Question

Find the discriminant of each of the following equations:

(i) $2x^2-7x+6=0$
(ii) $3x^2-2x+8=0$
(iii) $2x^2-5\sqrt{2}x+4=0$
(iv) $\sqrt{3}{x}^2+2\sqrt{2}x-2\sqrt{3}=0$
(v) $\left(x-1\right)\left(2x-1\right)=0$
(vi) $1-x=2x^2$2x2−7x+6=02x2−7x+6=0

Answer 1

$$\begin{gathered} \text{(i) 2}{x}^2-7x+6=0 \\ \text{Here,} \\ a=2, \\ b=-7, \\ c=6 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{diven}\mathrm{by}: \\ D=b^2-4ac \\ \text{=}{(-7)}^2-4\times 2\times 6 \\ =49-48 \\ \text{= 1} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)3x^2-2x+8=0 \\ \text{Here,} \\ a=3, \\ b=-2, \\ c=8 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=b^2-4ac \\ \text{=}{(-2)}^2-4\times 3\times 8 \\ \text{=}4-96 \\ =-92 \end{gathered}$$

​$$\begin{gathered} \left(\mathrm{iii}\right)2x^2-5\sqrt{2x}+4=0 \\ \text{Here,} \\ a=2, \\ b=-5\sqrt{2}, \\ c=4 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=b^2-4ac \\ ={(-5\sqrt{2})}^2-4\times 2\times 4 \\ =(\text{25}\times \text{2})-32 \\ =50-32 \\ =18 \end{gathered}$$

​$$\begin{gathered} \left(\mathrm{iv}\right)\sqrt{3}{x}^2+2\sqrt{2}x-2\sqrt{3}=0 \\ \text{Here,} \\ a=\sqrt{3}, \\ b=2\sqrt{2}, \\ c=-2\sqrt{3} \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=b^2-4ac \\ ={\left(2\sqrt{2}\right)}^2-4\times \sqrt{3}\times (-2\sqrt{3}) \\ =(4\times 2)+(8\times 3) \\ =\text{8 + 24} \\ =32 \end{gathered}$$

(v) $\left(x-1\right)\left(2x-1\right)=0$

$\Rightarrow 2x^2-3x+1=0$

Comparing it with $a{x}^2+bx+c=0$, we get

a = 2, b = −3 and c = 1

∴ Discriminant, D = $b^2-4ac={\left(-3\right)}^2-4\times 2\times 1=9-8=1$

​$$\begin{gathered} \left(\mathrm{vi}\right)1-x=2x^2 \\ \Rightarrow 2x^2+x-1=0 \\ \text{Here,} \\ a=2, \\ b=1, \\ c=-1 \\ \mathrm{Discriminant}D\mathrm{is}\mathrm{given}\mathrm{by}: \\ D=b^2-4ac \\ {\text{= 1}}^2-4\times 2(-1) \\ \text{= 1 + 8} \\ \text{= 9} \end{gathered}$$