Question 14 Marks
Solve, using formula :$x^2+x-(a+2)(a+1)=0$
AnswerGiven quadratic equation is $x^2+x-(a+2)(a+1)=0$
Using quadratic formula,
$\Rightarrow x=\frac{-1 \pm \sqrt{1^2+4(a+2)(a+1)}}{2}$
$ \Rightarrow x=\frac{-1 \pm \sqrt{1+4\left(a^2+3 a+2\right)}}{2} $
$ \Rightarrow x=\frac{-1 \pm \sqrt{4 a^2+12 a+9}}{2} $
$ \Rightarrow x=\frac{-1 \pm \sqrt{2 a+3}^2}{2} $
$ \Rightarrow x=\frac{-1 \pm(2 a+3)}{2} $
$ \Rightarrow x=\frac{-1 \pm(2 a+3)}{2} \text { or } x=\frac{-1 \pm(2 a+3)}{2} $
$ \Rightarrow x=\frac{2 a+2}{2} \text { or } x=\frac{-2 a-4}{2}$
$\Rightarrow x=\frac{2(a+1)}{2} \text { or } x=\frac{2(-a-2)}{2} $
$\Rightarrow x = a +1 \text { or } x =- a -2=-( a +2)$
View full question & answer→Question 24 Marks
If m and n are roots of the equation $\frac{1}{x}-\frac{1}{x-2}=3$ where x ≠ 0 and x ≠ 2; find m × n.
AnswerGiven quadratic equation is $\frac{1}{x}-\frac{1}{x-2}=3$
$ \begin{aligned} & \Rightarrow x-2-x=3 x(x-2) \\ & \Rightarrow-2=3 x^2-6 x \\ & \Rightarrow 3 x^2-6 x+2=0 \\ & a=3, b=-6, c=2 \\ & D=b^2-4 a c \\ & =(-6)^2-4(3)(2) \\ & =36-24 \\ & =12 \end{aligned} $
$\begin{aligned} & \Rightarrow x=\frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2-4 \mathrm{ac}}}{2 \mathrm{a}} \\ & \Rightarrow x=\frac{-6 \pm \sqrt{12}}{2 \times 3} \\ & \Rightarrow x=\frac{6 \pm \sqrt{2 \times 2 \times 3}}{6} \\ & \Rightarrow x=\frac{6 \pm(2 \sqrt{3})}{6} \\ & \Rightarrow x=\frac{6+2 \sqrt{3}}{6} \text { or } x=\frac{6-2 \sqrt{3}}{6}\end{aligned}$
Since, $m$ and $n$ are roots of the equation, we have
$ \begin{aligned} & \Rightarrow \mathrm{m}=\frac{6+2 \sqrt{3}}{6} n=\frac{6-2 \sqrt{3}}{6} \\ & \Rightarrow \mathrm{m} \times \mathrm{n}=\frac{6+2 \sqrt{3}}{6} \times \frac{6-2 \sqrt{3}}{6} \\ & =\frac{(6)^2-(2 \sqrt{3})^2}{36} \\ & =\frac{36-4 \times 3}{36} \\ & =\frac{36-12}{36} \\ & =\frac{24}{36} \\ & =\frac{2}{3} \end{aligned} $
View full question & answer→Question 34 Marks
One root of the quadratic equation $8 x^2+m x+15=0$ is $\frac{3}{4}$ Find the value of m. Also, find the other root of the equation.
