Question

Solve by completing the square. Round your answers to the nearest tenth and then locate the greater solution. tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align="absmiddle" class="latex-formula">+10x-7=0

Answer 1

Step-by-step explanation:

Step 1: Write our Givens

${x}^{2} + 10x - 7 = 0$

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

${x}^{2} + 10x = 7$

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

$( \frac{10}{2} ) {}^{2} = 25$

Then we add that to both sides

${x}^{2} + 10x + 25 = 7 + 25$

${ {x}^{2} } + 10x + 25 = 32$

Next, we factor the left,

$(x + 5)(x + 5) = 32$

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

so we get

$(x + 5) {}^{2} = 32$

This is called a perfect square trinomial.

Next, we take the square root of both sides

$x + 5 = ± \sqrt{32}$

± menas that we have a positive and negative solution.

Subtract 5 form both side so we get

$x = - 5± \sqrt{32}$

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

$\sqrt{32} - 5 = 0.7$