The roots of a quadratic equation $x^{2} - 4 x - {log}_{3} a = 0$ are real. Then what is the least value of $a$ ?

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\frac{1}{81}

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\frac{1}{64}

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Solution

The correct option is B $\frac{1}{81}$
We have, $x^{2} - 4 x - {log}_{3} a = 0$
If roots are real, then $D \geq 0$
$∴ 16 + 4 {log}_{3} a \geq 0$
$\Rightarrow {log}_{3} a \geq - \frac{16}{4} \Rightarrow {log}_{3} a \geq - 4$
$\Rightarrow a \geq 3^{- 4} \Rightarrow a \geq \frac{1}{81}$

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The roots of a quadratic equation $x^{2} - 4 x - a = 0$ are real. Then what is the least value of $a$ ?

If α,β are the roots of the quadratic equation $x^{2}$ – (a – 2)x –(a+1) =0, where 'a' is a variable then the least value of $α^{2}$ + $β^{2}$

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\frac{1}{α^{2}} + \frac{1}{β^{2}}

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