AnswerGiven quadratic equation is $8 x^2+m x+15=0 ....... (i)$
One of the roots of (i) is $\frac{3}{4}$ so it satisfies (i)
$\Rightarrow 8\left(\frac{3}{4}\right)^2+m\left(\frac{3}{4}\right)+15=0 $
$ \Rightarrow \frac{9}{2}+15+m\left(\frac{3}{4}\right)=0$
$\Rightarrow m\left(\frac{3}{4}\right)=-\frac{39}{2} $
$ \Rightarrow m=-26$
So, the equation (i) becomes $8 x^2-26 x+15=0$
$\Rightarrow 8 x^2-20 x-6 x+15=0 $
$ \Rightarrow 4 x(2 x-5)-3(2 x-5)=0$
$\Rightarrow(4 x-3)(2 x-5)=0$
$ \Rightarrow(4 x-3)=0 \text { or }(2 x-5)=0 $
$ \Rightarrow x=\frac{3}{4} \text { or } x=\frac{5}{2} $
$\Rightarrow x=\frac{3}{4}, \frac{5}{2}$
Hence, the other root is $\frac{5}{2}$
View full question & answer→Question 44 Marks
Solve: $2 x-3=\sqrt{2 x^2-2 x+21}$
AnswerGiven : $2 x-3=\sqrt{2 x^2-2 x+21}$
by squaring on both side
$(2 x-3)^2=2 x^2-2 x+21$
$ (2 x)^2+(3)^2-(2)(2 x)(3)=2 x^2-2 x+21 $
$ 4 x^2+9-12 x=2 x^2-2 x+21 $
$2 x^2=10 x+12$
$2 x^2-10 x-12=0$
by splitting the mid term
$2 x^2-12 x+2 x-12=0 $
$ 2 x(x-6)+(x-6)=0 $
$ (2 x+2)(x-6)=0$
here the values of $x$ are
$2x + 2 = 0$
$x = -1$
$x - 6 = 0$
$x = 6$
View full question & answer→Question 54 Marks
Solve $2\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)=11$
Answer$2\left(x^2+\frac{1}{x^2}\right)-\left(x+\frac{1}{x}\right)=11$
Let $x+\frac{1}{x}=y$
squaring on both side
$x^2+\frac{1}{x^2}=y^2-2$
Putting these values in the given equation
$2\left(y^2-2\right)-y=11 $
$\Rightarrow2 y^{\wedge} 2-4- y -11=0^`$
$ \Rightarrow 2 y^2-y-15=0$
$ \Rightarrow 2 y^2-6 y+5 y-15=0 $
$ \Rightarrow(y-3)(2 y+5)=0$
If $y - 3 = 0$ or $2y + 5 = 0$
then $y = 3$ or $y = (-5)/2$
$\Rightarrow x+\frac{1}{x}=3 \quad \text { or } x+\frac{1}{x}=\frac{-5}{2}$
$\Rightarrow \frac{x^2+1}{x}=3 \text { or } \frac{x^2+1}{x}=\frac{-5}{2}$
$ \Rightarrow x^2-3 x+1=0 \text { or } 2 x^2+5 x+2=0$
$\Rightarrow x=\frac{-3 \pm \sqrt{(-3)^2-4(1)(1)}}{2(1)}$ or $2 x^2+4 x+x+2=0$
$\Rightarrow x=\frac{-3 \pm \sqrt{5}}{2}$ or $2 x(x+2)+1(x+2)=0$
then $x =-2$ and $x=\frac{-1}{2}$
View full question & answer→Question 64 Marks
Solve $\left(x^2+\frac{1}{x^2}\right)-3\left(x-\frac{1}{x}\right)-2=0$
Answer$\left(x^2+\frac{1}{x^2}\right)-3\left(x-\frac{1}{x}\right)-2=0$
Let $x-\frac{1}{x}=\mathrm{y}$
squaring on both sides
squaring on both sides
$ \begin{aligned} & \left(x-\frac{1}{x}\right)^2=\mathrm{y}^2 \\ & x^2+\frac{1}{x^2}-2=y^2 \\ & \Rightarrow x^2+\frac{1}{x^2}=y^2+2 \\ & \mathrm{~m}^2+2-3(\mathrm{~m})-2=0 \\ & \mathrm{~m}^2-3 \mathrm{~m}=0 \\ & \mathrm{~m}(\mathrm{~m}-3)=0 \\ & \mathrm{~m}=0 \text { and } \mathrm{m}-3=0 \\ & x-\frac{1}{x}=0 \text { or } x-\frac{1}{x}=3 \\ & \frac{x^2-1}{x}=0 \text { or } \frac{x^2-1}{x}=3 \\ & =\mathrm{x}^2-1=0 \text { or } \mathrm{x}^2-1=3 \mathrm{x} \\ & \mathrm{x}^2=1 \text { or } \mathrm{x}^2-3 \mathrm{x}-1=0 \end{aligned} $
