Question 13 Marks
Find the value of m for which the equation $(m+4) x^2+(m+1) x+1=0$ has real and equal roots.
AnswerGiven quadratic equation is $(m+4) x^2+(m+1) x+1=0$
The quadratic equation has real and equal roots if its discriminant is zero.
$\Rightarrow D=b^2-4 a c=0$
$ \Rightarrow(m+1)^2-4(m+4)(1)=0 $
$ \Rightarrow m^2+2 m+1-4 m-16=0 $
$ \Rightarrow m^2-2 m-15=0 $
$ \Rightarrow m^2-5 m+3 m-15=0 $
$ \Rightarrow m(m-5)+3(m-5)=0 $
$ \Rightarrow(m-5)(m+3)=0 $
$ \Rightarrow m=5 \text { or } m=-3$
View full question & answer→Question 23 Marks
Solve the quadratic equation $8 x^2-14 x+3=0$
(i) When $x \in I$ (integers)
(ii) When $x \in Q$ (rational numbers)
AnswerGiven quadratic equation is $8 x^2-14 x+3=0$
$\Rightarrow 8 x^2-12 x-2 x+3=0$
$ \Rightarrow 4 x(2 x-3)-(2 x-3)=0$
$ \Rightarrow(2 x-3)(4 x-1)=0$
$ \Rightarrow(2 x-3)=0 \text { or }(4 x-1)=0$
$ \Rightarrow x=\frac{3}{2} \text { or } x=\frac{1}{4}$
(i) When $x \in I$ the equation $8 x^2-14 x+3=0$ has no roots
(ii) When $x \in Q$ the roots of $8 x^2-14 x+3=0$ are
$x=\frac{3}{2}$ or $x=\frac{1}{4}$
View full question & answer→Question 33 Marks
Find the solution of the equation $2 x^2-m x-25 n=0$ if $m + 5 = 0$ and $n - 1 = 0$
AnswerGiven quadratic equation is $2 x^2-m x-25 n=0 \ldots \ldots$ (i)
Also, given $m + 5 = 0$ and $n - 1 = 0$
$\Rightarrow m=-5$ and $n=1$
So, the equation (i) becomes
$2 x^2+5 x-25=0$
$ \Rightarrow 2 x^2+10 x-5 x-25=0$
$ \Rightarrow 2 x(x+5)-5(x+5)=0 $
$ \Rightarrow(x+5)(2 x-5)=0 $
$ \Rightarrow x=-5 \cdot \frac{5}{2}$
Hence, the solution of given quadratic equation are $-5$ and $\frac{5}{2}$
View full question & answer→Question 43 Marks
Solve $(x-10)\left(\frac{1200}{x}+2\right)=1260$ and $x<0$.
AnswerGiven quadratic equation is $(x-10)\left(\frac{1200}{x}+2\right)=1260$
$\Rightarrow(x-10)\left(\frac{1200}{x}+2\right)=1260$
$ \Rightarrow(x-10)(1200+2 x)=1260 x $
$ \Rightarrow 1200 x+2 x^2-12000-20 x+1260 x $
$ \Rightarrow 2 x^2-12000-8 x=0 $
$ \Rightarrow x^2-40 x-6000=0 $
$ \Rightarrow x^2-100 x+60 x-6000=0 $
$ \Rightarrow(x=100)(x+60)=0 $
$ \Rightarrow x =100 \text { or } x =-60$
But as $x < 0,$ so x can't be positive.
Hence, $x = -60$
View full question & answer→Question 53 Marks
Solve : $\frac{1}{18-x}-\frac{1}{18+x}=\frac{1}{24}$ and $x > 0$
Answer(i) Given quadratic equation is
$\frac{1}{18-x}-\frac{1}{18+x}=\frac{1}{24}$
$\Rightarrow \frac{(18+x)-(18-x)}{(18+x)(18-x)}=\frac{1}{24}$
$\Rightarrow \frac{2 x}{18^2-x^2}=\frac{1}{24}$
$\Rightarrow 48 x+324-x^2$
$\Rightarrow x^2+48 x-324=0$
$\Rightarrow x^2+54 x-6 x-324=0$
$\Rightarrow x ( x +54)-6( x +54)=0$
$\Rightarrow( x +54)( x -6)=0$
$\Rightarrow x =-54 \text { or } x =6$
But as $x > 0,$ so $x$ can't be negative.
