Question

How do you factor and solve $64{x^2} - 1 = 0$?

Answer 1

In this question, we will use the algebraic identity which relates as ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ to factorize the given quadratic equation. Then find the roots of the given quadratic equation to get the required answer.

Complete step by step answer:
We are given that we need to solve $64{x^2} - 1 = 0$ using the method of factorization as we have to factor the quadratic equation $64{x^2} - 1 = 0$.
We can write the given quadratic equation as follows:
${\left( {8x} \right)^2} - {\left( 1 \right)^2} = 0$
From the algebraic identities, we know that ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$.
By using this algebraic identity, we get the equation ${\left( {8x} \right)^2} - {\left( 1 \right)^2} = 0$ as follows:
${\left( {8x} \right)^2} - {\left( 1 \right)^2} = \left( {8x - 1} \right)\left( {8x + 1} \right) = 0$
So, we get the equation $64{x^2} - 1 = 0$ as $\left( {8x - 1} \right)\left( {8x + 1} \right) = 0$.
By this way, we get the factors of $64{x^2} - 1$ and they are $8x - 1$ and $8x + 1$.
Now, we need to solve $64{x^2} - 1 = 0 \Leftrightarrow \left( {8x - 1} \right)\left( {8x + 1} \right) = 0$.
So, we have

$$\left( {8x - 1} \right) = 0 \\\ \Rightarrow 8x = 0 + 1 = 1 \\\ \therefore x = \dfrac{1}{8}$$

And

$$\left( {8x + 1} \right) = 0 \\\ \Rightarrow 8x = 0 - 1 = - 1 \\\ \therefore x = - \dfrac{1}{8}$$

Thus, the values of $x$ are $\dfrac{{ - 1}}{8}$ and $\dfrac{1}{8}$ i.e., $x = - \dfrac{1}{8},\dfrac{1}{8}$.

Note: The students must note that solving an equation by using factorization helps us in many aspects because we do not have to use the formula for roots of a quadratic equation which itself involves a lot of calculations and can lead to mathematical errors.
The students must also note that in the last few steps, when we got the factors of the given quadratic equation, we did use a theorem to solve it. It follows as: If $a.b = 0$, then either $a = 0$ or $b = 0$ or both $a = b = 0$.