$\begin{aligned} & x=\sqrt{1} \text { or } x=\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-1)}}{2(1)} \\ & x= \pm 1 \text { or }=x=\frac{3 \pm \sqrt{13}}{2}\end{aligned}$
View full question & answer→Question 74 Marks
Solve $: \left(x^2+5 x+4\right)\left(x^2+5 x+6\right)=120$
Answer$\left(x^2+5 x+4\right)\left(x^2+5 x+6\right)=120$
Let $x^2+5 x=y$
then $(y + 4)(y + 6) = 120$
$\Rightarrow y^2+6 y+4 y+24-120=0$
$\Rightarrow y^2+10 y-96=0$
$\Rightarrow y^2+16 y-6 y-96=0$
$\Rightarrow y(y+16)-6(y+16)=0$
$\Rightarrow (y+16)(y-16)=0$
then $y = -16$ or $y = 6$
$\Rightarrow x^2+5 x+16=0 \text { or } x^2+5 x-6=0$
$\Rightarrow x=\frac{-5 \pm \sqrt{(5)^2-4(1)(16)}}{2(1)} \text { or } x^2+6 x-x-6=0$
$\Rightarrow x=\frac{-5 \pm \sqrt{-39}}{2} \text { or } x ( x +6)-1( x +6)=0$
$\Rightarrow x=\frac{-5 \pm \sqrt{-39}}{2} \text { or } x ( x +6)-1( x +6)=0$
(reject) or $(x + 6)(x - 1) = 0$
then $x = -6$ and $x = 1$
View full question & answer→Question 84 Marks
Solve the equation $2 x-\frac{1}{x}=7$ Write your answer correct to two decimal places.
Answer$2 x-\frac{1}{x}=7$
$\Rightarrow \frac{2 x^2-1}{x}=7$
$ \Rightarrow 2 x^2-1=7 x$
$\Rightarrow 2 x^2-7 x-1=0$
Here $a = 2, b = -7$ and $c = -1$
$\therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-7) \pm \sqrt{(-7)^2-4(2)(-1)}}{2(2)} $
$ =\frac{7 \pm \sqrt{57}}{4}=\frac{7 \pm 7.55}{4}$
$=\frac{7+7.55}{4}$ and $\frac{7-7.55}{4}=3.64$ and $-0.14$
View full question & answer→Question 94 Marks
Solve:
$\left(\frac{3 x+1}{x+1}\right)+\left(\frac{x+1}{3 x+1}\right)=\frac{5}{2}$
Answer$\frac{3 x+1}{x+1}=y$
then $y+\frac{1}{y}=\frac{5}{2}$
$\Rightarrow \frac{y^2+1}{y}=\frac{5}{2} $
$ \Rightarrow 2 y^2+2=5 y$
$ \Rightarrow 2 y^2+5 y+2=0 $
$ \Rightarrow 2 y^2-4 y-y+2=0$
$ \Rightarrow 2 y(y-2)-1(y-2)=0 $
$ \Rightarrow(y-2)(2 y-1) 0$
If $y - 2 = 0$ or $2y - 1 = 0$
then $y =2$ or $y=\frac{1}{2}$
$\Rightarrow \frac{3 x+1}{x+1}=2$ or $\frac{3 x+1}{x+1}=\frac{1}{2}$
$\Rightarrow 3 x+1=2 x+2$ or $6 x+2=x+1$
$\Rightarrow x=1$ or $5 x=-1$
$\Rightarrow x=1$ or $x=\frac{-1}{5}$
View full question & answer→Question 104 Marks
Solve:
$\left(\frac{2 x-3}{x-1}\right)-4\left(\frac{x-1}{2 x-3}\right)=3$
Answer$\left(\frac{2 x-3}{x-1}\right)-4\left(\frac{x-1}{2 x-3}\right)=3$
Let $\frac{2 x -3}{ x -1}= y$
Then $y -\frac{4}{ y }=3$
$\Rightarrow \frac{y^2-4}{y}=3$
