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step-by-step \sqrt{x+2}-\sqrt{x}=6 en $\begingroup$ Solve for x $$\sqrt{x^2-4}=x-2$$ My try: $$\sqrt{(x-2)(x+2)}=x-2$$ $$\sqrt{x+2}=\sqrt{x-2}$$ $$x+2=x-2$$ $$0x=4$$ This is not correct $x={4\over 0}$ How else can I solve this equation?
nonuser 87.6k18 gold badges101 silver badges195 bronze badges asked Mar 28, 2018 at 10:43 $\endgroup$ 1 $\begingroup$ You lost a solution $x=2$ when you divide it by $\sqrt{x-2}$. You can divide the equation with something only if you are sure it is not $0$. answered Mar 28, 2018 at 10:45
nonusernonuser 87.6k18 gold badges101 silver badges195 bronze badges $\endgroup$ $\begingroup$ $$\sqrt{x^2-4}=x-2$$ squaring both side, $${(x-2)(x+2)}=(x-2)^2$$ $$(x-2)((x+2)-(x-2))=0$$ $$(x-2)(4)=0$$ $$x = 2$$ In Your approach, you lost this root $x=2$ when you divided by $\sqrt{x-2}$ answered Mar 28, 2018 at 10:47
kayushkayush 2,40111 silver badges26 bronze badges $\endgroup$ $\begingroup$ Original equation: $$\sqrt{x^2-4}=x-2$$ Square on both sides: $$x^2-4=x^2-4x+4$$ "Move things": $$-4x=-8\\4x=8\\x=\frac{8}{4}=2$$ Now, let's test the solution: $$\sqrt{2^2-4}\stackrel{?}{=}2-2\\\sqrt{0}\stackrel{?}{=}0\\0=0$$ It does work, so $x=2$ answered Mar 28, 2018 at 10:55
$\endgroup$ 2 $\begingroup$ We can square both side and then divide by $(x-2)$ with the extra condition $x\neq 2$ $$\sqrt{x^2-4}=x-2\iff x^2-4 =(x-2)^2\iff x+2=x-2$$ then check directly the solution $x=2$ in the original equation. answered Mar 28, 2018 at 10:46
useruser 143k12 gold badges73 silver badges136 bronze badges $\endgroup$ Why is √ X 2 X and not X?The square root of x^2 does not have to equal x, because it is also -x! Suppose x = - 5. Then x^2 = 25, and the square root of x^2 is 5. So any value of x < 0 is a counterexample to the assertion that the square root of x^2 is x.
Why is √ X 2 X?The square root of a squared number removes the square and leaves the number itself with a positive and negative sign. ⇒√x2=±x . Hence, the square root of x squared is ±x.
How is √ X 2 X?The sqrt of x^2 is x because with simplifying square roots, you'd always find groups of two of the same character. In this case, x^2 is the same as saying x times x and this makes a perfect square root of x because this only group of x's goes fully out of the “house”.
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