JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

1 of 1

Application +4 | -1

The roots of the equation

{\left| {2x - 1} \right|^2} - 3\left| {2x - 1} \right| + 2 = 0

are

View answer and explanation

Correct option: D

\left\{ { - \frac{1}{2},0,1,\frac{3}{2}} \right\}

Option D (

\left\{ { - \frac{{1}}{{2}},0,1,\frac{{3}}{{2}}} \right\}

) is correct because these are the solutions obtained by solving the equation after substituting

t = \left| {2x - 1} \right|

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)