Question

If x^2-63x-64=0 and p and n are integers such that p^n=x which of the following CANNOT be a value of p ?
a) -8
b) -4
c) -1
d) 4
e) 64

Answer 1

Use the general solution to the quadratic to find the 2-member solution set. Those will be your values for x (do one at a time for the following): Plug in a) through e) one at a time for p and see if you can find an integer n so that p^n equals first the first value of x and then try the second. You will quickly start eliminating choices.

Answer 2

Start by solving the quadratic equation for the 2 values of x.

Answer 3

How did you get that

Answer 4

@leah Did you factor the quadratic? what did you get?

Answer 5

Thats the part im stuck on

Answer 6

to factor, 1st look at the -64. the minus means the roots are different signs
list the factors of 64
1,64
2,32
4,16
8,8
which pair, when added with different signs, gives -63 (from -63x term)

Answer 7

1 and 64

Answer 8

No, not 1 and 64, but -1 and 64.

Answer 9

specifically +1 and -64. this means the factors are
(x+1)(x-64)= 0 (use FOIL to check this matches x^2 - 63x -64 )

we now know either x+1 is zero or x-64 is zero. x+1=0 or x= -1
and x-64= 0 or x= +64

Answer 10

but to answer the question, x is either -1 or +64
let's try the values for p
a) p= -8 can you find an integer n so that (-8)^2 = 64 (or -1)

Answer 11

-1 and 64 added together is +63. you want -63. see post above

Answer 12

a) p= -8 can you find an integer n so that (-8)^2 = 64

Answer 13

n=2 make -8^n = -*^2= 64

p= -4: -4^3= -4*-4*-4= -64 this looks bad

p= -1: -1^1 = -1 matches -1

can you finish the last two?

Answer 14

Where did you get n=2

Answer 15

Never mind i got it

Answer 16

the question asks
p and n are integers such that p^n=x

with p= -8 (choice (a))
I know from the multiplication table that -8*-8= 64 so n=2 (square -8)

Answer 17

Yes i get it thank you