Question

Find the zeros of the function y = -4x^2 + 16

Answer 1

$-4x^2+16 & =0\\\\\implies -4(x^2-4) & =0 \\\\\implies x^2-4 =0\\ \\\implies x^2-2^2 =0\\\\\implies (x+2)(x-2)=0\\\\\implies x=2,~~x = -2$

Answer 2

Step-by-step explanation:

\begin{gathered}-4x^2+16 & =0\\\\\implies -4(x^2-4) & =0 \\\\\implies x^2-4 =0\\ \\\implies x^2-2^2 =0\\\\\implies (x+2)(x-2)=0\\\\\implies x=2,~~x = -2\end{gathered}

−4x

2

+16

⟹−4(x

2

−4)

⟹x

2

−4=0

⟹x

2

−2

2

=0

⟹(x+2)(x−2)=0

⟹x=2, x=−2

=0

=0