Solve −2x(x+(7/2))−(x−3)^2 ≤0


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Question Number 8071 by marcusvsoliveira last updated on 29/Sep/16

$S o l v e - 2 x (x + \frac{7}{2}) - {(x - 3)}^{2} ⩽ 0$
Commented by FilupSmith last updated on 29/Sep/16

$- 2 x (x + \frac{7}{2}) - {(x - 3)}^{2} = - (2 x^{2} + 7 x) - (x^{2} - 6 x + 9)$ $= - 2 x^{2} - x^{2} - 7 x - 6 x - 9$ $= - 3 x^{2} - 13 x - 9 ⩽ 0$ $∴ 3 x^{2} + 13 x + 9 ⩾ 0$ $x = \frac{- 13 \pm \sqrt{13^{2} - 4 (3) (9)}}{6} = \frac{- 13 \pm \sqrt{61}}{6}$ $w o r k i n g$
Commented by marcusvsoliveira last updated on 29/Sep/16

$w h y - 2 x^{2} - x^{2} - 7 x - 6 x - 9 ?$
	Answered by prakash jain last updated on 29/Sep/16

	$- 2 x (x + \frac{7}{2}) - {(x - 3)}^{2} ⩽ 0$ $- 2 x^{2} - 7 x - x^{2} + 6 x - 9 ⩽ 0$ $- 3 x^{2} - x - 9 ⩽ 0$ $- 3 x^{2} - x - 9 ⩽ 0$ $r o o t a r e \frac{1 \pm \sqrt{1 - 27 \times 4}}{- 6} \Rightarrow n o r e a l r o o t$ $as the quadratic has no real roots it$ $takes the same as coefficient of x^{2} for x \in R .$ $- 3 x^{2} - x - 9 ⩽ 0 for x \in R$ $Solution set R$
	Commented by FilupSmith last updated on 30/Sep/16

	$I made a silly error . no wonder i couldnt$ $figure it out$
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