#### State whether the following quadratic equations have two distinct real roots. Justify your answer.$(i)x^{2}-3x+4=0$         $(ii)2x^{2}+x-1=0$$(iii)2x^{2}-6x+\frac{9}{2}=0$$(iv)3x^{2}-4x+1=0$$(v)(x+4)^{2}-8x=0$$(vi)(x-\sqrt{2})^{2}-2(x+1)=0$$(vii)\sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+\frac{1}{2}=0$$(viii)x(1-x)-2=0$$(ix)(x-1)(x+2)+2=0$$(x)(x+1)(x-2)+x=0$

(i)     Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$  Where a, b and c are real numbers with $a\neq 0$

The given equation is $x^{2}-3x+4=0$

Here a=1,b=-3,c=4
$D=b^{2}-4ac\\ =(-3)^{2}-4(1)(4)\\ =9-16=-7\\ b^{2}-4ac<0$

Hence the equation has no real roots.

(ii)   Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $2x^{2}+x-1=0$

Here a=2,b=-1,c=-1
$D=b^{2}-4ac\\ =(1)^{2}-4(2)(-1)\\ =1+8=9\\ b^{2}-4ac>0$

Here for the equation $2x^{2}+x-1=0$$b^{2}-4ac>0$

Hence it is a quadratic equation with two distinct real roots

(iii)   Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $2x^{2}-6x+\frac{9}{2}=0$

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

$D=b^{2}-4ac\\ =(-6)^{2}-4(2)(\frac{9}{4})\\ =36-36=0\\ b^{2}-4ac=0$

here $b^{2}-4ac=0$

Hence the equation have two real and equal roots.

(iv)    Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $3x^{2}-4x+1=0$

Here a=3,b=-4,c=1
$D=b^{2}-4ac\\ =(-4)^{2}-4(3)(1)\\ =16-12=4\\ b^{2}-4ac>0$

Here  $b^{2}-4ac>0$

Hence the equation has distinct and real roots.

(v) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $(x+4)^{2}-8x=0$

$(x)^{2}+(4)^{2}+2(x)(4)-8x=0$                                              $\left (using (a+b)^{2}=(a)^{2}+(b)^{2}+2ab \right )$

$(x)^{2}+16+8x-8x=0\\ x^{2}+16=0$

$a=1,b=0,c=16\\ =(0)^{2}-4(1)(16)\\ =-64\\ b^{2}-4ac<0$

Here  $b^{2}-4ac<0$

Hence the equation has no real roots.

(vi) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

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

$(x)^{2}+(\sqrt{2})^{2}+2(x)(\sqrt{2})-2x-2=0$                        $\left (using (a+b)^{2}=(a)^{2}+(b)^{2}+2ab \right )$

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

Compare with $ax^{2}+bx+c=0$ where  $a\neq 0$

$a=1,b=-(2\sqrt{2}+2),c=0\\b^{2}-4ac =(-(2\sqrt{2}+2))^{2}-4(1)(0)\\ =(2\sqrt{2}+2)^{2}-0$

$=(2\sqrt{2})^{2}+(2)^{2}+2(2\sqrt{2})(2)$                            $\left (using (a+b)^{2}=(a)^{2}+(b)^{2}+2ab \right )$

$=8+4+8\sqrt{2}\\ =12+8\sqrt{2}\\$

$b^{2}-4ac>0$

Here  $b^{2}-4ac>0$

Hence the equation has two distinct real roots.

(vii)    Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $\sqrt{2}x^{2}-\frac{3}{\sqrt{2}}x+\frac{1}{2}=0$

Here $a=\sqrt{2},b=\frac{3}{\sqrt{2}},c=\frac{1}{2}$
$b^{2}-4ac =\left (- \frac{3}{\sqrt{2}} \right )^{2}-4(\sqrt{2})\left (\frac{1}{2} \right )\\ =\frac{9}{2}-2\sqrt{2}\\=4.5-3.8=1.7\\ b^{2}-4ac>0$

Here  $b^{2}-4ac>0$

Hence the equation has distinct real roots.

(viii) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

The given equation is $x(1-x)-2=0$

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

compare with $ax^{2}+bx+c=0$ where $a\neq 0$

$a=-1,b=1,c=-2\\ b^{2}-4ac=(1)^{2}-4(-1)(-2)\\1-8=-7\\ b^{2}-4ac<0$

Here  $b^{2}-4ac<0$

Hence the equation has no real roots.

(ix) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

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

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

$a=1,b=1,c=0\\b^{2}-4ac =(1)^{2}-4(1)(0)\\ =1$

$b^{2}-4ac>0$

Here  $b^{2}-4ac>0$

Hence the equation has two distinct real roots.

(x) Quadratic equation : A quadratic equation in x is an equation that can be written in the standard form $ax^{2}+bx+c=0$           Where a, b and c are real numbers with $a\neq 0$

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

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

Compare with $ax^{2}+bx+c=0$ where $a\neq 0$

$a=1,b=0,c=-2\\b^{2}-4ac =(0)^{2}-4(1)(-2)\\ =8$

$b^{2}-4ac>0$

Here  $b^{2}-4ac>0$

Hence the equation has two distinct real roots.