solve: (√(x^2 −4x+4)) +(√(x^2 +4x+4)) =(√(x^2 −6x+9))


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 172031 by Mikenice last updated on 23/Jun/22

$s o l v e :$ $\sqrt{x^{2} - 4 x + 4} + \sqrt{x^{2} + 4 x + 4} = \sqrt{x^{2} - 6 x + 9}$
	Answered by puissant last updated on 23/Jun/22
	$⇒(x−2)+(x+2)=(√(x^2 −6x+9)) ⇒ 4x^2 =x^2 −6x+9 ⇒ 3x^2 +6x−9=0 ⇒ x^2 +2x−3=0 Δ=4−4(1)(−3)=16 → (√Δ)=4. x_1 =−3 and x_2 = 1 take S={ −3 }$
	$\Rightarrow (x - 2) + (x + 2) = \sqrt{x^{2} - 6 x + 9}$ $\Rightarrow 4 x^{2} = x^{2} - 6 x + 9$ $\Rightarrow 3 x^{2} + 6 x - 9 = 0$ $\Rightarrow x^{2} + 2 x - 3 = 0$ $Δ = 4 - 4 (1) (- 3) = 16 \to \sqrt{Δ} = 4 .$ $x_{1} = - 3 a n d x_{2} = 1$ $t a k e S = {- 3}$
	Commented by Mikenice last updated on 23/Jun/22

	$t h a n k s s i r$
	Commented by Rasheed.Sindhi last updated on 23/Jun/22

	$x = 1 i s n o t v a l i d .$
	Answered by Rasheed.Sindhi last updated on 23/Jun/22

	$\sqrt{x^{2} - 4 x + 4} + \sqrt{x^{2} + 4 x + 4} = \sqrt{x^{2} - 6 x + 9}$ $(x + 2) + (x - 2) = x - 3$ $2 x = x - 3$ $x = - 3$
	Answered by kapoorshah last updated on 23/Jun/22

	$\sqrt{{(x - 2)}^{2}} + \sqrt{{(x + 2)}^{2}} = \sqrt{{(x - 3)}^{2}}$ $∣ x - 2 ∣ + ∣ x + 2 ∣ = ∣ x - 3 ∣$ $x < - 2$ $\Rightarrow - x + 2 - x - 2 = - x + 3$ $x = - 3$ $- 2 ⩽ x < 2$ $\Rightarrow - x + 2 + x + 2 = - x + 3$ $x = - 1$ $x ⩾ 3$ $\Rightarrow x - 2 + x + 2 = x - 3$ $x = - 3 (r e j e c t e d)$ $s o l u t i o n : - 3, 1$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum