Question 513 Marks
Solve the following equation by using quadratic equations for x and give your 5x(x + 2) = 3
Answer$
\begin{aligned}
& 5 x(x+2)=3 \\
& \Rightarrow 5 x^2+10 x=3 \\
& \Rightarrow 5 x^2+10 x-3=0
\end{aligned}
$
Here $a=5, b=10, c=-3$
$
\begin{aligned}
& D=b^2-4 a c \\
& =(10)^2-4 \times 5 \times(-3) \\
& =100+60 \\
& =160
\end{aligned}
$
$
\begin{aligned}
& \therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
& =\frac{10 \pm \sqrt{160}}{2 \times 5} \\
& =\frac{-10 \pm \sqrt{16 \times 10}}{10} \\
& =\frac{-10 \pm 4 \sqrt{10}}{10} \\
& =\frac{-10 \pm 4(3.162)}{10} \\
& =\frac{-10 \pm 12.648}{10}
\end{aligned}
$
$
\therefore \begin{array}{r}
x _1=\frac{-10+12.648}{10} \\
2.648
\end{array}
$
$
\begin{aligned}
& =\frac{2.648}{10} \\
& =0.2648 \\
& \times_2=\frac{-10-12.648}{10} \\
& =\frac{-22.648}{10} \\
& =-2.2648
\end{aligned}
$
$\therefore x=0.26,-2.26$.
View full question & answer→Question 523 Marks
Solve the following equation by using quadratic equations for x and give your $x^2 – 5x – 10 = 0$
Answer$
x^2-5 x-10=0
$
On comparing with, $a x^2+b x+c=0$
$
\begin{aligned}
& a=1, b=-5, c=-10 \\
& \because x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
& x=\frac{-(-5) \pm \sqrt{(-5)^2-4(1)(-10)}}{2 \times 1} \\
& \therefore x=\frac{5 \pm \sqrt{25+40}}{2} \\
& \Rightarrow x=\frac{5 \pm \sqrt{65}}{2} \\
& =\frac{5 \pm 8.06}{2}
\end{aligned}
$
Either $x=\frac{5+8.06}{2}$
$
\begin{aligned}
& =\frac{13.06}{2} \\
& =6.53
\end{aligned}
$
or
$
\begin{aligned}
& x=\frac{5-8.06}{2} \\
& =\frac{-3.06}{2} \\
& =1.53
\end{aligned}
$
$\therefore x=6.53, x=-1.53$.
View full question & answer→Question 533 Marks
Solve the following equation by factorization
$3(y^2 – 6) = y(y + 7) – 3$
Answer$
\begin{aligned}
& 3\left(y^2-6\right)=y(y+7)-3 \\
& \Rightarrow 3\left(y^2-6\right)=y^2+7 y-3 \\
& \Rightarrow 3 y^2-18=y^2+7 y-3 \\
& \Rightarrow 3 y^2-y^2-7 y-18+3=0 \\
& \Rightarrow 2 y^2-7 y-15=0 \\
& \Rightarrow 2 y^2-10 y+3 y-15=0 \\
& 2 y(y-5)+3(y-5)=0 \\
& \Rightarrow(y-5)(2 y+3)=0
\end{aligned}
$
Either y - $5=0$,
then $y=5$
or
$
2 y+3=0 \text {, }
$
then $2 y=-3$
$
\Rightarrow y =\frac{-3}{2}
$
Hence $y=\frac{-3}{2}, 5$.
View full question & answer→Question 543 Marks
Solve the following equation by factorization$\frac{1}{7}(3 x-5)^2=28$
Answer$
\begin{aligned}
& \frac{1}{7}(3 x-5)^2=28 \\
& (3 x-5)^2=28 \times 7 \\
& \Rightarrow 9 x^2-30 x+25=196 \\
& \Rightarrow 9 x^2-30 x+25-196=0 \\
& \Rightarrow 9 x^2-30 x-171=0 \\
& \Rightarrow 3 x^2-10 x-57=0 \\
& \Rightarrow 3 x^2-19 x+9 x-57=0 \\
& \Rightarrow x(3 x-19)+3(3 x-19)=0 \\
& \Rightarrow(3 x-19)(x+3)=0
\end{aligned}
$
Either $3 x-19=0$,
then $3 x=19$
$
\Rightarrow x =\frac{19}{3}
$
or
$
x+3=0 \text {, }
$
then $x=-3$
Hence $x=\frac{19}{3},-3$.
View full question & answer→Question 553 Marks
If $x = p$ is a solution of the equation $x(2x + 5) = 3,$ then find the value of $p.$
AnswerGiven, $x=p$ and $x(2 x+5)=3$
Substituting the value of $p$, we get
$p(2 p+5)=3$
$ \Rightarrow 2 p^2+5 p-3=0$
$ \Rightarrow 2 p^2+6 p-p-3=0 \ldots
\because 2 \times(-3)=-6$
$\therefore-6=6 \times(-1)$
$5=6-1$
$ \Rightarrow 2 p(p+3)-1(p+3)=9$
$ \Rightarrow(p+3)(2 p-1)=0$
$ \text { Either } p+3=0,$
$ \text { then } p=-3$
$ \text { or }$
$ 2 p-1=0$
$ \text { then } 2 p=1$
$ \Rightarrow p=\frac{1}{2}$
$ \therefore p=\frac{1}{2},-3 .$
View full question & answer→Question 563 Marks
Solve the following equation by factorization
$
\sqrt{x(x-7)}=3 \sqrt{2}
$
Answer$\sqrt{x(x-7)}=3 \sqrt{2}$
Squaring both sides,
$
\begin{aligned}
& x(x-7)=9 x 2 \\
& \Rightarrow x^2-7 x=18 \\
& \Rightarrow x^2-7 x-18=0 \\
& \Rightarrow x^2-9 x+2 x-18=0 \\
& \Rightarrow x(x-9)+2(x-9)=0 \\
& \Rightarrow(x-9)(x+2)=0
\end{aligned}
$
Either $x-9=0$,
then $x=9$
or
$x+2=0$,
then $x =-2$
$
\therefore x =9_i-2
$
check
(i) If $x=9$, then
L.H.S.
