solve: $16x^{2}-81 = 0$ \na $-\frac{4}{9},\frac{4}{9}$ \nb $-\frac{81}{16},\frac{81}{16}$ \nc…

solve: $16x^{2}-81 = 0$ \na $-\frac{4}{9},\frac{4}{9}$ \nb $-\frac{81}{16},\frac{81}{16}$ \nc $-\frac{16}{81},\frac{16}{81}$ \nd $-\frac{9}{4},\frac{9}{4}$

solve: $16x^{2}-81 = 0$ \na $-\frac{4}{9},\frac{4}{9}$ \nb $-\frac{81}{16},\frac{81}{16}$ \nc $-\frac{16}{81},\frac{16}{81}$ \nd $-\frac{9}{4},\frac{9}{4}$

Answer

Explanation:

Step1: Isolate the $x^{2}$ term

Add 81 to both sides of the equation $16x^{2}-81 = 0$. We get $16x^{2}=81$.

Step2: Solve for $x^{2}$

Divide both sides by 16. So $x^{2}=\frac{81}{16}$.

Step3: Find the values of $x$

Take the square - root of both sides. Remember that if $x^{2}=a$ ($a\geq0$), then $x=\pm\sqrt{a}$. So $x=\pm\sqrt{\frac{81}{16}}=\pm\frac{9}{4}$.

Answer:

D. $-\frac{9}{4},\frac{9}{4}$