Q. 285 - Intermediate Algebra

Question

Explain why multiplying $\sqrt{2x-3}$

by its conjugate results in an expression with no radicals.

Step-by-Step Solution

Answer

Consider the formula given below.

$\left(a+b\right)\left(a-b\right)=a^2-b^2$

The above formula shows that the variables $a$ and $b$ are squared. This is the reason why the radicals will vanish when an expression is multiplied by its conjugate.

Like wise, when $\sqrt{2x}-3$ is multiplied by its conjugate, there will be no radicals in its result.

Step 1 The given expression is,

$\sqrt{2x}-3$.

Let us consider the formula given below,

$\left(a+b\right)\left(a-b\right)=\left(a^2-b^2\right)$.

The above formula shows that the variables $a$ and $b$ are squared. This is the reason why the radicals will vanish when an expression is multiplied by its conjugate. 

Step 2 Multiply 2 x - 3 by its conjugate and it is shown below.

$$\begin{gathered} \left(\sqrt{2x}-3\right)\left(\sqrt{2x}+3\right)={\left(\sqrt{2x}\right)}^2-{\left(3\right)}^2 \\ =2x-9 \end{gathered}$$

Therefore, the reason for no radicals in the result is explained when the expression $\sqrt{2x}-3$is multiplied by its conjugate.