$=\sqrt{x(x-7)}$
$=\sqrt{9(9-7)}$
$=\sqrt{9 \times 2}$
$=\sqrt{18}$
$=\sqrt{9 \times 2}$
$=3 \sqrt{2}$
$=$ R.H.S.
$x=9$ is a root
(ii) If $x=-2$, then
L.H.S.
$
\begin{aligned}
& =\sqrt{x(x-7)} \\
& =\sqrt{-2(-2-7)} \\
& =\sqrt{-2 \times-9} \\
& =\sqrt{18} \\
& =\sqrt{9 \times 2} \\
& =3 \sqrt{2} \\
& =\text { R.H.S. }
\end{aligned}
$
$\therefore x =-2$ is also its root
Hence $x=9,-2$.
View full question & answer→Question 573 Marks
Solve the following equation by factorization$\sqrt{3 x+4}=x$
Answer$
\sqrt{3 x+4}=x
$Squaring on both sides
$
\begin{aligned}
& 3 x+4=x^2 \\
& \Rightarrow x^2-3 x-4=0 \\
& \Rightarrow 4 x+x-4=0 \\
& \Rightarrow x(x-4)+1(x-4)=0 \\
& \Rightarrow(x-4)(x+1)=0
\end{aligned}
$
Either $x-4=0$,
then $x=4$
or
$x+1=0$,
then $x =-1$
$\therefore x=4,-1$
Check
(i) If $x =4$, then
L.H.S.
$=\sqrt{3 x+4}$
$=\sqrt{3 \times 4+4}$
$=\sqrt{12+4}$
$=\sqrt{16}$
$=4$
R.H.S.
$= x$
$=4$
$\therefore$ L.H.S. $=$ R.H.S.
Hence $x=4$ is its root
(ii) If $x=-1$, then
L.H.S.
$
\begin{aligned}
& =\sqrt{3 x(-1)+4} \\
& =\sqrt{-3+4} \\
& =\sqrt{1} \\
& =1
\end{aligned}
$
R.H.S.
$= x$
$=-1$
$\because$ L.H.S. $\neq$ R.H.S.
$\therefore x =-1$ is not its root,
Hence x = 4.
View full question & answer→Question 583 Marks
Solve the following equation by factorization $\frac{1}{x+6}+\frac{1}{x-10}=\frac{3}{x-4}$
Answer$
\begin{aligned}
& \frac{1}{x+6}+\frac{1}{x-10}=\frac{3}{x-4} \\
& \Rightarrow \frac{x-10+x+6}{(x+6)(x-10)}=\frac{3}{x-4} \\
& \Rightarrow \frac{2 x-4}{(x+6)(x-10)}=\frac{3}{x-4} \\
& \Rightarrow(2 x-4)(x-4)=3(x+6)(x-10) \\
& \Rightarrow 2 x^2-8 x-4 x+16=3\left(x^2-4 x-60\right) \\
& \Rightarrow 2 x^2-8 x-4 x+16=3 x^2-12 x-180 \\
& \Rightarrow 2 x^2-12 x+16-3 x^2+12 x+180=0 \\
& \Rightarrow-x^2+196=0 \\
& \Rightarrow x^2-196=0 \\
& \Rightarrow(x)^2-(14)^2=0 \\
& \Rightarrow(x+14)(x-14)=0 \\
& \text { Either } x+14=0 \\
& \text { then } x=-14
\end{aligned}
$
or
$
x-14=0 \text {, }
$
then $x =14$
$
\therefore x=14,-14 .