$\Rightarrow y^2 - 4 = 3y$
$\Rightarrow y^2 - 3y - 4 = 0$
$\Rightarrow y^2 - 4y + y - 4 = 0$
$\Rightarrow y( y -4) +1 ( y -4) = 0$
$\Rightarrow ( y -4) (y + 1) = 0$
$If y -4 = 0$ or $y + 1 = 0$
Then $y = 4$ or $y = -1$
$\Rightarrow \frac{2 x-3}{x-1}=4$ or $\frac{2 x-3}{x-1}=-1$
$\Rightarrow 4x -4 = 2x -3$ or $2x -3 = -x +1$
$\Rightarrow 2x = 1$ or $3x = 4$
$\Rightarrow x =\frac{1}{2}$ or $x =\frac{4}{3}=1 \frac{1}{3}$
View full question & answer→Question 114 Marks
Solve:
$\sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}}=\frac{5}{2}$
Answer$\sqrt{\frac{x}{x-3}}+\sqrt{\frac{x-3}{x}}=\frac{5}{2}$
Let $\sqrt{\frac{x}{x-3}}=y$
Then $y +\frac{1}{ y }=\frac{5}{2}$
$\Rightarrow \frac{ y ^2+1}{ y }=\frac{5}{2}$
$\Rightarrow 2 y 2+2=5 y $
$ \Rightarrow 2 y 2-5 y+2=0$
$ \Rightarrow 2 y 2-4 y-y+2=0 $
$ \Rightarrow 2 y(y-2)-1(y-2)=0 $
$ \Rightarrow(y-2)(2 y-1)=0$
If $y-2=0$ or $2 y-1=0$ Then $y =2$ or $y =\frac{1}{2}$
$\Rightarrow \sqrt{\frac{x}{x-3}}=2$ or $\sqrt{\frac{x}{x-3}}=\frac{1}{2}$
$\Rightarrow \frac{x}{x-3}=4$ or $\frac{x}{x-3}=\frac{1}{4}$
$\Rightarrow x=4$ or $x=-1$
View full question & answer→Question 124 Marks
Solve :
$(x^2 - 3x)^2 - 16(x^2 - 3x) - 36 =0$
Answer$(x^2 - 3x)^2 - 16(x^2 - 3x) - 36 =0$ Let $x^2 - 3x = y$
then $y^2 - 16y - 36 = 0$
$\Rightarrow y^2 - 18y + 2y - 36 = 0$
$\Rightarrow y(y - 18) + 2(y - 18) = 0$
$\Rightarrow ( y - 18) (y + 2) = 0$
If $y -18=0$ or $y +2=0$
$\Rightarrow x^2 - 3x - 18 = 0$ or $x^2 - 3x + 2 = 0$
$\Rightarrow x^2 - 6x + 3x - 18 = 0 or x^2 -2x x + 2 = 0$
$\Rightarrow x( x -6) + 3(x - 6) = 0 or x( x -2) -1 (x -2) = 0$
$\Rightarrow (x -6) ( x + 3) = 0 or (x -2)(x -1) = 0$
If $x-6=0$ or $x+3=0$ or $x-2=0$ or $x-1=0$
then $x=6$ or $x=-3$ or $x=2$ or $x=1$
View full question & answer→Question 134 Marks
Solve : $(x^2 – x)^2 + 5(x^2 – x)+ 4=0$
Answer$\left(x^2-x\right)^2+5\left(x^2-x\right)+4=0$
Let $x^2-x=y$
Then $y^2+5 y+4=0$
$\Rightarrow y^2+4 y+y+4=0 $
$ \Rightarrow y(y+4)+1(y+4)=0$
$\Rightarrow(y+4)(y+1)=0$
if $y+4=0$ or $y+1=0$
$\Rightarrow x^2-x+4=0$ or $x^2-x+1=0$
$\Rightarrow x=\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(4)}}{2(1)}$ or $\frac{-(-1) \pm \sqrt{(-1)^2-4(1)(1)}}{2(1)}$
$\Rightarrow x=\frac{1 \pm \sqrt{-15}}{2}$ (reject) or $x=\frac{1 \pm \sqrt{-3}}{2}$ (reject)
$\therefore$ Given equation has no real solution
View full question & answer→Question 144 Marks
Solve the following equation for $x$ and give your answer correct to $3$ decimal places:
$x^2 - 16x +6= 0$
Answer$x^2-16 x+6=0$
Here a = 1, b = - 16 and c = 6
