Question

4m² – 9 = 0, m = $2,\frac{2}{3},\frac{3}{2}$

Answer 1

Given: 4m² – 9 = 0 and m = $2,\frac{2}{3},\frac{3}{2}$

On substituting m = 2 in L.H.S. of the given equation, we get:

4(2)² – 9

= 16 – 9

= 7

$\Rightarrow \mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}.$

Thus, m = 2 does not satisfy the given equation.

Therefore, m = 2 is not a root of the given quadratic equation.

On substituting m = $\frac{2}{3}$ in L.H.S. of the given equation, we get:
$4{\left(\frac{2}{3}\right)}^2-9$

$$\begin{gathered} =4\times \frac{4}{9}-9 \\ =\frac{16}{9}-9 \\ =\frac{16-81}{9} \\ =\frac{-65}{9} \\ \Rightarrow \mathrm{L}.\mathrm{H}.\mathrm{S}.\ne \mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$
.
Thus, m = $\frac{2}{3}$ does not satisfy the given equation.

Therefore, m = $\frac{2}{3}$ is not a root of the given quadratic equation.

On substituting m = $\frac{3}{2}$ in L.H.S. of the given equation, we get:
$$\begin{gathered} 4{\left(\frac{3}{2}\right)}^2-9 \\ =4\times \frac{9}{4}-9 \\ =9-9=0 \\ \Rightarrow \mathrm{L}.\mathrm{H}.\mathrm{S}.=\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

Thus, m = $\frac{3}{2}$ satisfies the given equation.
Therefore, m =$\frac{3}{2}$ is a root of the given quadratic equation.