Question

Find the zeroes of the polynomial 100x^2 - 81

Answer 1

• x = ±0.9

Explanation:

The algebraic identity to be used here :

• a² - b² = (a + b)(a - b)

We know that :

• 100x² = (10x)²
• 81 = 9²

Therefore,

• 100x² - 81
• (10x + 9)(10x - 9)

The zeros are :

• 10x + 9 = 0 ⇒ 10x = -9 ⇒ x = -9/10 ⇒ x = -0.9
• 10x - 9 = 0 ⇒ 10x = 9 ⇒ x = 9/10 ⇒ x = 0.9

Answer 2

`x=-(9)/(10) \quad x=(9)/(10)`

Explanation:

`\begin{aligned}100x^2-81 & =(10^2)x^2-9^2\\ & = (10x)^2-9^2\end{aligned}`

`\textsf{Difference of Two Squares Formula}: \quad a^2-b^2=\left(a+b\right)\left(a-b\right)`

`\implies a=10x \: \textsf{ and } \: b=9`

`\implies 100x^2-81=(10x+9)(10x-9)`

`\begin{aligned}\textsf{To find the zeros}: \quad 100x^2-81 & = 0\\\\\implies (10x+9)(10x-9) & =0\\\implies 10x+9 & =0 \implies x=-(9)/(10)\\\implies 10x-9 & =0 \implies x=(9)/(10)\end{aligned}`