# State whether the following quadratic

Question:

(i) $x^{2}-3 x+4=0$

(ii) $2 x^{2}+x-1=0$

(iii) $2 x^{2}-6 x+\frac{9}{2}=0$

(iv) $3 x^{2}-4 x+1=0$

(v) $(x+4)^{2}-8 x=0$

(vi) $(x-\sqrt{2})^{2}-\sqrt{2}(x+1)=0$

(vii) $\sqrt{2} x^{2}-\frac{3}{\sqrt{2}} x+\frac{1}{\sqrt{2}}=0$

(viii) $x(1-x)-2=0$

(ix) $(x-1)(x+2)+2=0$

(x) $(x+1)(x-2)+x=0$

Solution:

(i) Given equation is $x^{2}-3 x+4=0$.

On comparing with $a x^{2}+b x+c=0$, we get

$a=1, b=-3$ and $c=4$

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-3)^{2}-4(1)(4)$

$=9-16=-7<0$ i.e., $D<0$

Hence, the equation $x^{2}-3 x+4=0$ has no real roots .

(ii) Given equation is, $2 x^{2}+x-1=0$

On comparing with $a x^{2}+b x+c=0$, we get

$a=2, b=1$ and $c=-1$

$\therefore$ Discriminant,  $D=b^{2}-4 a c=(1)^{2}-4(2)(-1)$

$=1+8=9>0$ i.e., $D<0$

Hence, the equation $2 x^{2}+x-1=0$ has two distinct real roots .

(iii) Given equation is $2 x^{2}-6 x+\frac{9}{2}=0$.

On comparing with $a x^{2}+b x+c=0$, we get

$a=2, b=-6$ and $c=\frac{9}{2}$

$\therefore$ Discriminant, $\quad D=b^{2}-4 a c$

$=(-6)^{2}-4(2)\left(\frac{9}{2}\right)=36-36=0$ i.e., $D=0$

Hence, the equation $2 x^{2}-6 x+\frac{9}{2}=0$ has equal and real roots.

(iv) Given equation is $3 x^{2}-4 x+1=0$

On comparing with $a x^{2}+b x+c=0$, we get

$a=3, b=-4$ and $c=1$

$\therefore$ Discriminant. $\quad D=b^{2}-4 a c=(-4)^{2}-4(3)(1)$

$=16-12=4>0$ i.e., $D>0$

Hence, the equation $3 x^{2}-4 x+1=0$ has two distinct real roots.

(v) Given equation is $(x+4)^{2}-8 x=0$.

$\Rightarrow \quad x^{2}+16+8 x-8 x=0 . \quad\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$

$\Rightarrow \quad x^{2}+16=0$

$\Rightarrow \quad x^{2}+0 \cdot x+16=0$

On comparing with $a x^{2}+b x+c=0$, we get

$a=1, b=0$ and $c=16$

$\therefore$ Discriminant, $D=b^{2}-4 a c=(0)^{2}-4(1)(16)=-64<0$ i.e., $D<0$

Hence, the equation $(x+4)^{2}-8 x=0$ has imaginary roots, i.e., no real roots.

(vi) Given equation is $(x-\sqrt{2})^{2}-\sqrt{2}(x+1)=0$

$\Rightarrow \quad x^{2}+(\sqrt{2})^{2}-2 x \sqrt{2}-\sqrt{2} x-\sqrt{2}=0 \quad\left[\because(a-b)^{2}=a^{2}-2 a b+b^{2}\right]$

$\Rightarrow \quad x^{2}+2-2 \sqrt{2} x-\sqrt{2} x-\sqrt{2}=0$

$\Rightarrow \quad x^{2}-3 \sqrt{2} x+(2-\sqrt{2})=0$

On comparing with $a x^{2}+b x+c=0$, we get

$a=1, b=-3 \sqrt{2}$ and $c=2-\sqrt{2}$

$\therefore$ Discriminant, $D=b^{2}-4 a c$

$=(-3 \sqrt{2})^{2}-4(1)(2-\sqrt{2})=9 \times 2-8+4 \sqrt{2}$

$=18-8+4 \sqrt{2}=10+4 \sqrt{2}>0$ i.e., $D>0$

Hence, the equation $(x-\sqrt{2})^{2}-\sqrt{2}(x+1)=0$ has two distinct real roots.

(vii) Given, equation is $\sqrt{2} x^{2}-\frac{3}{\sqrt{2}} x+\frac{1}{\sqrt{2}}=0$.

On comparing with $a x^{2}+b x+c=0$, we get

$a=\sqrt{2}, b=-\frac{3}{\sqrt{2}}$ and $c=\frac{1}{\sqrt{2}}$

$\therefore$ Discriminant, $\quad D=b^{2}-4 a c$

$=\left(-\frac{3}{\sqrt{2}}\right)^{2}-4 \sqrt{2}\left(\frac{1}{\sqrt{2}}\right)$

$=\frac{9}{2}-4=\frac{9-8}{2}=\frac{1}{2}>0$ i.e., $D>0$

Hence, the equation $\sqrt{2} x^{2}-\frac{3}{\sqrt{2}} x+\frac{1}{\sqrt{2}}=0$ has two distinct real roots.

(viii) Given equation is $x(1-x)-2=0$.

$\Rightarrow \quad x-x^{2}-2=0$

$\Rightarrow \quad x^{2}-x+2=0$

On comparing with $a x^{2}+b x+c=0$, we get

$a=1, b=-1$ and $c=2$

$\therefore$ Discriminant, $D=b^{2}-4 a c$

$=(-1)^{2}-4(1)(2)=1-8=-7<0$ i.e., $D<0$

Hence, the equation $x(1-x)-2=0$ has imaginary roots i.e., no real roots.

(ix) Given equation is

$(x-1)(x+2)+2=0$

$\Rightarrow \quad x^{2}+x-2+2=0$

$\Rightarrow \quad x^{2}+x+0=0$

On comparing the equation with $a x^{2}+b x+c-0$, We have

$a=1, b=1$ and $c=0$

$\therefore$ Discriminant, $D=b^{2}-4 a c$

$=1-4(1)(0)=1>0$ i.e., $D>0$

Hence, equation has two distinct real roots.

(x) Given equation is $(x+1)(x-2)+x=0$

$\Rightarrow \quad x^{2}+x-2 x-2+x=0$

$\Rightarrow \quad x^{2}-2=0$

$\Rightarrow \quad x^{2}+0 \cdot x-2=0$

On comparing with $a x^{2}+b x+c=0$, we get

$a=1, b=0$ and $c=-2$

$\therefore$ Discriminant, $D=b^{2}-4 a c=(0)^{2}-4(1)(-2)=0+8=8>0$

Hence, the equation $(x+1)(x-2)+x=$ O has two distinct real roots.