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

Question

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!!!! :) 

Answer 1

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.

Answer 2

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.

 

Answer 3

Turn the word problem into an equation.  Call the unknown number x. The square of that number is x². The square of 5 more than that number is (x+5)². The 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.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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.