The number of real solutions of $x^2-4|x|-2=0$ is

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

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Application +4 | -1

The number of real solutions of

x^2-4|x|-2=0

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Correct option: B

Option B (2) is correct because substituting |x| = y results in a quadratic equation with two positive solutions for |x|, leading to two real solutions for x.

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The solution of the equation

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

and

\alpha < 1 < \beta ,

then If

\alpha ,\;\beta

be the roots of the equation

{x^2} - px + q = 0

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

then the equation whose roots are

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

is If

{x^2} + 4ax + 2 > 0

for all values of

x

, then

a

lies in the interval If

f\left( x \right) = a{x^2} + bx + c\;

and

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

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\;a \ne 0,

then Root(s) of the equation

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

belonging to the domain of definition of the function

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is (are)