation having

\frac{\alpha }{\beta }

and

\frac{\beta }{\alpha }

as its roots is

View answer and explanation

Correct option: B

3{x^2} - 19x + 3 = 0

Option B (

3{x^2} - 19x + 3 = 0

) is correct because it represents the quadratic equation with roots

\frac{\alpha}{\beta}

and

\frac{\beta}{\alpha}

derived from the given conditions.

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)