$
View full question & answer→Question 593 Marks
Solve the following equation by factorization$\frac{1}{2 a+b+2 x}=\frac{1}{2 a}+\frac{1}{b}+\frac{1}{2 x}$
Answer$
\begin{aligned}
& \frac{1}{2 a+b+2 x}=\frac{1}{2 a}+\frac{1}{b}+\frac{1}{2 x} \\
& \Rightarrow \frac{1}{2 a+b+2 x}=\frac{1}{2 x}+\frac{1}{2 a}+\frac{1}{b} \\
& \Rightarrow \frac{2 x-(2 a+b+2 x)}{(2 a+b+2 x) 2 x}=\frac{b+2 a}{2 a b} \\
& \Rightarrow \frac{-(2 a+b)}{(2 a+b+2 x) 2 x}=\frac{(2 a+b)}{2 a b} \\
& \Rightarrow \frac{-1}{(2 a+b+2 x) 2 x}=\frac{1}{2 a b} \\
& \Rightarrow-2 a b=(2 a+b+2 x) 2 x \\
& \Rightarrow 4 a x+2 x b+4 x^2=-2 a b \\
& \Rightarrow 4 x^2+2 b x+4 a x+2 a b=0 \\
& \Rightarrow 2 x(2 x+b)+2 a(2 x+b)=0 \\
& \Rightarrow(2 x+2 a)(2 x+b)=0 \\
& \Rightarrow 2 x+2 a=0 \text { or } 2 x + b =0 \\
& x=-a \text { or } x=-\frac{b}{2}
\end{aligned}
$
Hence, the roots of the given equation are
$-a$ and $-\frac{b}{2}$
View full question & answer→Question 603 Marks
Solve the following equation by factorization $\frac{a}{a x-1}+\frac{b}{b x-1}=a+b, a+b \neq 0, a b \neq 0$
Answer$
\begin{aligned}
& \frac{a}{a x-1}+\frac{b}{b x-1}=a+b, a+b \neq 0, a b \neq 0 \\
& \Rightarrow\left(\frac{a}{a x-1}-b\right)+\left(\frac{b}{b x-1}-a\right)=0 \\
& \Rightarrow \frac{a-a b x+b}{(a x-1)}+\frac{b-a b x+a}{(b x-1)}=0 \\
& \Rightarrow(a+b-a b x)\left[\frac{1}{a x-1}+\frac{1}{b x-1}\right]=0 \\
& \Rightarrow(a+b-a b x)\left[\frac{b x-1+a x-1}{(a x-1)(b x-1)}\right]=0 \\
& \Rightarrow \frac{(a+b-a b x)(a x+b x-2)}{(a x-1)(b x-1)}=0 \\
& \Rightarrow(a+b-a b x)(a x+b x-2)=0 \\
& \Rightarrow \text { Either a + b-abx =0, } \\
& \text { then a+b }=\text { abx } \\
& \text { x }=\frac{a+b}{a b}
\end{aligned}
$
or
$
a x+b x-2=0 \text {, }
$
then $x(a+b)=2$
$
x=\frac{2}{a+b}
$
Hence $x =\frac{a+b}{a b}, \frac{2}{a+b}$.
View full question & answer→Question 613 Marks
Solve the following equation by factorization $\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6}$
Answer$
\begin{aligned}
& \frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6} \\
& \frac{x+5-x+3}{(x-3)(x+5)}=\frac{1}{6} \\
& \Rightarrow \frac{8}{x^2+2 x-15}=\frac{1}{6} \\
& \Rightarrow x^2+2 x-15=48 \\
& \Rightarrow x^2+2 x-15-48=0 \\
& \Rightarrow x^2+2 x-63=0 \\
& \Rightarrow x^2+9 x-7 x-63=0 \\
& \Rightarrow x(x+9)-7(x+9)=0 \\
& \Rightarrow(x+9)(x-7)=0
\end{aligned}
$
Either $x+9=0$,
then $x=-9$
or
$x-7=0$,
then $x=7$
Hence $x=-9,7$.
View full question & answer→Question 623 Marks
Solve the following equation by factorization $\frac{x+1}{x-1}+\frac{x-2}{x+2}=3$
Answer$
\begin{aligned}
& \frac{x+1}{x-1}+\frac{x-2}{x+2}=3 \\
& \Rightarrow \frac{(x+1)(x+2)+(x-2)(x-1)}{(x-1)(x+2)}=3 \\
& \Rightarrow \frac{x^2+2 x+x+2+x^2-x-2 x+2}{x^2+2 x-x-2} \\
& \Rightarrow \frac{x^2+3 x+2+x^2-3 x+2}{x^2+x-2}=\frac{3}{1} \\
& \Rightarrow 2 x^2+4=3 x^2+3 x-6 \\
& \Rightarrow 2 x^2+4-3 x^2-3 x+6=0 \\
& \Rightarrow-x^2-3 x+10=0 \\
& \Rightarrow x^2+3 x-10=0 \\
& \Rightarrow x^2+5 x-2 x-10=0 \\
& \Rightarrow x(x+5)-2(x+5)=0 \\
& \Rightarrow(x+5)(x-2)=0
\end{aligned}
$
Either $x+5=0$,
then $x=-5$
or
$
x-2=0 \text {, }
$
then $x =2$
Hence $x=-5,2$.
View full question & answer→Question 633 Marks
Solve the following equation by factorization $\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15}$
Answer$
\begin{aligned}
& \frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15} \\
& \frac{x^2+x^2+2 x+1}{x(x+1)}=\frac{34}{15} \\
& \Rightarrow \frac{2 x^2+2 x+1}{x^2+x}=\frac{34}{15} \\
& \Rightarrow 30 x^2+30 x+15=34 x^2+34 x \\
& \Rightarrow 30 x^2+30 x+15-34 x^2-34 x=0 \\
& \Rightarrow-4 x^2-4 x+15=0 \\
& \Rightarrow 4 x^2+4 x-15=0 \\
& \Rightarrow 4 x^2+10 x-6 x-15=0 \\
& \Rightarrow 2 x(2 x+5)-3(2 x+5)=0 \\
& \Rightarrow(2 x+5)(2 x-3)=0
\end{aligned}
$
Either $2 x+5=0$,
then $2 x=-5$
$
\Rightarrow x=\frac{-5}{2}
$
or
$
2 x-3=0 \text {, }
$
then $2 x=3$
$
\Rightarrow x =\frac{3}{2}
$
Hence $x=\frac{-5}{2}, \frac{3}{2}$.