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-16) \pm \sqrt{(-16)^2-4(1)(6)}}{2(1)}$
$=\frac{16 \pm \sqrt{232}}{2}$
$=\frac{16 \pm 15.23]}{2}$
$x=\frac{16+15.231}{2}$ or $x=\frac{16-15.231}{2}$
$\therefore x=\frac{16+15.231}{2}$
x = 15.616
$\therefore x=\frac{16-15.231}{2}$
$x=0.384$
View full question & answer→Question 154 Marks
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :
$x^2-5 x-10=0$
Answer$x^2-5 x-10=0$
Here a = 1, b = -5 and c = -10
$\therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(5) \pm \sqrt{(-5)^2-4 \times 1 \times(-10)}}{2 \times 1}$
$=\frac{5 \pm \sqrt{25+40}}{2}$
$=\frac{5 \pm \sqrt{65}}{2}$
$=\frac{5 \pm 8.06}{2}$
$\therefore x=\frac{5+8.06}{2}$ or $x=\frac{5-8.06}{2}$
$\Rightarrow x=\frac{13.06}{2}$ or $x=-\frac{3.06}{2}$
$\Rightarrow x=6.53$ or $x=-1.53$
View full question & answer→Question 164 Marks
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places : $x^2 – 3x – 9 =0$
Answer$x^2-3 x-9=0$
Here a = 1, b = -3 and c = -9
$\therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-9)}}{2(1)}$
$=\frac{3 \pm \sqrt{45}}{2}$
$=\frac{3 \pm 6.70}{2}$
$=\frac{3+6.70}{2}$ and $\frac{3-6.70}{2}$
= 4.85 and -1.85
View full question & answer→Question 174 Marks
Solve the quadratic equation $x^2 - 3(x + 3) = 0$; Give your answer correct two significant figures
Answer$x^2-3(x+3)=0$
$\Rightarrow x^2-3 x-9=0$
Comparing with $a x^2+b y+c$ we get
$\therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} $
$ \Rightarrow x=\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-9)}}{2(1)} $.
$ \Rightarrow x=\frac{3 \pm \sqrt{9+36}}{2} $
$ \Rightarrow x=\frac{3 \pm \sqrt{45}}{2}$
$ \Rightarrow x=\frac{3 \pm \sqrt{9 \times 5}}{2}$
$ \Rightarrow x=\frac{3 \pm 3 \sqrt{5}}{2}$
$ \Rightarrow x=\frac{3+3 \sqrt{5}}{2} \text { or } x=\frac{3-3 \sqrt{5}}{2}$
$\Rightarrow x=\frac{3+3 \times 2.236}{2} \text { or } x=\frac{3-3 \times 2.236}{2}$
$\Rightarrow x=\frac{3+6.708}{2} \text { or } x=\frac{3-6.708}{2}$
$ \Rightarrow x=\frac{9.708}{2} \text { or } x=\frac{-3.708}{2}$
$ \Rightarrow x =4.85 \text { or } x =-1.85$
$ \Rightarrow x =4.9 \text { or } x =-1.9$
View full question & answer→Question 184 Marks
Solve each of the following equations using the formula:
$\frac{x-1}{x-2}+\frac{x-3}{x-4}=3 \frac{1}{3}$
Answer$\frac{x-1}{x-2}+\frac{x-3}{x-4}=3 \frac{1}{3}$
$\Rightarrow \frac{(x-1)(x-4)+(x-2)(x-3)}{(x-2)(x-4)}=\frac{10}{3} $
$ \Rightarrow \frac{x^2-4 x-x+4+x^2-3 x-2 x+6}{x^2-4 x-2 x+8}=\frac{10}{3}$
$\Rightarrow \frac{12 x^2-10+10}{x^2-6 x+8}=\frac{10}{3}$
$\Rightarrow 10 x^2-60 x+80=6 x^2-30 x+30 $
$\Rightarrow 4 x^2-30 x+50=0 $
