206/MR00. problem / Math - Solved Math Problems

MR-206 / 20613. problem

Solve the following equation:

$\sqrt{2 x - 3} = \sqrt{x + 10} - 1$

$\sqrt{2 x - 3} = \sqrt{x + 10} - 1$

{(\sqrt{2 x - 3})}^{2} = {(\sqrt{x + 10} - 1)}^{2}

2 x - 3 = x + 10 - 2 \sqrt{x + 10} + 1

2 \sqrt{x + 10} = x + 10 + 1 - 2 x + 3

2 \sqrt{x + 10} = 14 - x \Rightarrow x \leq 14

{(2 \sqrt{x + 10})}^{2} = {(14 - x)}^{2}

4 (x + 10) = 196 - 28 x + x^{2}

4 x + 40 = 196 - 28 x + x^{2}

196 - 28 x + x^{2} - 4 x - 40 = 0

a = 1, b = - 32, c = 156

x_{1, 2} = \frac{32 \pm \sqrt{{(- 32)}^{2} - 4 \cdot 1 \cdot 156}}{2} = \frac{32 \pm \sqrt{400}}{2} = \frac{32 \pm 20}{2}

$x_{1} = 26; x_{2} = 6$

x \leq 14 \Rightarrow x_{1} = 26 ⩽ 14 ⊥, x_{2} = 6 ⩽ 14 ⊤

$x = 6$

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