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Nature of Roots

For what valu...

Question

For what value of $k, (4 - k) x^{2} + (2 k + 4) x + (8 k + 1) = 0$ , is a perfect square.

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Solution

The given quadratic will be a perfect square if it has two real and equal roots.
To have two real and equal roots the discriminant must be zero.
i.e., $b^{2} - 4 a c = 0$
$\Rightarrow (2 k + 4) ² - 4 (4 - k) (8 k + 1) = 0$
$\Rightarrow (4 k^{2} + 16 k + 16) - 4 (32 k - 8 k^{2} + 4 - k) = 0$
[Since, $(a + b)^{2} = a^{2} + b^{2} + 2 a b$ ]
$\Rightarrow 4 k^{2} + 16 k + 16 - 128 k + 32 k^{2} - 16 + 4 k = 0$
$\Rightarrow 36 k^{2} - 108 k = 0$
$\Rightarrow 36 k (k - 3) = 0$
either $36 k = 0$ or $k - 3 = 0$
$k = 0$ or, $k = 3$
Therefore, $k = 0$ or $3$

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