$\Rightarrow 2 x^2-15 x+25=0$
Here $a = 2, b = − 15$ and $c = 25$
Then $x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$\frac{=-(-15) \pm \sqrt{(-15)^2-4(2)(25)}}{2(2)}$
$=\frac{15 \pm \sqrt{25}}{4}=\frac{15 \pm 5}{4}$
$=\frac{15+5}{4}$ and $\frac{15-5}{4}=5$ and $\frac{5}{2}$
View full question & answer→Question 194 Marks
Solve the following equation using the formula:
$\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=8 \frac{1}{3}$
Answer$\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=8 \frac{1}{3}$
$\Rightarrow \frac{2 x(x-3)+(x-4)(2 x-5)}{(x-4)(x-3)}=\frac{25}{3}$
$\Rightarrow \frac{2 x^2-6 x+2 x^2-5 x-8 x+20}{x^2-3 x-4 x+12}=\frac{25}{3}$
$\Rightarrow \frac{4 x^2-19 x+20}{x^2-7 x+12}=\frac{25}{3}$
$\Rightarrow 25 x^2-175 x+300=12 x^2-57 x+60 $
$\Rightarrow 13 x^2-118 x+240=0$
Here $a = 13, b = − 118$ and $c = 240$
Then $x =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-118) \pm \sqrt{(-118)^2-4(13)(240)}}{2(13)}$
$=\frac{118 \pm \sqrt{1444}}{26}$
$=\frac{118 \pm 38}{26}$
$x=\frac{118+38}{26}$ or $x=\frac{118-38}{26}$
$\therefore x=\frac{118+38}{26}$
$\Rightarrow x =6$
$ x=\frac{118-38}{26} $
$ \therefore x =\left(\frac{ 4 0 }{ 1 3 }\right)$
View full question & answer→Question 204 Marks
If -1 and 3 are the roots of $x^2+px+q=0$
then find the values of $p$ and $q$
AnswerSince $-1$ is a root of $x2 + px + q = 0,$ we have $(-1)^2 + p(-1) + q = 0$
$\Rightarrow 1 - p + q = 0$
$\Rightarrow -p + q = -1.......(i)$
Also, 3 is a root of $x^2 + px + q = 0$
$\Rightarrow (3)^2 + p(3) + q = 0$
$\Rightarrow 9 + 3p + q = 0$
$\Rightarrow 3p + q = -9.......(ii)$
Subtracting equation $(ii)$ from $(i)$, we get
$-4p = 8$
$\Rightarrow p = -2$
$\Rightarrow -(-2) + q = -1 ...........[From (i)]$
$\Rightarrow 2 + q = -1$
$\Rightarrow q = -3$
Hence, $p = -2$ and $q = -3$
View full question & answer→Question 214 Marks
Solve : $\left(\frac{1200}{x}+2\right)(x-10)-1200=60$
Answer$\left(\frac{1200}{x}+2\right)(x-10)-1200=60$
$\Rightarrow 2\left(\frac{600}{x}+1\right)(x-10)=1260 $
$ \Rightarrow\left(\frac{600}{x}+1\right)(x-10)=630$
$\Rightarrow\left(\frac{600+x}{x}\right)(x-10)=630$
$\Rightarrow 600 x-6000+x^2-10 x=630 x $
$ \Rightarrow x^2-40 x-6000=0 $
$ \Rightarrow x^2-100 x+60 x-6000=0 $
$\Rightarrow x(x-100)+60(x-100)=0$
$\Rightarrow(x-100)(x+60)=0$
$ \Rightarrow x-100=0 \text { or } x+60=0$
$ \Rightarrow x=100 \text { or } x=-60$
View full question & answer→Question 224 Marks
Given that 2 is a root of the equation 3x² – p(x + 1) = 0 and that the equation px² – qx + 9 = 0 has equal roots, find the values of p and q.