View full question & answer→Question 643 Marks
Solve the following equation by factorization $\frac{x}{x-1}+\frac{x-1}{x}=2 \frac{1}{2}$
Answer$\frac{x}{x-1}+\frac{x-1}{x}=2 \frac{1}{2}$
$\frac{x}{x-1}+\frac{x-1}{x}=\frac{5}{2}$
$\Rightarrow \frac{x^2+x^2-2 x+1}{x(x-1)}=\frac{5}{2}$
$\Rightarrow \frac{2 x^2-2 x+1}{x(x-1)}=\frac{5}{2}$
$\Rightarrow 4x^2 - 4x + 2 = 5x^2 - 5x$
$\Rightarrow 4x^2 - 4x + 2 - 5x^2 + 5x = 0$
$\Rightarrow -x^2 + x + 2 = 0$
$\Rightarrow x^2 - x - 2 = 0$
$\Rightarrow x^2 - 2x + x - 2 = 0$
$\Rightarrow x(x - 2) + 1(x - 2) = 0$
$\Rightarrow (x - 2) (x + 1) = 0$
Either $x - 2 = 0$,
then $x = 2$
or
$x + 1 = 0$,
then $x = -1$
Hence $x = 2, -1$.
View full question & answer→Question 653 Marks
Solve the following equation by factorization $\frac{8}{x+3}-\frac{3}{2-x}=2$
Answer$
\begin{aligned}
& \frac{8}{x+3}-\frac{3}{2-x}=2 \\
& \frac{16-8 x-3 x-9}{(x+3)(2-x)}=2 \\
& \Rightarrow \frac{-11 x+7}{2 x-x^2+6-3 x}=2 \\
& \Rightarrow-11 x+7=4 x-2 x^2+12-6 x \\
& \Rightarrow-11 x+7-4 x+2 x^2-12+6 x=0 \\
& \Rightarrow 2 x^2-9 x-5=0 \\
& \Rightarrow 2 x^2-10 x+x-5=0 \\
& \Rightarrow 2 x(x-5)+1(x-5)=0 \\
& \Rightarrow(x-5)(2 x+1)=0
\end{aligned}
$
Either $x-5=0$,
$
\text { then } x=5
$
or
$
2 x+1=0 \text {, }
$
then $2 x=-1$
$
\Rightarrow x =-\frac{1}{2}
$
Hence $x=5,-\frac{1}{2}$.
View full question & answer→Question 663 Marks
Solve the following equation by factorization$\frac{x+2}{x+3}=\frac{2 x-3}{3 x-7}$
Answer$
\begin{aligned}
& \frac{x+2}{x+3}=\frac{2 x-3}{3 x-7} \\
& (x+2)(3 x-7)=(2 x-3)(x+3) \\
& \Rightarrow 3 x^2-7 x+6 x-14=2 x^2+6 x-3 x-9 \\
& \Rightarrow 3 x^2-x-14=2 x^2+3 x-9 \\
& \Rightarrow 3 x^2-x-14-2 x^2-3 x+9=0 \\
& \Rightarrow x^2-4 x-5=0 \\
& \Rightarrow x^2-5 x+x-5=0 \\
& x(x-5)+1(x-5)=0 \\
& \Rightarrow(x-5)(x+1)=0
\end{aligned}
$
Either $x-5=0$,
then $x=5$
or
$
x+1=0 \text {, }
$
then $x=-1$
Hence $x=5,-1$.
View full question & answer→Question 673 Marks
Solve the following equation by factorization$4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0$
Answer$
\begin{aligned}
& 4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0 \\
& \{4 \sqrt{3} \times(-2 \sqrt{3})=8 \times(-3)=-24\} \\
& 4 \sqrt{3} x^2+8 x-3 x-2 \sqrt{3}=0 \\
& \Rightarrow 4 x(\sqrt{3} x+2)-\sqrt{3}(\sqrt{3} x+2)=0 \\
& \Rightarrow(\sqrt{3} x+2)(4 x-\sqrt{3})=0
\end{aligned}
$
Either $\sqrt{3} x+2=0$,
then $\sqrt{3} x=-2$
$
\begin{aligned}
& \Rightarrow x =-\frac{2}{\sqrt{3}} \\
& \Rightarrow x =\frac{-2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\
& =\frac{-2 \sqrt{3}}{3}
\end{aligned}
$
or
$
4 x-\sqrt{3}=0 \text {, }
$
then $4 x=\sqrt{3}$
$
\Rightarrow x =\frac{\sqrt{3}}{4}
$
Hence $x=\frac{-2 \sqrt{3}}{3}, \frac{\sqrt{3}}{4}$.
View full question & answer→Question 683 Marks
Solve the following equation by factorization $\sqrt{3} x^2+10 x+7 \sqrt{3}=0$
Answer$
\begin{aligned}
& \sqrt{3} x^2+10 x+7 \sqrt{3}=0 \\
& {[\because \sqrt{3} \times 7 \sqrt{3}=7 \times 3=21]} \\
& \Rightarrow \sqrt{3} x(x+\sqrt{3}+7(x+\sqrt{3}=0 \\
& \Rightarrow(x+\sqrt{3})(\sqrt{3} x+7)=0
\end{aligned}
$
Either $x+\sqrt{3}=0$,
then $x=-\sqrt{3}$
or
$
\sqrt{3} x+7=0
$
then $\sqrt{3} x=-7$
$
\begin{aligned}
& \Rightarrow x =\frac{-7}{\sqrt{3}} \\
& \Rightarrow x =\frac{-7 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\
& =\frac{-7 \sqrt{3}}{3}
\end{aligned}
$
Hence $x =-\sqrt{3},-\frac{7 \sqrt{3}}{3}$.
