Question

How do you find all the solutions to the equation 5x^2-27=2x^2+48 using the zero product property?

Answer 1

set to zero both sides since if we have
xy=0 then assume x and y=0
so

5x^2-27=2x^2+48
minus 2x^2+48 from both sides
3x^2-75=0
factor
3(x^2-25)=0
remember diffrence fo 2 perpfect squares (a^2-b^2=(a-b)(a+b))

3(x-5)(x+5)=0
set to zero
x-5=0
x=5

x+5=0
x=-5


x=5 and -5

Answer 2

First, you have to get the equation equal to 0. To do that, you need to subtract the
`2 x^(2)` and 48 from the right side of the equation and put it on the left side. The new equation will read