Question

Solve each of the following equations using a method other than the Quadratic Formula.

A.y^2 - 6y=0
B. n^2+5n +7= 7
C. 2t^2 -14t +3=3
D. 1/3x^2 +3x– 4=-4
E. Zero is a solution to each of the above equations. What do all of the above equations have in common that causes them to have zero as a solution?​

Answer 1

A)

y^2-6y = 0

or, y(y-6) = 0

or, y = 0 or y = 6

B)

n^2+5n+7 = 7

or, n^2+5n+7-7 = 7-7 ( Subtracting 7 from both sides)

or, n^2+5n = 0

or, n(n+5) = 0

or, n=0 or n= -5

C)

2t^2-14t+3 = 3

or, 2t^2-14t = 0

or, 2t(t-7) = 0

or, t=0 or t=7

D)

1/3x^2+3x-4 = -4

or, 1/3x^2+3x = 0

or, 1/3x(x+9) = 0

or, x=0 or x= -9

E)

Zero is a common solution to each of the equations. This is because each of the equations had a variable outside the parenthesis with an operation of multiplication.

THANK YOU FOR READING.