#### Find whether the following equations have real roots. If real roots exist, find them.$(i)8x^{2}+2x-3=0$$(ii)-2x^{2}+3x+2=0$$(iii)5x^{2}-2x-10=0$$(iv)\frac{1}{2x-3}+\frac{1}{x-5}=1,x\neq \frac{3}{2},5$$(v)x^{2}+5\sqrt{5}x-70=0$

(i) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0  in the case of imaginary roots.

For the equation $8x^{2}+2x-3=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq0$

a=8,b=2,c=-3

$\\b^{2} - 4ac =(2)^{2} - 4(8)(-3)\\ =4+96\\ =100\\ b^{2}-4ac>0$

Hence the equation has real roots

$\\8x^{2}+2x-3=0\\ 8x^{2}+6x-4x-3=0\\ 2x(4x+3)-1(4x+3)=0\\ (4x+3)(2x-1)=0\\ 4x+3=0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2x-1=0\\ 4x=-3 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2x=1\\ x=\frac{-3}{4} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x= \frac{1}{2}\\$

Roots of the equation $8x^{2}+2x-3=0$  are $\frac{-3}{4}, \frac{1}{2}\\$

(ii) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0  in the case of imaginary roots.

For the equation $-2x^{2}+3x+2=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq0$

a=-2,b=3,c=2

$\\b^{2} - 4ac =(3)^{2} - 4(-2)(2)\\ =9+16\\ =25\\ b^{2}-4ac>0$

Hence the equation has real roots

$\\-2x^{2}+3x+2=0\\ -2x^{2}+4x-x+2=0\\-2x(x-2)-1(x-2)=0\\(x-2)(-2x-1)=0\\ -2x-1=0 \;\;\;\;\;\;\;\;\;\;\;\;\; x-2=0\\ -2x=1 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x=2\\ x=\frac{-1}{2}$

Roots of the equation $-2x^{2}+3x+2=0$  are $2, \frac{-1}{2}\\$

(iii) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0  in the case of imaginary roots.

For the equation $5x^{2}-2x-10=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq0$

a=5,b=-2,c=-10

$\\b^{2} - 4ac =(-2)^{2} - 4(5)(-10)\\ =4+200\\ =204\\ b^{2}-4ac>0$

Hence the equation has real roots

$\\5x^{2}-2x-10=0\\ x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\\ x=\frac{2 \pm \sqrt{4-4(5)(-10)}}{2(5)}\\ x=\frac{2 \pm \sqrt{4+200}}{10}\\ x=\frac{2 \pm \sqrt{204}}{10}\\ x=\frac{2(1 \pm \sqrt{51})}{10}\\ x=\frac{1 \pm \sqrt{51}}{5}\\ x=\frac{1 + \sqrt{51}}{5}, x=\frac{1 - \sqrt{51}}{5}\\$

Roots of the equation $5x^{2}-2x-10=0$  are $\frac{1 + \sqrt{51}}{5}, \frac{1 - \sqrt{51}}{5}$

(iv) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0  in the case of imaginary roots.

For the equation $\frac{1}{2x-3}+\frac{1}{x-5}=1$

$\\\frac{(x-5)+(2x-3)}{(2x-3)(x-5)}=1\\ x-5+2x-3=2x(x-5)3(x-5)\\ 3x-8=2x^{2}-10x-3x+15\\ 2x^{2}-13x+15-3x+8=0\\ 2x^{2}-16x+23=0\\$

Compare with $ax^{2}+bx+c=0$ where $a\neq0$

a=2,b=-16,c=23

$\\b^{2}-4ac=(-16)^{2}-4(2)(23)\\ =256-184\\ =72\\ b^{2}-4ac>0$

Hence the equation has real roots

$\\x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\\ x=\frac{16 \pm \sqrt{72}}{2(2)}\;\;\;\;\;\;\;\;\;\;(\because b^{2}-4ac=72)\\ x=\frac{16 \pm 6\sqrt{2}}{4}=\frac{8 \pm 3\sqrt{2}}{2}\\ x=\frac{8 + 3\sqrt{2}}{2},\frac{8 - 3\sqrt{2}}{2}$

Roots of the equation $\frac{1}{2x-3}+\frac{1}{x-5}=1$ are $\frac{8 + 3\sqrt{2}}{2},\frac{8 - 3\sqrt{2}}{2}$

(v) We Know that b2-4ac > 0 in the case of real roots and b2 - 4ac < 0  in the case of imaginary roots.

For the equation $x^{2}+5\sqrt{5}x-70=0$

Compare with $ax^{2}+bx+c=0$ where $a\neq0$

$\\a=1, b=5\sqrt{5},c=-70\\ b^{2} - 4ac =(5\sqrt{5})^{2} - 4(1)(-70)\\ =125+280\\ =405\\ b^{2}-4ac>0$

Hence the equation has real roots

$\\x^{2}+5\sqrt{5}x-70=0\\x^{2}+7\sqrt{5}x-2\sqrt{5}x-70=0\\x(x+7\sqrt{5})-2\sqrt{5}(x+7\sqrt{5})=0\\(x+7\sqrt{5})(x-2\sqrt{5}) \\ x+7\sqrt{5}=0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x+7\sqrt{5}=0\\ x=-7\sqrt{5} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; x=2\sqrt{5}\\$

Roots of the equation $x^{2}+5\sqrt{5}x-70=0$  are $-7\sqrt{5}, 2\sqrt{5}\\$