Search 83,739 tutors
FIND TUTORS
Ask a question
0 0

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

Tutors, please sign in to answer this question.

7 Answers

SHEESH!  All of the above answers do the math correctly and get you to the right answer - eventually.  I primarily tutor for the SAT and couldn't care less about anything other than getting the answer right as quickly as possible.  Thus, I teach The Down and Dirty Bill method: I know that 10 squared is 100, and that 5 squared is 25, so they don't work but are close.  Moving up a notch, I try 6 and 11.  BINGO!  Next SAT question, please.  I ADMIT: this is NOT the method your math teacher wants you to use.  It IS the method the real world rewards.  Just ask the folks at the College Board.

Turn the word problem into an equation.  Call the unknown number xThe square of that number is x2The square of 5 more than that number is (x+5)2The sum of those two quantities, x2 + (x+5)2, equals 157.

OK, let's solve it.

x2 + (x+5)2 = 157

x2 + (x2 + 10x + 25) = 157

2x2 + 10x + 25 = 157

2x2 + 10x - 132 = 0

This is the standard quadratic form, ax2 + 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 ± √(b2 - 4ac) ] / 2a

x = [-10 ± √(102 - 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

x2 + (x + 5)2 = 157                       Equation derived from problem

x2 + x2 + 10x + 25 = 157              Square the binomial (FOIL)

2x2 + 10x + 25 = 157                   Simplify

2x2 + 10x + 25 - 157 = 157 - 157  Subtract 157 from each side

2x2 + 10x - 132 = 0                      Simplify

(2x2 + 10x - 132)/2 = 0/2             Divide each side by 2

x2 + 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.

 

Answering! Was just making sure I could actually post because it was giving me trouble the other day. Editing now.

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.

x2 + (x+5)2 = 157

Now we'll do some algebra to expand the parentheses and combine like terms.

2x2 + 10x + 25 = 157

Alright, now we want to get it into the form Ax2 + Bx + C = 0 so we can easily find the answers by factoring it. Let's move the 157 over to the other side.

2x2 + 10x - 132 = 0

I don't like that 2 out front, so I'm going to divide the whole equation by 2.

x2 + 5x - 66 = 0

Now 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.

Let's call the number x. 

x2 + (x + 5)2 = 157

x2 + (x2 + 10x + 25) = 157

2x2 + 10x + 25 - 157 = 157 - 157

2x2 + 10x - 132 = 0 now divide both sides by 2

x2 + 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...

x2 + (x + 5)2 = 157

62 + (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:

n2 + (n + 5)2 = 157

Expanding this, we get:

n2 + (n + 5)(n + 5) = 157

which equals:

n2 + n2 + 10n + 25 = 157

or:

2n2 + 10n - 132 = 0

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

n2 + 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.