Answered

Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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!

We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.