Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: }\text{Write the discriminant of the following quadratic equations:}$
$\displaystyle \text{(i) }(x-1)(2x-1)=0$
$\displaystyle \text{(ii) }x^2-2x+k=0,\quad k\in R$
$\displaystyle \text{(iii) }\sqrt{3}x^2+2\sqrt{2}x-2\sqrt{3}=0$
$\displaystyle \text{(iv) }(x+5)^2=2(5x-3)\hfill\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }(x-1)(2x-1)=0$
$\displaystyle \Rightarrow 2x^2-3x+1=0$
$\displaystyle \text{Here, }a=2,\ b=-3,\ c=1.$
$\displaystyle D=b^2-4ac=(-3)^2-4(2)(1)=9-8=1$
$\displaystyle \therefore D=1$

$\displaystyle \text{(ii) }x^2-2x+k=0$
$\displaystyle \text{Here, }a=1,\ b=-2,\ c=k.$
$\displaystyle D=b^2-4ac=(-2)^2-4(1)(k)$
$\displaystyle \therefore D=4-4k=4(1-k)$

$\displaystyle \text{(iii) }\sqrt{3}x^2+2\sqrt{2}x-2\sqrt{3}=0$
$\displaystyle \text{Here, }a=\sqrt{3},\ b=2\sqrt{2},\ c=-2\sqrt{3}.$
$\displaystyle D=b^2-4ac=(2\sqrt{2})^2-4(\sqrt{3})(-2\sqrt{3})$
$\displaystyle =8+24=32$
$\displaystyle \therefore D=32$

$\displaystyle \text{(iv) }(x+5)^2=2(5x-3)$
$\displaystyle \Rightarrow x^2+10x+25=10x-6$
$\displaystyle \Rightarrow x^2+31=0$
$\displaystyle \text{Here, }a=1,\ b=0,\ c=31.$
$\displaystyle D=b^2-4ac=0^2-4(1)(31)=-124$
$\displaystyle \therefore D=-124$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{In the following, determine whether the given quadratic equations have real}$
$\displaystyle \text{roots and if so, find the roots:}$
$\displaystyle \text{(i) }3x^2+2\sqrt{5}x-5=0$
$\displaystyle \text{(ii) }2x^2+5\sqrt{3}x+6=0$
$\displaystyle \text{(iii) }\sqrt{2}x^2+7x+5\sqrt{2}=0\hfill\text{[CBSE 2013]}$
$\displaystyle \text{(iv) }2x^2-2\sqrt{2}x+1=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }3x^2+2\sqrt{5}x-5=0$
$\displaystyle \text{Here, }a=3,\ b=2\sqrt{5},\ c=-5.$
$\displaystyle D=b^2-4ac=(2\sqrt{5})^2-4(3)(-5)=20+60=80>0$
$\displaystyle \therefore \text{The equation has two distinct real roots.}$
$\displaystyle x=\frac{-b\pm\sqrt{D}}{2a}$
$\displaystyle =\frac{-2\sqrt{5}\pm\sqrt{80}}{6}$
$\displaystyle =\frac{-2\sqrt{5}\pm4\sqrt{5}}{6}$
$\displaystyle \therefore x=\frac{\sqrt{5}}{3}\text{ or }x=-\sqrt{5}$

$\displaystyle \text{(ii) }2x^2+5\sqrt{3}x+6=0$
$\displaystyle \text{Here, }a=2,\ b=5\sqrt{3},\ c=6.$
$\displaystyle D=b^2-4ac=(5\sqrt{3})^2-4(2)(6)=75-48=27>0$
$\displaystyle \therefore \text{The equation has two distinct real roots.}$
$\displaystyle x=\frac{-b\pm\sqrt{D}}{2a}$
$\displaystyle =\frac{-5\sqrt{3}\pm\sqrt{27}}{4}$
$\displaystyle =\frac{-5\sqrt{3}\pm3\sqrt{3}}{4}$
$\displaystyle \therefore x=-\frac{\sqrt{3}}{2}\text{ or }x=-2\sqrt{3}$

$\displaystyle \text{(iii) }\sqrt{2}x^2+7x+5\sqrt{2}=0$
$\displaystyle \text{Here, }a=\sqrt{2},\ b=7,\ c=5\sqrt{2}.$
$\displaystyle D=b^2-4ac=7^2-4(\sqrt{2})(5\sqrt{2})=49-40=9>0$
$\displaystyle \therefore \text{The equation has two distinct real roots.}$
$\displaystyle x=\frac{-b\pm\sqrt{D}}{2a}$
$\displaystyle =\frac{-7\pm3}{2\sqrt{2}}$
$\displaystyle \therefore x=-\frac{4}{2\sqrt{2}}\text{ or }x=-\frac{10}{2\sqrt{2}}$
$\displaystyle \therefore x=-\sqrt{2}\text{ or }x=-\frac{5\sqrt{2}}{2}$