View full question & answer→Question 693 Marks
If $\frac{2}{3}$ and -3 are the roots of the equation $p x^2+7 x+q=0$, find the values of $p$ and $q$.
Answer$\frac{2}{3}$ and -3 are the roots of the equation $p x^2+7 x+q=0$
Substituting the value of $x=$ and -3 respectively, we get
$
\begin{aligned}
& p\left(\frac{2}{3}\right)^2+7\left(\frac{2}{3}\right)+q=0 \\
& \Rightarrow \frac{4}{9} p+\frac{14}{3}+q=0 \\
& \Rightarrow 4 p+42+9 q=0 \\
& \Rightarrow 4 p+9 q=-42.....(1) \\
& \text { and } \\
& p(-3)^2+7(-3)+q=0 \\
& 9 p-21+q=0 \\
& \Rightarrow 9 p+q=21.....(2) \\
& q=21-9 p
\end{aligned}
$
Substituting the value of $q$ in (1)
$
\begin{aligned}
& 4 p+9(21-9 p)=-42 \\
& 4 p+189-81 p=-42 \\
& -77 p=-42-189=-231 \\
& p=\frac{-231}{-77}=3 \\
& \therefore q \\
& =21-9 \times 3 \\
& =21-27 \\
& =-6 \\
& \therefore p=3, q=-6 .
\end{aligned}
$
View full question & answer→Question 703 Marks
If $\frac{2}{3}$ is a solution of the equation $7 x^2+k x-3=0$, find the value of $k$.
Answer$
7 x^2+k x-3=0, x=\frac{2}{3}
$
$\because x=\frac{2}{3}$ is its solution
$
\begin{aligned}
& \because 7\left(\frac{2}{3}\right)^2+k\left(\frac{2}{3}\right)-3=0 \\
& \Rightarrow 7 \times \frac{4}{9}+\frac{2}{3} k-3=0 \\
& \Rightarrow \frac{28}{9}-3+\frac{2}{3} k=0 \\
& \Rightarrow \frac{2}{3} k=3-\frac{28}{9} \\
& \Rightarrow \frac{2}{3} k=\frac{27-28}{9} \\
& \Rightarrow \frac{2}{3} k=\frac{-1}{9} \\
& \Rightarrow k =\frac{-1}{9} \times \frac{3}{2} \\
& =-\frac{1}{6}
\end{aligned}
$
Hence $k=-\frac{1}{6}$.
View full question & answer→Question 713 Marks
In each of the following, determine whether the given numbers are roots of the given equations or not;$6 x^2-x-2=0 ; \frac{-1}{2}, \frac{2}{3}$
Answer$
6 x^2-x-2=0 ; \frac{-1}{2}, \frac{2}{3}
$
If $x=\frac{-1}{2}$, then
$
\begin{aligned}
& =6\left(-\frac{1}{2}\right)^2-\left(-\frac{1}{2}\right)-2 \\
& =6 \times \frac{1}{4}+\frac{1}{2}-2 \\
& =\frac{3}{2}+\frac{1}{2}-2 \\
& =\frac{4}{2}-2=0
\end{aligned}
$
$\therefore x=\frac{-1}{2}$ is its root
If $x=\frac{2}{3}$, then
$
\begin{aligned}
& =6 \times \frac{4}{9}-\frac{2}{3}-2 \\
& =\frac{8}{3}-\frac{2}{3}-2 \\
& =\frac{6}{3}-2=0
\end{aligned}
$
$\therefore x =\frac{2}{3}$ is also its root.
Hence $\frac{-1}{2}, \frac{2}{3}$ are both its root.
View full question & answer→Question 723 Marks
In each of the following, determine whether the given numbers are roots of the given equations or not; $3 x^2-13 x-10=0 ; 5$, $\frac{-2}{3}$
Answer$
\begin{aligned}
& 3 x^2-13 x-10=0 ; 5, \frac{-2}{3} \\
& x=5 \\
& 3(5)^2-13 \times 5-10 \\
& =75-65-10 \\
& =75-75 \\
& =0
\end{aligned}
$
$\therefore x =5$ is its root
if $x=\frac{-2}{3}$, then
$
\begin{aligned}
& 3\left(-\frac{2}{3}\right)^2-13 \times \frac{-2}{3}-10 \\
& =\frac{3 \times 4}{9}+\frac{26}{3}-10 \\
& =\frac{4}{3}+\frac{26}{3}-10 \\
& =\frac{30}{3}-10 \\
& =10-10 \\
& =0
\end{aligned}
$
$\therefore x=\frac{-2}{3}$ is also its root.
Hence both $5, \frac{-2}{3}$ are its roots.