Hence, $x = 6$
View full question & answer→Question 63 Marks
Solve $3 x^2-2 \sqrt{6} x+2=0$
AnswerGiven : $3 x^2-2 \sqrt{6} x+2=0$
$\Rightarrow 3 x^2-\sqrt{6} x-\sqrt{6} x+2=0 $
$\Rightarrow \sqrt{3} x(\sqrt{3} x-\sqrt{2})-\sqrt{2}(\sqrt{3} x-\sqrt{2})=0$
$\Rightarrow(\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})=0 $
$ \Rightarrow(\sqrt{3} x-\sqrt{2})=0 \text { or }(\sqrt{3} x-\sqrt{2})=0$
$ \Rightarrow x=\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}$
View full question & answer→Question 73 Marks
Find, using quadratic formula, the roots of the following quadratic equations, if they exist
$3 x^2-5 x+2=0$
AnswerGiven quadratic equation is
$3 x^2-5 x+2=0$
$D=b^2-4 a c={ }^`(-5)^{\wedge} 2-4(3)(2)=25-24=1$
Since D > 0, the roots of the given quadratic equation are real and distinct. Using quadratic formula, we have
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$\Rightarrow x=\frac{5 \pm \sqrt{(-5)^2-4(3)(2)}}{2(3)}$
$ \Rightarrow x=\frac{5 \pm \sqrt{25-24}}{6}$
$\Rightarrow x=\frac{5 \pm 1}{6} $
$ \Rightarrow x=\frac{5+1}{6} \text { or } x=\frac{5-1}{6} $
$ \Rightarrow=\frac{6}{6} \text { or } x=\frac{4}{6} $
$ \Rightarrow x=1 \text { or } x=\frac{2}{3}$
View full question & answer→Question 83 Marks
Find the values of m for which equation $3 x^2+m x+2=0$ has equal roots. Also, find the roots of the given equation.
AnswerGiven quadratic equation is $3 x^2+m x+2=0 \ldots (i)$
The quadratic equation has equal roots if its discriminant is zero
$\Rightarrow D=b^2-4 a c=0 $
$ \Rightarrow m^2-4(2)(3)=0$
$ \Rightarrow m^2=-24$
When $m=2 \sqrt{6}$, equation $(i)$ becomes
$3 x^2+2 \sqrt{6} x+2=0$
$\Rightarrow(\sqrt{3} x+\sqrt{2})^2=0 $
$ \Rightarrow x=\frac{\sqrt{2}}{\sqrt{3}}=-\frac{\sqrt{2}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=-\frac{\sqrt{6}}{3}$
When $m=-2 \sqrt{6}$, equation $(i)$ becomes
$x=-\frac{\sqrt{6}}{3}, \frac{\sqrt{6}}{3}$
View full question & answer→Question 93 Marks
Solve: $x^4 – 2x^2 – 3 = 0.$
Answer$x^4-2 x^2-3=0$
$\Rightarrow x^4-3 x^2+x^2-3=0 $
$ \Rightarrow x^2\left(x^2-3\right)+1\left(x^2-3\right)=0 $
$ \Rightarrow\left(x^2-3\right)\left(x^2+1\right)=0$
If $x^2-3=0$ or $x^2+1=0$
then $x^2=3$ or $x^2=-1$ (reject)
$\Rightarrow x= \pm \sqrt{3}$
View full question & answer→Question 103 Marks
Solve: $2x^4 – 5x^2 + 3 = 0$
Answer$2 x^4-5 x^2+3=0$
$\Rightarrow 2 x^4-3 x^2-2 x^2+3=0 $
$ \Rightarrow x^2\left(2 x^2-3\right)-1\left(2 x^2-3\right)=0$
$ \Rightarrow\left(2 x^2-3\right)\left(x^2-1\right)=0$
If $2 x^2-3=0$ or $x^2-1=0$
then $x^2=\frac{3}{2}$ or $x^2=1$
