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Question Number 147011 by mathdanisur last updated on 17/Jul/21

$2 x - \sqrt{2 x - 3} - 9 = 0$ $i f {t h e r e}^{'} s a s o l u t i o n t o e q u a t i o n,$ $f i n d 4 a + 3 = ?$
Commented by 7770 last updated on 17/Jul/21

$4 x^{2} - 36 x + 81 - 2 x + 3 = 0$ $4 x^{2} - 38 x + 84 = 0$ $x_{1} = 6 x_{2} = 3.5$ $∴ 4 (3.5) + 3 = 17$ $4 (6) + 3 = 27$
Commented by otchereabdullai@gmail.com last updated on 18/Jul/21

$nice question$
Commented by mathdanisur last updated on 18/Jul/21

$t h a n k y o u S e r$
	Answered by gsk2684 last updated on 17/Jul/21

	${(\sqrt{2 a - 3})}^{2} - \sqrt{2 a - 3} - 6 = 0$ $t^{2} - t - 6 = 0$ $(t - 3) (t + 2) = 0$ $t = 3$ $\sqrt{2 a - 3} = 3$ $2 a = 12$ $4 a + 3 = 27$
	Commented by mathdanisur last updated on 17/Jul/21

	$t h a n k y o u S e r, b u t a n s w e r 17$
	Commented by gsk2684 last updated on 17/Jul/21

	$i f s o t h e n a = \frac{17 - 3}{4} = \frac{14}{4} = \frac{7}{2}$ $2 a - \sqrt{2 a - 3} - 9$ $7 - 2 - 9$ $\neq 0$
	Commented by mathdanisur last updated on 18/Jul/21

	$t h a n k y o u S e r$
	Answered by Rasheed.Sindhi last updated on 17/Jul/21

	$2 x - \sqrt{2 x - 3} - 9 = 0$ $\underline{4 a + 3 = ? a : s o l u t i o n t o t h e e q u a t i o n}$ $4 x^{2} - 36 x + 81 = 2 x - 3$ $4 x^{2} - 38 x + 84 = 0$ $2 x^{2} - 19 x + 42 = 0$ $(x - 6) (2 x - 7) = 0$ $x = 6 \lor x = 7 / 2 (E x t a n e o u s r o o t)$ $a = 6$ $4 a + 3 = 4 (6) + 3 = 27$ $(I f x = 7 / 2 w e r e s a t i s f i e d$ $t h e n 4 a + 3 = 4 (\frac{7}{2}) + 3 = 17)$
	Commented by mathdanisur last updated on 17/Jul/21

	$t h a n k y o u S e r$
	Commented by mathdanisur last updated on 18/Jul/21

	$t h a n k y o u S e r$
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