AnswerSince $2$ is a root of the equation $3 x^2-p(x+1)=0$
$\Rightarrow 3(2)^2-p(2+1)=0$
$\Rightarrow 3 \times 4-3 p=0$
$\Rightarrow 12-3 p=0$
$\Rightarrow3 p=12$
$\Rightarrow p=4$
Now the other equation become $4 x^2-q x+9=0$ Here $a=4, b=-q$ and $c=9$
Since then root are equal we have
$b^2-4 a c=0$
$\Rightarrow(-q)^2-4 \times 4 \times 9=0$
$\Rightarrow q^2-144=0$
$\Rightarrow q^2=144$
$\Rightarrow q=12$
Hence $p = 4$ and $q = 12 $
View full question & answer→Question 234 Marks
If quadratic equation $x^2 - (m + 1)x + 6 = 0$ has one root as $x = 3$; find the value of m and the root of the equation.
Answer$x^2 - (m + 1)x + 6 = 0$Put x = 3 in the given equation
$(3)^2 - (m + 1)(3) + 6 = 0$
$\Rightarrow 9 - 3m - 3 + 6 = 0$
$\Rightarrow -3m = 12$
$\Rightarrow m = 4$
Put this value of m in the given equation we get
$x^2 – 5x + 6 = 0$
$\Rightarrow x^2 – 3x – 2x + 6 = 0$
$\Rightarrow x(x – 3) – 2(x – 3) = 0$
$\Rightarrow (x – 3) (x – 2) = 0$
$If x – 3 = 0 Or x – 2 = 0$
Then $x = 3$ Or $x = 2$
$\therefore 2$ is the other root of the given equation
View full question & answer→Question 244 Marks
Solve equation using factorisation method:
$\frac{5}{x-2}-\frac{3}{x+6}=\frac{4}{x}$
Answer$\frac{5}{x-2}-\frac{3}{x+6}=\frac{4}{x}$
$\Rightarrow \frac{5( x +6)-3( x -2)}{( x -2)( x +6)}=\frac{4}{ x } $
$ \Rightarrow \frac{5 x +30-3 x +6}{ x ^2+6 x -2 x -12}=\frac{4}{ x }$
$ \Rightarrow \frac{2 x +36}{ x ^2+4 x -12}=\frac{4}{ x }$
$\Rightarrow 4 x^2+16 x-48=2 x^2+36 x$
$ \Rightarrow 2 x^2-20 x-48=0 $
$ \Rightarrow x^2-10 x-24=0 $
$ \Rightarrow x^2-12 x+2 x-24=0 $
$ \Rightarrow x(x-12)+2(x-12)=0 $
$ \Rightarrow(x-12)(x+2)=0$
If $x - 12 = 0$ or $x + 2 = 0$
then $x = 12$ or $x = -2$
View full question & answer→Question 254 Marks
Solve equation using factorisation method:
$\frac{4}{x+2}-\frac{1}{x+3}=\frac{4}{2 x+1}$
Answer$\frac{4}{x+2}-\frac{1}{x+3}=\frac{4}{2 x+1}$
$\Rightarrow \frac{4(x+3)-1(x+2)}{(x+2)(x+3)}=\frac{4}{2 x+1} $
$\Rightarrow \frac{4 x+12-x-2}{x^2+2 x+3 x+6}=\frac{4}{2 x+1} $
$\Rightarrow \frac{3 x+10}{x^2+5 x+6}=\frac{4}{2 x+1}$
$\Rightarrow(3 x+10)(2 x+1)=4\left(x^2+5 x+6\right)$
$\Rightarrow 6 x^2+3 x+20 x+10=4 x^2+20 x+24$
$\Rightarrow 2 x^2+3 x-14=0$
$\Rightarrow 2 x^2+7 x-4 x-14=0$
$\Rightarrow 2 x ^2+7 x -4 x -14=0$
$\Rightarrow x(2 x+7)-2(2 x+7)=0$
$\Rightarrow(2 x+7)(x-2)=0$
If $2 x+7=0$ or $x-2=0$
then $x=\frac{-7}{2}$ or $x=2$
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