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solve for x
[tex] \sqrt{x^2-4x+8} +x=2 - x[/tex]
please show all workings​

Sagot :

xKelvin

Answer:

[tex]\displaystyle x=-\frac{2}{3}[/tex]

Step-by-step explanation:

We want to solve the equation:

[tex]\displaystyle \sqrt{x^2-4x+8}+x=2-x[/tex]

We can isolate the square root. Subtract x from both sides:

[tex]\sqrt{x^2-4x+8}=2-2x[/tex]

And square both sides:

[tex](\sqrt{x^2-4x+8})^2=(2-2x)^2[/tex]

Expand:

[tex]x^2-4x+8=4-8x+4x^2[/tex]

Isolate the equation:

[tex]3x^2-4x-4=0[/tex]

Factor:

[tex]\displaystyle (3x+2)(x-2)=0[/tex]

Zero Product Property:

[tex]3x+2=0\text{ or } x-2=0[/tex]

Solve for each case. Hence:

[tex]\displaystyle x=-\frac{2}{3}\text{ or } x=2[/tex]

Now, we need to check for extraneous solutions. To do so, we can substitute each value back into the original equation and check whether or not the resulting statement is true.

Testing x = -2/3:

[tex]\displaystyle \begin{aligned} \sqrt{\left(-\frac{2}{3}\right)^2-4\left(-\frac{2}{3}\right)+8}+\left(-\frac{2}{3}\right)&\stackrel{?}{=}2-\left(-\frac{2}{3}\right)\\ \\ \sqrt{\frac{4}{9}+\frac{8}{3}+8}-\frac{2}{3}&\stackrel{?}{=}2+\frac{2}{3} \\ \\ \sqrt{\frac{100}{9}}-\frac{2}{3}& \stackrel{?}{=} \frac{8}{3}\\ \\ \frac{10}{3}-\frac{2}{3} =\frac{8}{3}& \stackrel{\checkmark}{=}\frac{8}{3}\end{aligned}[/tex]

Since the resulting statement is true, x = -2/3 is indeed a solution.

Testing x = 2:

[tex]\displaystyle \begin{aligned}\sqrt{(2)^2-4(2)+8}+(2) &\stackrel{?}{=}2-(2) \\ \\ \sqrt{4-8+8}+2&\stackrel{?}{=}0 \\ \\ \sqrt{4}+2&\stackrel{?}{=}0 \\ \\ 2+2=4&\neq 0\end{aligned}[/tex]

Since the resulting statement is not true, x = 2 is not a solution.

Therefore, our only solution to the equation is x = -2/3.  

Sueraiuka

Step-by-step explanation:

Hey there!

Given;

[tex] \sqrt{ {x}^{2} - 4x + 8} + x = 2 - x[/tex]

Take "X" in right side.

[tex] \sqrt{ {x - 4 + 8}^{2} } = 2 - 2x[/tex]

Squaring on both sides;

[tex] {( \sqrt{ {x}^{2} - 4x + 8 } )}^{2} = {(2 - 2x)}^{2} [/tex]

Simplify;

[tex] {x}^{2} - 4x + 8 = {(2)}^{2} - 2.2.2x + {(2x)}^{2} [/tex]

[tex] {x }^{2} - 4x + 8 = 4 - 8x + 4 {x}^{2} [/tex]

[tex]3 {x}^{2} - 4x - 4 = 0[/tex]

[tex]3 {x}^{2} - (6 - 2)x - 4 = 0[/tex]

[tex] 3 {x}^{2} - 6x + 2x - 4 = 0[/tex]

[tex]3x(x - 2) + 2(x - 2) = 0[/tex]

[tex](3x + 2)(x - 2) = 0[/tex]

Either;

3x+2 = 0

x= -2/3

Or;

x-2 = 0

x= 2

Check:

Keeping X= -2/3,

√(x²-4x+8 ) +X = 2-x

√{(-2/3)²-4*-2/3+8}+(-2/3) = 2+2/3

8/3 = 8/3 (True)

Now; Keeping X= 2

√{(2)²-4*2+8}+2 = 2-2

8 ≠0 (False)

Therefore, the value of X is -2/3.

Hope it helps!

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