Question

Solve for x-

$\sqrt{ {x}^{2} + 2x } + \sqrt{ {x}^{2} + 2x + 3 } = 4$
​

Answer 1

Step-by-step explanation:

$let \: {x}^{2} + 2x = t</p><p>$

$\sqrt{t} + \sqrt{t + 3} = 4$

$\sqrt{t + 3} = 4 - \sqrt{t}$

squaring both sides

$t + 3 = 16 + t - 8 \sqrt{t}$

$8 \sqrt{t} = 13$

$\sqrt{t} = \frac{13}{8} \: \: i.e \: t = \frac{169}{64}$

$so \: \: {x}^{2} + 2x = \frac{169}{64}$

$64 {x}^{2} + 128x - 169 = 0$

solve this quadratic & get x.