$\Rightarrow x= \pm \sqrt{\frac{3}{2}}$ or $x = \pm 1$
View full question & answer→Question 113 Marks
Solve $x+\frac{4}{x}=-4 ; x \neq 0$
Answer$x+\frac{4}{x}=-4$
$\Rightarrow \frac{x^2-4}{x}=-4 $
$ \Rightarrow x^2+4=-4 x $
$\Rightarrow x^2+4 x+4=0 $
$ \Rightarrow(x+2)^2=0 $
$ \Rightarrow(x+2)=0 $
$ \Rightarrow x=-2$
View full question & answer→Question 123 Marks
Solve $x^2-\frac{11}{4} x+\frac{15}{8}=0$
Answer$x^2-\frac{11}{4} x+\frac{15}{8}=0$
$\Rightarrow \frac{8 x^2-22 x+15}{8}=0$
$ \Rightarrow 8 x^2-22 x+15=0$
$ \Rightarrow 8 x^2-12 x-10 x+15=0$
$ \Rightarrow 4 x(2 x-3)-5(2 x-3)=0 $
$ \Rightarrow(2 x-3)(4 x-5)=0 $
$ \Rightarrow 2 x-3=0 \text { or } 4 x-5=0$
$ \Rightarrow x=\frac{3}{2} \text { or } x=\frac{5}{4}$
View full question & answer→Question 133 Marks
Without solving the following quadratic equation, find the value of $m$ for which the given equation has equation has real and equal roots.
$x^2+2(m-1) x+(m+5)=0$
AnswerGiven Quadratic Equation is
$x^2+2(m-1) x+(m+5)=0$
Here $a = 1,b = 2(m - 1)$ and $c = (m + 5)$
Discriminant isgiven by $D^2=b^2-4 a c$
For Real and equal roots, $D = 0$
$\Rightarrow b^2-4 a c=0$
$\Rightarrow[2(m-1)]^2-4(m+5)=0$
$\Rightarrow m^2+1-2 m-m-5=0$
$\Rightarrow m^2-3 m-4=0$
Factorising we get $(m + 1)(m - 4) = 0$
$\Rightarrow m=-1$ or $m=4$
View full question & answer→Question 143 Marks
Solve
$\frac{2 x}{x-3}+\frac{1}{2 x+3}+\frac{3 x+9}{(x-3)(2 x+3)}=0 ; x \neq 3, x \neq-\frac{3}{2}$
Answer$\frac{2 x}{x-3}+\frac{1}{2 x+3}+\frac{3 x+9}{(x-3)(2 x+3)}=0 ; x \neq 3, x \neq-\frac{3}{2}$
$\Rightarrow \frac{2 x(2 x+3)+1(x-3)+3 x+9}{(x-3)(2 x+3)}=0 $
$\Rightarrow 4 x^2+6 x+x-3+3 x+9=0$
$ \Rightarrow 4 x^2+10 x+6=0 $
$ \Rightarrow 4 x(x+1)+6(x+1)=0$
$ =(x+1)(4 x+6)=0 $
$ \Rightarrow x+1=0 \text { or } 4 x+6=0$
$ \Rightarrow x=-1 \text { or } x=\frac{-6}{4}=\frac{-3}{2} \text { (reject) }$
View full question & answer→Question 153 Marks
For each equation given below find the values of m so that the equation has equal roots. Also find the solution of equation
$x^2-(m+2) x+(m+5)=0$
Answer$x^2-(m+2) x+(m+5)=0$
Here $a = 1, b = -(m + 2)$ and $c = m + 5$
Given equation has equa;l roots then$ D = 0$
$\Rightarrow b^2-4 a c=0 $
$ \Rightarrow[-(m-2)]^2-4(1)(m+5)=0$
$ \Rightarrow m^2+4 m+4-4 m-20=0 $
$\Rightarrow m^2-16=0 $
$ \Rightarrow m^2=16 $
$ \Rightarrow m \pm 4$
Put value of m in given equation
$x^2-6 x+9=0 \text { or } x^2+2 x+1=0 $
$\Rightarrow(x-3)^2=0 \text { or }(x+1)^2=0$
$ \Rightarrow x-3=0 \text { or } x +1=0$
$\Rightarrow x =3 \text { or } x =-1$
