Answered

At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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]