Question

$\sqrt{6x + 7} - \sqrt{2x - 2} = 3$

Answer 1

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Here is ur Ans ➡


✔
$\sqrt{6x + 7} - \sqrt{2x - 2} = 3$

$6x + 7 = 9 + 6 \sqrt{2x - 2} + 2x - 2$

$6x + 7 = 7 + 6 \sqrt{2x - 2} + 2x$

$7 \: are \: \: reduced$

$6x = 6 \sqrt{2x - 2} + 2x$

$- 6 \sqrt{2x - 2} = 2x - 6x$

$- 6 \sqrt{2x - 2} = - 4x$

$3 \sqrt{2x - 2} = 2x$

$9(2x - 2) = 4x {}^{2}$

$18x - 18 = 4x {}^{2}$
$9x - 9 = 2x {}^{2}$

$9x - 9 - 2x {}^{2} = 0$

$- 2x {}^{2} + 9x - 9 = 0$

$2x {}^{2} - 9x + 9 = 0$

using formula ....

$x = \frac{ - ( - 9) + - \sqrt{( - 9) {}^{2} - 4 \times 2 \times 9} }{2 \times 2}$

$x = \frac{9 + - \sqrt{81 - 72} }{4}$

$x = \frac{9 + - 9}{4}$


$x = \frac{9 + 3}{4} \\ \\ or \\ \\ x = \frac{9 - 3}{4}$
means

$x = 3 \\ \\ or \\ \\ x = \frac{3}{2}$
in decimal
$x = 3 \\ \\ x = 1.5$
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