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Question Number 209633 by Abdullahrussell last updated on 17/Jul/24

	Answered by som(math1967) last updated on 17/Jul/24

	$(x + 1) (3 x + 2) {(6 x + 5)}^{2} = 6$ $\Rightarrow (3 x^{2} + 5 x + 2) (36 x^{2} + 60 x + 25) = 6$ $\Rightarrow (a + 2) (12 a + 25) = 6$ $[l e t 3 x^{2} + 5 x = a]$ $\Rightarrow 12 a^{2} + 49 a + 50 - 6 = 0$ $\Rightarrow 12 a^{2} + 49 a + 44 = 0$ $\Rightarrow 12 a^{2} + 16 a + 33 a + 44 = 0$ $\Rightarrow 4 a (3 a + 4) + 11 (3 a + 4) = 0$ $\Rightarrow (3 a + 4) (4 a + 11) = 0$ $(9 x^{2} + 15 x + 4) (12 x^{2} + 20 x + 11) = 0$ $i f (9 x^{2} + 15 x + 4) = 0$ ${(3 x)}^{2} + 2.3 x . \frac{5}{2} + \frac{25}{4} = \frac{25}{4} - 4$ $∴ {(3 x + \frac{5}{2})}^{2} = \frac{9}{4}$
	Answered by MM42 last updated on 17/Jul/24

	$\frac{1}{6} [(6 x + 5) + 1) \frac{1}{2} [(6 x + 5) - 1) {(6 x + 5)}^{2} = 6$ $(u + 1) (u - 1) {(u)}^{2} = 72$ $(u^{2} - 1) u^{2} - 72 =$ $u^{4} - u^{2} - 72 = 0$ $(u^{2} - 9) (u^{2} + 8) = 0$ $u = \pm 3 & u = \pm 2 \sqrt{2} i$ ${(3 x + \frac{5}{2})}^{2} = \frac{1}{4} {(6 x + 5)}^{2} = \frac{9}{4} ✓ & - 2 ✓$
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