1 math prob. please help! thanks so much!?

Question

1 Math Question Please!?
Find an equation in standard form with integral coefficients that has the given solution set.
1. (-1,5)

a. x^2 - 4x - 5 = 0
b. x^2 + 4x - 5 =0
c. x^2 - 4x +5 = 0
d. x^2 - x+5 =0

Thanks so much!! :-)

Answer 1

The easy way to do this is to write your solutions as:
(x - x1)(x - x2) = 0

You have x1 = -1 and x2 = 5

So that becomes:
(x + 1)(x - 5) = 0

When you multiply using FOIL you get:
x² + -5x + x - 5 = 0

Simplifying you get:
x² - 4x - 5 = 0

This matches with a.

Answer 2

If (-1, 5) is the solution set, then the equation is:

(x + 1)(x - 5)
x² - 5x + x - 5
x² - 4x - 5

Answer a.

Another way you could do it is just by plugging in.

-1:
(-1)² - 4(-1) - 5 = 0
1 + 4 - 5 = 0...check!

5:
5² - 4(5) - 5 = 0
25 - 20 - 5 = 0...check!

so no need to check any of the rest.

Answer 3

Notice that 1/4 = (1/2)^2 = (2^-1)^2 = 2^-2. Since 2^x equals 2^-2, then that implies x = -2. For the second problem, remember that any non-zero number raised to the power of 0 is one. That's because, for example: 3^0 = 3^(5-5) = 3^5 / 3^5 = 1

Answer 4

Right Answer is a)

You just have to susbtitute both values (one at a time) in order to check the answer:

(-1)² - 4(-1) - 5 = 0
1+4-5 = 0
5-5=0
0=0 o.k.

(5)²-4(5)-5=0
25-20-5=0
25-25=0
0=0 o.k.

Good luck!

Answer 5

x^2 +4x - 5 = (x + 5) (x - 1)

Answer 6

-1, 5
(x+1)(x-5)=0 use FOIL
x^2-5x+x-5=0
x^2-4x-5=0

a. x^2 - 4x - 5 = 0 is correct

Answer 7

a.x^2-5x+x-5=0
x(x-5)+1(x-5)=0
(x+1)(x-5)=0

b.similarly as a. ans (x-1)(x+5)
c. " " " " (x+1)(x^2-25)

Answer 8

The answer is A. Hope that helped ;-)

Answer 9

a

Answer 10

b