View full question & answer→Question 733 Marks
Solve for x using the quadratic formula. Write your answer correct to two significant figures: $(x – 1)^2 – 3x + 4 = 0.$
Answer$
\begin{aligned}
& (x-1)^2-3 x+4=0 \\
& x^2-2 x+1-3 x+4=0 \\
& x^2-5 x+5=0
\end{aligned}
$
Here $a=1, b=-5$ and $c=5$
$
\begin{aligned}
& x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\
& x=\frac{-(5) \pm \sqrt{(-5)^2-4(1)(5)}}{2} \\
& =\frac{5 \pm \sqrt{25-20}}{2} \\
& =\frac{5 \pm \sqrt{5}}{2} \\
& =\frac{5+2.236}{2} \text { or } \frac{5-2.236}{2} \\
& =\frac{7.236}{2} \text { or } \frac{2.764}{2} \\
& =3.618 \text { or } 1.382 \\
& \therefore x=3.618 \approx 3.6 \text { or } 1.382 \approx 1.4
\end{aligned}
$
View full question & answer→Question 743 Marks
Solve the following equation by using formula :
$x^2 + (4 – 3a)x – 12a = 0$
Answer$
x^2+(4-3 a) x-12 a=0
$Here $a=1, b=4-3 a, c=-12 a$
$
\begin{aligned}
& \therefore D=b^2-4 a c \\
& =(4-3 a)^2-4 \times 1 \times(-12 a) \\
& =16-24 a+9 a^2+48 a \\
& =16+24 a+9 a^2=(4+3 a)
\end{aligned}
$
$
\begin{aligned}
& \therefore x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{-(4-3 a) \pm \sqrt{4+3 a^2}}{2 \times 1} \\
& =\frac{3 a-4 \pm 3 a+4}{2}
\end{aligned}
$
$\therefore x _1=\frac{3 a-4+3 a+4}{2}$
$
=\frac{6 a}{2}
$
$
=3 a
$
and
$
\begin{aligned}
& x_2=\frac{3 a-4-3 a-4}{2} \\
& =\frac{-8}{2} \\
& =-4
\end{aligned}
$
$\therefore$ Roots are $3 a,-4$.
View full question & answer→Question 753 Marks
Solve the following equation by using formula :
$
\frac{2}{x+2}-\frac{1}{x+1}=\frac{4}{x+4}-\frac{3}{x+3}
$
Answer$
\begin{aligned}
& \frac{2}{x+2}-\frac{1}{x+1}=\frac{4}{x+4}-\frac{3}{x+3} \\
& \frac{(x+2-x-2}{(x+2)(x+1)}=\frac{4 x+12-3 x-12}{(x+4)(x+3)} \\
& \Rightarrow \frac{x}{(x+2)(x+1)}=\frac{1}{(x+4)(x+3)} \\
& \left.\Rightarrow \frac{1}{(x+2)(x+1)}=\frac{1}{(x+4)(x+3)} \ldots \text { [Dividing by } x \text { if } x \neq 0\right] \\
& \Rightarrow \frac{1}{x^2+3 x+2}=\frac{1}{x^2+7 x+12} \\
& \Rightarrow x^2+7 x+12-x 2-3 x-2=0 \\
& \Rightarrow 4 x+10 \\
& \Rightarrow 2 x+5=0 \\
& \Rightarrow 2 x=-5 \\
& \Rightarrow x=\frac{-5}{2} \\
& \text { If } x=0, \text { then } \\
& \frac{0}{(x+2)(x+1)}=\frac{0}{(x+4)(x+3)}
\end{aligned}
$
Which is correct
Hence $x=0, \frac{-5}{2}$.
View full question & answer→Question 763 Marks
Solve the following equation by using formula :
$
\frac{2 x+5}{3 x+4}=\frac{x+1}{x+3}
$
Answer$\begin{aligned} & \frac{2 x+5}{3 x+4}=\frac{x+1}{x+3} \\ & (2 x+5)(x+3)=(x+1)(3 x+4) \\ & 2 x^2+6 x+5 x+15=3 x^2+4 x+3 x+4 \\ & \Rightarrow 3 x^2+7 x+4-2 x^2-11 x-15=0 \\ & \Rightarrow x^2-4 x-11=0 \\ & \text { Here } a=1, b=-4, c=-11 \\ & D=b^2-4 a c \\ & =(-4)^2-4 x 1 x(-11) \\ & =16+44 \\ & =60 \\ & \because x=\frac{-b \pm \sqrt{D}}{2 a} \\ & =\frac{-(-4) \pm \sqrt{60}}{2 \times 1} \\ & =\frac{4 \pm \sqrt{4 \times 15}}{2} \\ & =\frac{4 \pm 2 \sqrt{15}}{2} \\ & =2 \pm \sqrt{15} \\ & \therefore x=2+\sqrt{15}, 2-\sqrt{15} .\end{aligned}$
View full question & answer→Question 773 Marks
Solve the following equation by using formula :
$
x\left(3 x+\frac{1}{2}\right)=6
$
Answer$
\begin{aligned}
& x\left(3 x+\frac{1}{2}\right)=6 \\
& 3 x^2+\frac{x}{2}=6 \\
& \Rightarrow 6 x^2+x=12 \\
& \Rightarrow 6 x^2+x-12=0
\end{aligned}
$
Here $a=6, b=1, c=-12$
$
\begin{aligned}
& D=b^2-4 a c \\
& =(1)^2-4 \times 6 \times(-12) \\
& =1+288 \\
& =289
\end{aligned}
$
$
\begin{aligned}
& \because x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{-1 \pm \sqrt{289}}{2 \times 6} \\
& =\frac{-1 \pm 17}{12} \\
& \therefore x_1=\frac{-1+17}{12} \\
& =\frac{16}{12} \\
& =\frac{4}{3} \\
& x_2=\frac{-1-17}{12} \\
& =\frac{-18}{12} \\
& =-\frac{3}{2}
\end{aligned}
$
$\therefore x=\frac{4}{3}, \frac{3}{2}$.
