How do you solve #x^ { 2} - 11x + 19= - 5#?

Question

8443 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-05T23:48:02

#x = 3 , 8#

STEP ONE: Add #5# to each side. You want all of the terms to be on one side before you start factoring!

#x^2 -11x+19=-5#

#x^2 -11x+24=0#

STEP TWO: Factor the equation! Two number that equal #24# when multiplied and #-11# when added together are #-8# and #-3#

#x^2 -11x+24=0#

#(x-3)(x-8)#

THEREFORE, #x = 3 , 8#

Answer 2 · 2017-12-05T23:48:51

#3=x=8#

First, we turn the equation so that it equals zero.
#x^2-11x+19=-5#
#x^2-11x+24=0#

Let's see whether we could factor this.

We find the factors for 24:
1,24
2,12
3,8
4,6
-1,-24
-2,-12
-3,-8
-4,-6

We see that -3 and -8 add up to -11.

Therefore, #x^2-11x+24=0# becomes #(x-3)(x-8)=0#
Therefore, our answers are 3 and 8.

You could have used the quadratic formula, but it is unnecessary in this case.

Answer 3 · 2017-12-05T23:53:56

#2# real solutions:

#x=3\quad,\quad x=8#

Rearrange the equation so it becomes a standard form quadratic equation, #ax^2+bx+c#.

#x^2-11x+24#, where:

Plug those values into the quadratic formula:

#x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}#

#\implies x=\frac{11\pm\sqrt{(-11)^2-4(1)(24)}}{2(1)}#

#\implies x=\frac{11\pm\sqrt{121-96}}{2}#

#\implies x-\frac{11\pm5}{2}#

#\implies x=3\quad,\quad x=8#