The values for which $x^{4} - 2 a x^{2} + a^{2} - a = 0$ has all real roots are

- 1

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Solution

The correct option is B $1$
$x^{4} - 2 a x^{2} + a^{2} - a = 0$
This equation can also be written in quadratic form as
${(x^{2})}^{2} = 2 a (x^{2}) + a^{2} - a = 0$
Substituting $x^{2} = t$ where $t > 0$
we get
$t^{2} - 2 A a t + a^{2} - a = 0$
Now, for quadratic equation to have real rpots
${(- 2 a)}^{2} - 4 \times 1 \times (a^{2} - a) \geq 0$
(...applied the condition for real roots of quadratic equation $a x^{2} + b x + c = 0$ $b^{2} - 4 a c \geq 0$ )
Hence
$4 A^{2} - 4 (a^{2} - a) \geq 0$
$\Rightarrow$ $4 a^{2} - 4 a^{2} + 4 a \geq 0$
$4 a \geq 0$
$a \geq 0$
Also we have $t \geq 0$
$\frac{- (2 a) \pm \sqrt{4 a}}{2.1} \geq 0$
$\frac{- (2 a) \pm 2 \sqrt{a}}{2.1} \geq 0$
$- a \pm \sqrt{a} \geq 0$
This gives us the only solution possible from options given as $a = 1$

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