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2 years ago

8

Select the expression that is equivalent to (x-7)^2

1 answer:

Paladinen [302]2 years ago

7 0

<em><u>THE </u></em><em><u>ANSWER </u></em><em><u>IS </u></em><em><u> </u></em><em><u>></u></em><em><u> </u></em><em><u>C </u></em><em><u> </u></em><em><u>(</u></em><em><u> </u></em><em><u>7</u></em><em><u>×</u></em><em><u>+</u></em><em><u>1</u></em><em><u>4</u></em><em><u>)</u></em><em><u> </u></em>

<em><u>HOPE </u></em><em><u>IT </u></em><em><u>HELPS </u></em>

<h2><em><u>#</u></em><em><u> </u></em><em><u>CARRY </u></em><em><u>ON </u></em><em><u>LEARNING</u></em></h2>

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Show all work to solve the equation for x. If a solution is extraneous, be sure to identify it in your final answer.

IRISSAK [1]

Answer:

There is one extraneous solution at $x = 4$

The only solution is $x=7$

Step-by-step explanation:

Given:

The equation to solve is given as:

$\sqrt{x-3}+5=x$

In order to solve, we will first isolate the radical part.

<em>Adding -5 on both the sides, we get:</em>

$\sqrt{x-3}+5-5=x-5\\\sqrt{x-3}=x-5$

<em>Now, squaring both the sides, we get:</em>

$(\sqrt{x-3})^2=(x-5)^2\\\\x-3=x^2+25-10x\\\\x^2-10x-x+25+3=0\\\\x^2-11x+28=0\\\\(x-4)(x-7)=0\\\\x=4\ or\ x=7$

Now, let us check whether the calculated values satisfy the equations or not.

Plug in 4 and 7 for 'x' and check the value of the equation.

At $x=4$ ,

$\sqrt{4-3}+5=4\\\sqrt1+5=4\\1+5=4\\6=4$

But, 6 is not equal to 4 and hence 4 is not a solution to the given equation. It is an extraneous solution for the given equation.

Now, at $x=7$

$\sqrt{7-3}+5=7\\\sqrt4+5=7\\2+5=7\\7=7(True)$

Therefore, $x=7$ is a solution to the equation.

Hence, there is one extraneous solution at $x = 4$

The only solution is $x=7$

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4 years ago