Question

2. Solve for X: 2
$\sqrt{2x + 9} - \sqrt{x - 4} = 3$

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Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

$\bf \: Solve \: for \: x: \sqrt{2x + 9} - \sqrt{x - 4} = 3</p><p>$

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$\large\underline\purple{\bold{Solution :- }}$

$\bf \:  ⟼ \sqrt{2x + 9} - \sqrt{x - 4} = 3$

$\sf \:  ⟼\sqrt{2x + 9} \: = \: \sqrt{x - 4} \: + 3$

☆ On squaring both sides, we get

$\sf \:  ⟼2x + 9 = { \bigg(3 + \sqrt{x - 4} \bigg) }^{2}$

$\sf \:  ⟼2x + \cancel9 = \cancel9 + x - 4 + 6 \sqrt{x - 4}$

$\sf \:  ⟼x + 4 = 6 \sqrt{x - 4}$

☆ On squaring both sides, we get

$\sf \:  ⟼ {(x + 4)}^{2} = 36(x - 4)$

$\sf \:  ⟼ {x}^{2} + 16 + 8x = 36x - 144$

$\sf \:  ⟼ {x}^{2} - 28x + 160 = 0$

$\sf \:  ⟼ {x}^{2} - 20x - 8x + 160 = 0$

$\sf \:  ⟼x(x - 20) - 8(x - 20) = 0$

$\sf \:  ⟼(x - 20)(x - 8) = 0$

$\bf\implies \:x \: = 20 \: or \: x \: = \: 8$

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☆ Verification :-

☆ Case :- 1 When x = 20

☆ Consider LHS

$\sf \:  ⟼\sqrt{2x + 9} - \sqrt{x - 4}$

$\sf \:  ⟼\sqrt{2 \times 20 + 9} - \sqrt{20 - 4}$

$\sf \:  ⟼ \sqrt{49} - \sqrt{16}$

$\sf \:  ⟼7 - 4$

$\sf \:  ⟼3$

$\bf \:  ➦ LHS = RHS$

$\bf \:  ⟼ \boxed{\bf \: Hence, Verified }✔$

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☆Case :- 2 When x = 8

☆Consider LHS

$\bf \:  ⟼ \sqrt{2x + 9} - \sqrt{x - 4}$

$\sf \:  ⟼\sqrt{2 \times 8 + 9} - \sqrt{8 - 4}$

$\sf \:  ⟼ \sqrt{25} - \sqrt{4}$

$\sf \:  ⟼5 - 2$

$\sf \:  ⟼3$

$\bf \:  ➦ LHS = RHS$

$\bf \:  ⟼ \boxed{\bf \:  Hence, Verified} ✔$

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$\large{\boxed{\boxed{\bf{Hence, x \: = \: 20 \: or \: 8}}}}$