solve x^2-7x-8=0?

Question

using the quadratic formula

Answer 1

This can be solved by factoring into the form
(x - 8) (x + 1) = 0

Notice that for a quadratic it must equal zero
given the form
ax^2 + bx + c = 0
Your a = 1
b = -7
c = -8

Your a =1 so
Since c = a negative
you can only multiply two numbers to get - if one is + and the other is -
+(-) = -
So you need two numbers which
when multiplied will equal -8, but
when add will give you -7
the factors of -8 are
-1(8) but -1 + 8 = +7
2(-4) but 2 - 4 = -2
1(-8) and 1 - 8 = -7

hence
(x + 1)(x - 8) = 0

now since the multiplicative property of zero says any number times zero = 0
When (x + 1) = 0 then your equation will equal 0
x = -1
and when (x - 8) = 0 your equation will also equal 0
x = 8

These are the x-intercepts

Answer 2

using quadratic formula makes a = 1 b = -7 and c = -8 the quadratic formula states: -b +/- sqrt(b^2 - 4ac) / 2a
ERGO
7 +/- sqrt(49 + 32) / 2
7 +/- sqrt(81) / 2
7 +/- 9 / 2
16 / 2 = 8
-2 / 2 = -1

x = -1, 8

Answer 3

x^2 - 7x - 8 = 0
(x - 8)(x + 1) = 0

x - 8 = 0 or x + 1 = 0
x = 8 or x = -1

Answer 4

x^2-7x-8=0
(x+1)(x-8)=0
x= -1 or x=8 =0

Answer 5

I didn't use the quadratic formula; factoring is SO much simpler in this instance. It's x = 8 or -1.

Answer 6

x = [ 7 ± √(49 + 32)] /2 ]
x = [ 7 ± √81] / 2
x = 8, x = - 1
Using factors:-
(x - 8).(x + 1) = 0
x = 8 , x = - 1

Answer 7

(x - 8)(x + 1) = 0
x = 8, x = -1

Answer 8

x1=8
x2=-1