View full question & answer→Question 783 Marks
Solve the following equation by factorisation :
$
\frac{6}{x}-\frac{2}{x-1}=\frac{1}{x-2}
$
Answer$
\begin{aligned}
& \frac{6}{x}-\frac{2}{x-1}=\frac{1}{x-2} \\
& \Rightarrow \frac{6 x-6-2 x}{x(x-1)}=\frac{1}{x-2} \\
& \Rightarrow \frac{4 x-6}{x^2-x}=\frac{1}{x-2} \\
& \Rightarrow(4 x-6)(x-2)=x 6-x \\
& \Rightarrow 4 x^2-8 x-6 x+12=x^2-x \\
& \Rightarrow 4 x^2-14 x+12-x^2+x=0 \\
& \Rightarrow 3 x^2-13 x+12=0 \\
& \Rightarrow 3 x^2-9 x-4 x+12=0 \\
& \Rightarrow 3 x(x-3)-4(x-3)=0 \\
& \Rightarrow(x-3)(3 x-4)=0
\end{aligned}
$
Either $x-3=0$,
then $x=3$
or
$
3 x-4=0 \text {, }
$
then $3 x =4$
$
\Rightarrow x =\frac{4}{3}
$
Hence $x=3, \frac{4}{3}$.
View full question & answer→Question 793 Marks
Forty years hence, Mr. Pratap’s age will be the square of what it was $32$ years ago. Find his present age.
AnswerLet Partap’s present age = x years
40 years hence his age = x + 40
and 32 years ago his age = x – 32
According to the condition
$x + 40 = (x - 32)^2$
$\Rightarrow x + 40 = x^2 – 64x + 1024$
$\Rightarrow x^2 – 64x + 1024 – x – 40 = 0$
$\Rightarrow x^2 – 65x + 1024 – x – 40 = 0$
$\Rightarrow x^2 – 65x + 984 = 0$
$\Rightarrow x^2 – 24x – 41x + 984 = 0$
$\Rightarrow x(x – 24) – 41(x – 24) = 0$
$\Rightarrow (x – 24)(x – 41) = 0$
EIther x – 24 = 0,
then x = 24
but it is not possible as it is less than 32
or
x – 41 = 0,
then x = 41
Hence present age = 41 years.
View full question & answer→Question 803 Marks
By selling an article for Rs. 21, a trader loses as much per cent as the cost price of the article. Find the cost price.
AnswerS.P. of an article $=$ Rs. 21
Let cost price $=$ Rs. $x$
Then loss $=x \%$
$
\begin{aligned}
& \therefore \text { S.P. }=\frac{\text { C.P. }(100-\text { loss } \%)}{100} \\
& 21=\frac{x(100-x)}{100} \\
& 2100=100 x-x^2 \\
& \Rightarrow x^2-100 x+2100=0 \\
& \Rightarrow x^2-30 x-70 x+2100=0 \\
& \Rightarrow x(x-30)-70(x-30)=0 \\
& \Rightarrow(x-30)(x-70)=0
\end{aligned}
$
Elther $x-30=0$,
then $x=30$
or
$
x-70=0 \text {, }
$
then $x =70$
$\therefore$ Cost price $=$ Rs. 30 or Rs. 70 .
View full question & answer→Question 813 Marks
The hypotenuse of a right-angled triangle is $1$ m less than twice the shortest side. If the third side is $1$ m more than the shortest side, find the sides of the triangle.
AnswerLet the length of shortest side $= x m$
Length of hypotenuse $= 2x – 1$
and third side $= x + 1$
Now according to the condition,
$(2x – 1)^2 = (x)^2 + (x + 1)^2$ ...(By Pythagorus Theorem)
$\Rightarrow 4x^2 – 4x + 1 = x^2 + x^2 + 2x + 1$
$\Rightarrow 4x^2 – 4x + 1 = 2x^2 - 2x – 1 = 0$
$\Rightarrow 2x^2 – 6x = 0$
$\Rightarrow x^2 – 3x = 0$
$\Rightarrow x(x – 3) = 0$ ...(Dividing by $2$)
Either $x = 0$,
but it is not possible
or
$x – 3 = 0$,
then $x = 3$
Shortest side $= 3m$
Hypotenuse $= 2 \times 3 – 1 = 6 – 1 – 5$
Third side $= x + 1 = 3 + 1 = 4$
Hence sides are $3, 4, 5$ (in m).
View full question & answer→Question 823 Marks
The length of a rectangular garden is $12 m$ more than its breadth. The numerical value of its area is equal to $4$ times the numerical value of its perimeter. Find the dimensions of the garden.
AnswerLet breadth $= x m$
then length $= (x + 12) m$
Area $= l \times b = x (x + 12) m^2$^
and perimeter
$= 2(l + b)$
$= 2(x + 12 + x)$
$= 2 (2x + 12) m$
According to the condition.
