Question

(1)
State whether the following quadratic equation have two distinct real roots. Justify your answer.
(i) $x^2-3x+4=0$

(2)
State whether the following quadratic equation have two distinct real roots. Justify your answer.
(ii) $2x^2+x-1=0$

(3)
State whether the following quadratic equation have two distinct real roots. Justify your answer.
(iii) $2x^2-6x+\frac{9}{2}=0$

(4)
State whether the following quadratic equation have two distinct real roots. Justify your answer.
(iv) $3x^2-4x+1=0$

(5)
State whether the following quadratic equation have two distinct real roots. Justify your answer.
(v) $(x+4{)}^2-8x=0$

(6)
State whether the following quadratic equation have two distinct real roots. Justify your answer.​
(vi) $(x-\sqrt{2}{)}^2-2(x+1)=0$

(7)
State whether the following quadratic equation have two distinct real roots. Justify your answer.
(vii) $\sqrt{2}{x}^2-\frac{3}{\sqrt{2}}x+\frac{1}{\sqrt{2}}=0$

(8)
State whether the following quadratic equation have two distinct real roots. Justify your answer.
(viii) $x(1-x)-2=0$

(9)
State whether the following quadratic equation have two distinct real roots. Justify your answer
(ix) $(x-1)(x+2)+2=0$

(10)
State whether the following quadratic equation have two distinct real roots. Justify your answer.
(x) $(x+1)(x-2)+x=0$

Answer 1

(1)
Step : Finding the discriminant.
Given equation is $x^2-3x+4=0$
On comparing with $a{x}^2+bx+c=0,$ we get
$a=1,b=-3$ and $c=4$
$\therefore \text{Discriminant,}D=b^2-4ac$
$=(-3{)}^2-4(1\left)\right(4)$
$=9-16$
$=-7$
As, $D<0,$ the equation $x^2-3x+4=0$ has no real roots.

(2)
Step : Finding the siscriminant.
Given equation is $2x^2+x-1=0$
On comparing with $a{x}^2+bx+c=0,$ we get
$a=2,b=1$ and $c=-1$
$\therefore \text{Discriminant,}D=b^2-4ac$
$=(1{)}^2-4(2\left)\right(-1)$
$=1+8$
$=9$
As $D>0,$ the equation $2x^2+x-1=0$ has two distinct real roots.

(3)
Step : Finding the discriminant.
Compare equation $2x^2-6x+\frac{9}{2}=0$ with general form of quadratic equation
$a{x}^2+bx+c=0,$ we get
$a=2,b=-6$ and $c=\frac{9}{2}$
$\therefore \text{Discriminant}D=b^2-4ac$
$=(-6{)}^2-4\times 2\times \frac{9}{2}$
$=36-36$
$=0$
i.e., $D=0.$
Hence, the equation $2x^2-6x+\frac{9}{2}=0$ has equal and real roots.

(4)
Step : Finding the discriminant.
Compare given equation $3x^2-4x+1=0$ with general form of quadratic equation $a{x}^2+bx+c=0,$
we get
$a=3,b=-4$ and $c=1$
$\therefore \text{Discriminant,}D=b^2-4ac$
$=(-4{)}^2-4\times (3)\times (1)$
$=16-12=4>0$
As $D>0,$ the equation $3x^2-4x+1=0$ has two distinct real roots.

(5)
Step : Finding the discriminant.
Given equation is $(x+4{)}^2-8x=0$
$x^2+16+2\left(4\right)\left(x\right)-8x=0$
$\because (a+b{)}^2=a^2+b^2+2ab$
$\Rightarrow x^2+8x-8x+16=0$
$\Rightarrow x^2+16=0$
$\Rightarrow x^2+0.x+16=0$

On comparing with $a{x}^2+bx+c=0,$ we get
$a=1,b=0$ and $c=16$
$\therefore \text{Discriminant,}D=b^2-4ac$
$=(0{)}^2-4\times 1\times (16)$
$=-64<0$
As $D<0$, the equation $(x+4{)}^2-8x=0$ has no real roots.

(6)
Step : Finding the discriminant.
Given equation is $(x-\sqrt{2}{)}^2-2(x+1)=0$
$\Rightarrow x^2+(\sqrt{2}{)}^2-2(x\left)\right(\sqrt{2})-2x-2=0$
$[\because (a+b{)}^2=a^2+b^2+2ab]$
$\Rightarrow x^2+2-2\sqrt{2}x-2x-2=0$
$\Rightarrow x^2-(2\sqrt{2}+2)x=0$
On comparing with $a{x}^2+bx+c=0,$ we get
$a=1,b=-(2\sqrt{2}+2)$ and $c=0$
$\therefore \text{Discriminant,}D=b^2-4ac$
$=[-(2\sqrt{2}+2){]}^2-4\times 1\times 0$
$=\left[\right(2\sqrt{2}+2){]}^2>0$
As $D>0,$ the given equation has real and distinct roots.

(7)
Step : Finding the discriminant.
Given equation is $\sqrt{2}{x}^2-\frac{3}{\sqrt{2}}x+\frac{1}{\sqrt{2}}=0$
On comparing with $a{x}^2+bx+c=0,$ we get
$a=\sqrt{2},b=-\frac{3}{\sqrt{2}}\text{and}c=\frac{1}{\sqrt{2}}$
$\therefore \text{Discriminant,}D=b^2-4ac$
$={(-\frac{3}{\sqrt{2}})}^2-4\times \sqrt{2}\times \frac{1}{\sqrt{2}}$
$=\frac{9}{2}-4=\frac{9-8}{2}=\frac{1}{2}>0$
As, $D>0,$ the roots of the equation
$\sqrt{2}{x}^2-\frac{3}{\sqrt{2}}x+\frac{1}{\sqrt{2}}=0$ are real and distinct.

(8)
Step : Finding the discriminant.
Given equation is $x(1-x)-2=0$
$\Rightarrow x-x^2-2=0$
$x^2-x+2=0$
compare the given equation $x^2-x+2=0$
with general form of quadratic equation.
$a{x}^2+bx+c=0.$ we get
$a=1,b=-1$ and $c=2$
$\therefore \text{Discriminant,}D=b^2-4ac$
$=(-1{)}^2-4\times (1)\times (2)$
$=1-\left(4\right)\times \left(2\right)$
$=1-8$
$=-7<0$
As, $D<0,$ the equation $x(1-x)-2=0$ has no real roots.

(9)
Step : Finding the discriminant.
Given equation is :
$(x-1)(x+2)+2=0$
$\Rightarrow x^2+x=0$
Compare the given equation $x^2+x=0$ with general form of quadratic equation $a{x}^2+bx+c=0$
we get
$a=1,b=1$ and $c=0$
$\therefore \text{Discriminant,}D=b^2-4ac$
$=(1{)}^2-4(1\left)\right(0)$
$=1-0$
$=1>0$
As $D>0,$ the equation $(x-1)(x+1)+2=0$ has two distinct roots.

(10)
Step : Finding the discriminant.
Given equation is $(x+1)(x-2)+x=0$
$\Rightarrow x^2-2x+x-2+x=0$
$\Rightarrow x^2-2=0$
$\Rightarrow x^2+0\times x-2=0...\left(1\right)$
Compare the given equation $x^2-2=0$
with general form of quadratic equation.
$a{x}^2+bx+c=0.$ we get
$a=1,b=0$ and $c=-2$
$\therefore \text{Discriminant,}D=b^2-4ac$
$=(0{)}^2-4\times (1)\times (-2)$
$=8>0$
As, $D>0$ the equation $(x+1)(x-2)+x=0$ has two distinct real roots.3