Let x,y be two different numbers
suppose x^2=y^2
then x^2-y^2=0
which yields (x+y)(x-y)=0
so either x=y or x=-y
In any case, x and y must be the same value
also when a vairable is squared like y=x^2
we must note that there are 2 possible solutions
x=(+/-)sqrt(y)
Answer:
x = {-4, 4}
Step-by-step explanation:
Subtracting 3 gives ...
5x^2 = 80
Dividing by 5, we get ...
x^2 = 16
Taking the square root gives the solutions ...
x = ±4
The smaller value of x is -4; the larger value is 4.
Answer:
A
Step-by-step explanation:
y= x+1 at the same time y=4x+5
x+1= 4x+5, so answer is A
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