$x(x + 12) = 4 x 2(2x + 12)$
$\Rightarrow x^2 + 12x = 16x + 96$
$\Rightarrow x^2 + 12x - 16x - 96 = 0$
$\Rightarrow x^2 - 4x - 96 = 0$
$\Rightarrow x^2 - 12x + 8x - 96 = 0$
$\Rightarrow x(x - 12) + 8(x - 12) = 0$
$\Rightarrow (x - 12)(x + 8) = 0$
Either $x - 12 = 0,$
then $x = 12$
or
$x + 8 = 0,$
then $x = -8,$
but it is not possible as it is in negative.
$\therefore $ Breadth $= 12m$
and length $= 12 + 12 = 24m.$
View full question & answer→Question 833 Marks
Two squares have sides A cm and $(x + 4)$ cm. The sum of their areas is $656$ sq. cm.Express this as an algebraic equation and solve it to find the sides of the squares.
AnswerSide of first square $= x ~cm $.
and side of second square $= (x + 4) cm$
Now according to the condition,
$(x)^2 + (x + 4)^2 = 656$
$\Rightarrow x^2 – x^2 + 8x + 16 = 656$
$\Rightarrow 2x^2 + 8x + 16 – 656 = 0$
$\Rightarrow 2x^2 + 8x – 640 = 0$
$\Rightarrow x^2+ 4x – 320 = 0$ ...(Dividing by 2)
$\Rightarrow x^2 + 20x – 16x – 320 = 0$
$\Rightarrow x(x + 20) – 16(x + 20) = 0$
$\Rightarrow (x + 20)(x – 16) = 0$
EIther $x + 20 = 0$,
then $x = –20$,
but it not possible as it is in negative.
or
$x – 16 = 0$ then $x = 16$
Side of first square $= 16 cm$
and side of second square $= 16 + 4 – 4 = 20 cm$.
View full question & answer→Question 843 Marks
Divide 16 into two parts such that the twice the square of the larger part exceeds the square of the smaller part by $164$.
AnswerLet larger part = x
then smaller part = 16 – x
(∵ sum = 16)
According to the condition
$2x^2 – (116 – x)^2 = 164$
$\Rightarrow 2x^2 – (256 – 32x + x^2) = 164$
$\Rightarrow 2x^2 – 256 + 32x – x^2 = 164$
$\Rightarrow x^2 + 32x – 256 – 164 = 0$
$\Rightarrow x^2 + 32x – 420 = 0$
$\Rightarrow x^2 + 42x – 10x – 420 = 0$
$\Rightarrow x(x + 42) – 10(x + 42) = 0$
$\Rightarrow (x + 42)(x – 10) = 0$
Either $x + 42 = 0$,
then $x = –42$,
but it is not possible.
or
$x - 10 = 0$,
then $x = 10$
∴ Larger part = 10
and smaller part = 16 - 10 = 6.
View full question & answer→Question 853 Marks
Find the value(s) of k for which each of the following quadratic equation has equal roots: $3kx^2 = 4(kx – 1)$
Answer$
\begin{aligned}
& 3 k x^2=4(k x-1) \\
& \Rightarrow 3 k x^2=4 k x-4 \\
& \Rightarrow 3 k x^2-4 k x+4=0
\end{aligned}
$
Here $a=3 k, b=-4 k, c=4$
$
\begin{aligned}
& \therefore D=b^2-4 a c \\
& =(-4 k)^2-4 \times 3 k \times 4 \\
& =16 k^2-48 k
\end{aligned}
$
$\therefore$ Roots are equal
$
\begin{aligned}
& \therefore D=0 \\
& \Rightarrow 16 k ^2-48 k =0 \\
& \Rightarrow k ^2-3 k =0 \\
& \Rightarrow k ( k -3)=0
\end{aligned}
$
Either $k=0$
or
$
k -3=0
$
$\Rightarrow$ then $k =3$
$\therefore x=\frac{-b \pm \sqrt{ D }}{2 a}=\frac{-b}{2 a} \ldots(\because D =0)$
$4 k$
$2 \times 3 k$
$=\frac{4 \times 3}{2 \times 3 \times 3}$
$=\frac{12}{18}$
$=\frac{2^2}{3}$
$\therefore x=\frac{2}{3}, \frac{2}{3}$.
View full question & answer→Question 863 Marks
Find the values of k so that the quadratic equation $(4 – k) x^2 + 2 (k + 2) x + (8k + 1) = 0$ has equal roots.
Answer$(4 – k) x^2 + 2 (k + 2) x + (8k + 1) = 0$
Here $a = (4 – k), b = 2 (k + 2), c = 8k + 1$
$\therefore D = b^2 – 4ac$
$= [2(k + 2)]^2 – 4 x (4 – k)(8k + 1) = 0$
$= 4(k + 2)^2 - 4(32k + 4 – 8k^2 – k)$
$= 4(k^2 + 4k + 4) –4(32k + 4 – 8k^2 – k)$
$= 4k^2 + 16k + 16 - 128k – 16 + 32k^2 + 4k$
$= 36k^2 – 108k$
$= 36k(k – 3)$
$\because$ Roots are equal
$\therefore D = 0$
$\Rightarrow 36k(k – 3) = 0$
$\Rightarrow k(k – 3) = 0$
Either $k = 0$
or
$k – 3 = 0$,
then $k= 3$
$k = 0, 3$.
View full question & answer→