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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.
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!