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Solve by completing the square. Round your answers to the nearest tenth and then locate the greater solution. [tex]x^{2}[/tex]+10x-7=0

Sagot :

Step-by-step explanation:

Step 1: Write our Givens

[tex] {x}^{2} + 10x - 7 = 0[/tex]

Move the constant term ,(the term with no variable) to the right side.

Here we have a negative 7, so we add 7 to both sides

[tex] {x}^{2} + 10x = 7[/tex]

Next, we take the linear coeffeicent and divide it by 2 then square it.

[tex]( \frac{10}{2} ) {}^{2} = 25[/tex]

Then we add that to both sides

[tex] {x}^{2} + 10x + 25 = 7 + 25[/tex]

[tex] { {x}^{2} } + 10x + 25 = 32[/tex]

Next, we factor the left,

[tex](x + 5)(x + 5) = 32[/tex]

we got 5 because 5 add to 10 and multiply to 25 as well.

so we get

[tex](x + 5) {}^{2} = 32[/tex]

This is called a perfect square trinomial.

Next, we take the square root of both sides

[tex]x + 5 = ± \sqrt{32} [/tex]

± menas that we have a positive and negative solution.

Subtract 5 form both side so we get

[tex]x = - 5± \sqrt{32} [/tex]

The greater solution is when sqr root of 32 is positive so the answer to that is

[tex] \sqrt{32} - 5 = 0.7[/tex]

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