Question

Determine whether the values given against each of the quadratic equation are the roots of the equation.
(1) x² + 4x – 5 = 0 , x = 1, –1
(2) 2m² – 5m = 0, $m=2,\frac{5}{2}$

Answer 1

(1) x² + 4x – 5 = 0 , x = 1, –1
$$\begin{gathered} Forx=1 \\ {\left(1\right)}^2+4\left(1\right)-5=0 \\ \Rightarrow 1+4-5=0 \\ \Rightarrow 5-5=0 \end{gathered}$$
So, x = 1 is a solution of the given equation.
For x = –1
$$\begin{gathered} {\left(-1\right)}^2+4\left(-1\right)-5=0 \\ \Rightarrow 1-4-5=0 \\ \Rightarrow -3-5=0 \\ \Rightarrow -8\ne 0 \end{gathered}$$
So, x = –1 is not a solution of the given equation.
Thus, only x = 1 is root of the given equation.

(2) 2m² – 5m = 0, $m=2,\frac{5}{2}$
$$\begin{gathered} Whenm=2 \\ 2{\left(2\right)}^2-5\left(2\right)=0 \\ \Rightarrow 8-10=0 \\ \Rightarrow -2\ne 0 \end{gathered}$$
So, m = 2 is not a solution of the given equation.
$$\begin{gathered} \mathrm{When}m=\frac{5}{2} \\ 2{\left(\frac{5}{2}\right)}^2-5\left(\frac{5}{2}\right)=0 \\ \Rightarrow \frac{25}{2}-\frac{25}{2}=0 \end{gathered}$$
So, $m=\frac{5}{2}$ is a solution of the given quadratic equation.
Thus, only $m=\frac{5}{2}$ is a root of the given quadratic equation.