the sum of the square of a positive number and the square of 5 more than the number is 157. what is the number?

please help, thanks!!!! :)

the sum of the square of a positive number and the square of 5 more than the number is 157. what is the number?

please help, thanks!!!! :)

Tutors, please sign in to answer this question.

Turn the word problem into an equation. Call the unknown number
**x**. The square of that number is **x**** ^{2}**. The square of 5 more than that number is

OK, let's solve it.

x^{2} + (x+5)^{2} = 157

x^{2} + (x^{2} + 10x + 25) = 157

2x^{2} + 10x + 25 = 157

2x^{2} + 10x - 132 = 0

This is the standard quadratic form, ax^{2} + bx + c = 0. There will be two solutions. You can solve the equation by factoring, or by the quadratic formula. If I don't see an easy factorization pattern right away, I just go straight to the quadratic formula myself.

x = [-b ± √(b^{2} - 4ac) ] / 2a

x = [-10 ± √(10^{2} - 4 · 2 · -132) ] / (2 · 2)

x = [-10 ± √(100+1056) ] / 4

x = [-10 ± √1156] / 4

x = [-10 ± 34] / 4

x = -44 / 4 , +24 / 4

x = -11, 6

But the problem tells us one more thing: x is positive. Therefore, x = 6.

x = First number

x + 5 = Second number

x^{2} + (x + 5)^{2} = 157 Equation derived from problem

x^{2} + x^{2} + 10x + 25 = 157 Square the binomial (FOIL)

2x^{2} + 10x + 25 = 157 Simplify

2x^{2} + 10x + 25 - 157 = 157 - 157 Subtract 157 from each side

2x^{2} + 10x - 132 = 0 Simplify

(2x^{2} + 10x - 132)/2 = 0/2 Divide each side by 2

x^{2} + 5x - 66 = 0 Simplify

(x + 11)(x - 6) = 0 Factor

x + 11 = 0 or x - 6 = 0 Set each factor equal to 0

x = -11 or x = 6 Solve each factor for x

The answer is 6, because you are looking for the positive number.

Let's call the number x.

x^{2} + (x + 5)^{2} = 157

x^{2} + (x^{2} + 10x + 25) = 157

2x^{2} + 10x + 25 - 157 = 157 - 157

2x^{2} + 10x - 132 = 0 now divide both sides by 2

x^{2} + 5x - 66 = 0

You can use the quadratic formula here, but it looks like we can factor this. We want to find two factors that multiply to equal -66 and add up to 5. They are 11 and -6, so we can write

(x + 11)(x - 6) = 0

The number has to be positive, so ignore x + 11 = 0 as that'll give us a negative

x - 6 = 0

x = 6

Now we can check that 6 works in the original equation...

x^{2} + (x + 5)^{2} = 157

6^{2} + (6 + 5)^{2} = 157

36 + 121 = 157

the sum of the square of a positive number and the square of 5 more than the number is 157. what is the number?

Let n = the positive number

then

from: "the sum of the square of a positive number and the square of 5 more than the number is 157" we get

n^2 + (n+5) ^2 = 157

n^2 + (n+5)(n+5) = 157

n^2 + n^2+10n+25 = 157

2n^2+10n+25 = 157

2n^2+10n-132 = 0

n^2+5n-66 = 0

(n+11)(n-6) = 0

n ={-11, 6}

since we're looking for a POSITIVE number, throw out the -11 leaving

n = 6

Let us call the unknown number "n". The sum of (the square of a positive number) PLUS (the square of 5 more than the number is equal to 157. Mathematically, this can be represented as:

n^{2} + (n + 5)^{2} = 157^{
}

Expanding this, we get:

n^{2} + (n + 5)(n + 5) = 157

which equals:

n^{2} + n^{2} + 10n + 25 = 157

or:

2n^{2} + 10n - 132 = 0

We can divide each side of the equation by 2 and get the following:

n^{2} + 5n - 66 = 0

Factoring this, we get (n + 11)(n - 6) = 0, and n can equal -11 or 6. However, because the original question states that the unknown "n" is positive, we know that it must therefore be 6 and not -11.

Tara W.

High school physics & engineering teacher, 17 yrs experience

New York, NY

5.0
(16 ratings)

Kevin S.

Personalized Tutoring Services

Brooklyn, NY

5.0
(231 ratings)

## Comments

Well it's not letting me edit again, so I'll answer it in the comment section.

The usual approach to these problems is setting up an equation that provides the same details as the statement of the problem. We don't know what the number is yet so let's call it x.

x

^{2 }+ (x+5)^{2 }= 157Now we'll do some algebra to expand the parentheses and combine like terms.

2x

^{2 }+ 10x + 25 = 157Alright, now we want to get it into the form Ax

^{2}+ Bx + C = 0 so we can easily find the answers by factoring it. Let's move the 157 over to the other side.2x

^{2}+ 10x - 132 = 0I don't like that 2 out front, so I'm going to divide the whole equation by 2.

x

^{2 }+ 5x - 66 = 0Now we can factor this into

(x+11)(x-6) = 0, which means x=-11 or x=6.

Since the original statement of the problem specifies a positive number, our answer is x=6!

Well, nevermind apparently. It won't let me edit and if I try to answer in the comment, it doesn't post.