What is the solution of x=2+sqrt x-2

What is the solution of x=2+sqrt x-2

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What is the solution of x=2+sqrt x-2

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What is the solution of x=2+sqrt x-2

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\sqrt{x+2}-\sqrt{x}=6

en

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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

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asked Mar 28, 2018 at 10:43

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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

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$$\sqrt{x^2-4}=x-2$$

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

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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

What is the solution of x=2+sqrt x-2

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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

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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”.