$(2 x^{2} + 3 x + 5)^{1 / 2} + (2 x^{2} + 3 x + 20)^{1 / 2} = 15.$ Then find the value of $x .$

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Solution

$(2 x^{2} + 3 x + 5)^{1 / 2} + (2 x^{2} + 3 x + 20)^{1 / 2} = 15$

First let us assume that,
$2 x^{2} + 3 x + 5 = t$
$\Rightarrow (t)^{1 / 2} + (t + 15)^{1 / 2} = 15$
$\Rightarrow (t + 15)^{1 / 2} = (15 - t^{1 / 2})$

Now, on squaring both sides of above equation,
$t + 15 = 225 + t - \sqrt{t} (30)$
$\Rightarrow 210 = 30 \sqrt{t}$
$\Rightarrow \sqrt{t} = 7$
$\Rightarrow t = 49$

Now,
$2 x^{2} + 3 x + 5 = 49$
$\Rightarrow 2 x^{2} + 3 x - 44 = 0$
$\Rightarrow 2 x^{2} + 11 x - 8 x - 44 = 0$
$\Rightarrow x (2 x + 11) - 2 (2 x + 11) = 0$
$\Rightarrow x = 2, - \frac{11}{2}$

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