Question

25x² – 36 = 0, x = 2$,\frac{6}{5},\frac{-6}{5}$

Answer 1

Given: 25x² – 36 = 0 and x = 2$,\frac{6}{5},\frac{-6}{5}$
On substituting x = 2 in L.H.S. of the given equation, we get:
25(2)² – 36
= 100 – 36
= 64
=> L.H.S.$\ne$ R.H.S.
Thus, x = 2 does not satisfy the given equation.
Therefore, x = 2 is not a root of the given quadratic equation.

On substituting x= $\frac{6}{5}$ in L.H.S. of the given equation, we get:

$$\begin{gathered} =25{\left(\frac{6}{5}\right)}^2-36 \\ =\frac{25\times 36}{25}-36 \end{gathered}$$

= 36 – 36
= 0
=> L.H.S. = R.H.S.
Thus, x =$\frac{6}{5}$ satisfies the given equation.
Therefore, x =$\frac{6}{5}$ is a root of the given quadratic equation.

On substituting x = –$\frac{6}{5}$ in L.H.S. of the given equation, we get:
$$\begin{gathered} 25{\left(\frac{-6}{5}\right)}^2-36 \\ =\frac{25\times 36}{35}-36 \end{gathered}$$

= 36 – 36
= 0
=> L.H.S. = R.H.S.
Thus, x = –$\frac{6}{5}$ satisfies the given equation.
Therefore, x = –$\frac{6}{5}$ is a root of the given quadratic equation.