$\displaystyle \text{(iv) }2x^2-2\sqrt{2}x+1=0$
$\displaystyle \text{Here, }a=2,\ b=-2\sqrt{2},\ c=1.$
$\displaystyle D=b^2-4ac=(-2\sqrt{2})^2-4(2)(1)=8-8=0$
$\displaystyle \therefore \text{The equation has two equal real roots.}$
$\displaystyle x=\frac{-b}{2a}=\frac{2\sqrt{2}}{4}=\frac{\sqrt{2}}{2}$
$\displaystyle \therefore \text{Both roots are equal and are }\frac{\sqrt{2}}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Solve for }x:$
$\displaystyle \text{(i) }\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3},\quad x\ne2,4\hfill\text{[CBSE 2005]}$
$\displaystyle \text{(ii) }\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6},\quad x\ne3,-5\hfill\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3}$
$\displaystyle \Rightarrow 3(x-1)(x-4)+3(x-3)(x-2)=10(x-2)(x-4)$
$\displaystyle \Rightarrow 3(x^2-5x+4)+3(x^2-5x+6)=10(x^2-6x+8)$
$\displaystyle \Rightarrow 2x^2-15x+25=0$
$\displaystyle \text{Here, }a=2,\ b=-15,\ c=25.$
$\displaystyle D=b^2-4ac=(-15)^2-4(2)(25)=225-200=25$
$\displaystyle x=\frac{-b\pm\sqrt{D}}{2a}$
$\displaystyle =\frac{15\pm\sqrt{25}}{4}=\frac{15\pm5}{4}$
$\displaystyle \therefore x=5\text{ or }x=\frac{5}{2}$

$\displaystyle \text{(ii) }\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6}$
$\displaystyle \Rightarrow \frac{(x+5)-(x-3)}{(x-3)(x+5)}=\frac{1}{6}$
$\displaystyle \Rightarrow \frac{8}{(x-3)(x+5)}=\frac{1}{6}$
$\displaystyle \Rightarrow (x-3)(x+5)=48$
$\displaystyle \Rightarrow x^2+2x-15=48$
$\displaystyle \Rightarrow x^2+2x-63=0$
$\displaystyle \text{Here, }a=1,\ b=2,\ c=-63.$
$\displaystyle D=b^2-4ac=2^2-4(1)(-63)=4+252=256$
$\displaystyle x=\frac{-b\pm\sqrt{D}}{2a}$
$\displaystyle =\frac{-2\pm\sqrt{256}}{2}=\frac{-2\pm16}{2}$
$\displaystyle \therefore x=7\text{ or }x=-9$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Solve for }x:$
$\displaystyle \text{(iii) }x+\frac{1}{x}=3,\quad x\ne0\hfill\text{[NCERT]}$
$\displaystyle \text{(iv) }\frac{16}{x}-1=\frac{15}{x+1},\quad x\ne0,-1\hfill\text{[CBSE 2014]}$
$\displaystyle \text{(v) }\frac{1}{x}+\frac{2}{2x-3}=\frac{1}{x-2},\quad x\ne0,\frac{3}{2},2\hfill\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iii) }x+\frac{1}{x}=3$
$\displaystyle \Rightarrow x^2+1=3x$
$\displaystyle \Rightarrow x^2-3x+1=0$
$\displaystyle \text{Here, }a=1,\ b=-3,\ c=1.$
$\displaystyle D=b^2-4ac=(-3)^2-4(1)(1)=9-4=5$
$\displaystyle x=\frac{-b\pm\sqrt{D}}{2a}$
$\displaystyle =\frac{3\pm\sqrt{5}}{2}$
$\displaystyle \therefore x=\frac{3+\sqrt{5}}{2}\text{ or }x=\frac{3-\sqrt{5}}{2}$

$\displaystyle \text{(iv) }\frac{16}{x}-1=\frac{15}{x+1}$
$\displaystyle \Rightarrow \frac{16-x}{x}=\frac{15}{x+1}$
$\displaystyle \Rightarrow (16-x)(x+1)=15x$
$\displaystyle \Rightarrow 16x+16-x^2-x=15x$
$\displaystyle \Rightarrow x^2-16=0$
$\displaystyle \text{Here, }a=1,\ b=0,\ c=-16.$
$\displaystyle D=b^2-4ac=0^2-4(1)(-16)=64$
$\displaystyle x=\frac{-b\pm\sqrt{D}}{2a}$
$\displaystyle =\frac{\pm8}{2}$
$\displaystyle \therefore x=4\text{ or }x=-4$

$\displaystyle \text{(v) }\frac{1}{x}+\frac{2}{2x-3}=\frac{1}{x-2}$
$\displaystyle \Rightarrow (2x-3)(x-2)+2x(x-2)=x(2x-3)$
$\displaystyle \Rightarrow 2x^2-7x+6+2x^2-4x=2x^2-3x$
$\displaystyle \Rightarrow 2x^2-8x+6=0$
$\displaystyle \Rightarrow x^2-4x+3=0$
$\displaystyle \text{Here, }a=1,\ b=-4,\ c=3.$
$\displaystyle D=b^2-4ac=(-4)^2-4(1)(3)=16-12=4$
$\displaystyle x=\frac{-b\pm\sqrt{D}}{2a}$
$\displaystyle =\frac{4\pm2}{2}$
$\displaystyle \therefore x=3\text{ or }x=1$
$\displaystyle \\$