View full question & answer→Question 163 Marks
For each equation given below find the values of $m$ so that the equation has equal roots. Also find the solution of equation $3 x^2+12 x+(m+7)=0$
Answer$3 x^2+12 x+(m+7)=0$
Here $a = 3, b = 12$ and $c= m + 7$
Given equation has equal roots then $D = 0$
$\Rightarrow b^2-4 a c=0$
$\Rightarrow(12)^2-4(3)(m+7)=0$
$\Rightarrow144-12 m-84=0$
$ \Rightarrow -12 m=-60 $
$ \Rightarrow m=5$
Put value of m in given equation
$3 x^2+12 x+12=0 $
$ \Rightarrow x^2+4 x+4=0 $
$ \Rightarrow(x+2)^2=0 $
$ \Rightarrow x=-2$
View full question & answer→Question 173 Marks
For each equation given below find the values of m so that the equation has equal roots. Also find the solution of equation
$(m-3) x^2-4 x+1=0$
Answer$(m-3) x^2-4 x+1=0$
Here $a = (m - 3), b = -4$ and $c = 1$
Given equation has equation roots then $D = 0$
$\Rightarrow b^2-4 a c=0$
$\Rightarrow(-4)^2-4(m-3)(1)=0$
$\Rightarrow 16-4 m+12=0$
$\Rightarrow-4 m=-28$
$\Rightarrow m =7$
Put value of $m$ in given equation
$4 x^2-4 x+1=0$
$\Rightarrow(2 x-1)^2=0$
$\Rightarrow 2 x-1=0$
$\Rightarrow x=\frac{1}{2}$
View full question & answer→Question 183 Marks
Solve $\frac{1}{x-1}-\frac{2}{x+2}=\frac{3}{x+3}-\frac{4}{x+4}$
Answer$\frac{1}{x-1}-\frac{2}{x+2}=\frac{3}{x+3}-\frac{4}{x+4}$
$\Rightarrow \frac{1(x+2)-2(x+1)}{(x+1)(x+2)}=\frac{3(x+4)-4(x+3)}{(x+3)(x+4)}$
$\Rightarrow \frac{-x}{x^2+3 x+2}=\frac{-x}{x^2+7 x+12} $
$ \Rightarrow-x\left[x^2+3 x+2=x^2+7 x+12\right] $
$ \Rightarrow-x[-4 x=10]$
$ \Rightarrow x=0 \text { and } x=\frac{-10}{4}=-2.5$
View full question & answer→Question 193 Marks
Solve $x(x + 1) + (x + 2)(x + 3) = 42$
Answer$x(x + 1) + (x + 2)(x + 3) = 42$
$\Rightarrow x^2+x+x^2+3 x+2 x+6-42=0$
$\Rightarrow 2 x^2+6 x-36=0$
$\Rightarrow 2 x^2+12 x-6 x-36=0$
$\Rightarrow 2 x(x+6)-6(x+6)=0$
$\Rightarrow (x+6)(2 x-6)=0$
if $x + 6 = 0$ or $2x - 6 = 0$
then $x = -6$ or $x = 3$
View full question & answer→Question 203 Marks
Solve $\frac{1}{p}+\frac{1}{q}+\frac{1}{x}=\frac{1}{x+p+q}$
Answer$\frac{1}{p}+\frac{1}{q}+\frac{1}{x}=\frac{1}{x+p+q}$
$\Rightarrow \frac{1}{p}+\frac{1}{q}+\frac{1}{x}-\frac{1}{x+p+1}=0 $
$ \Rightarrow\left((q+p)\left[\frac{1}{p q}+\frac{1}{x^2+p x+q x}\right]=0\right. $
$ \Rightarrow(p+q)\left[\frac{x^2+p x+q x+p q}{p q\left(x^2+p x+q x\right)}\right]=0$
$\Rightarrow x^2+p x+q x+p q=0$
$ \Rightarrow x(x+p)+q(x+p)=0$
$\Rightarrow(x+p)(x+q)=0 $
$ \Rightarrow x=-p \text { and } x=-q$
View full question & answer→Question 213 Marks
Solve $\left(\frac{x}{x+2}\right)^2-7\left(\frac{x}{x+2}\right)+12=0 ; x \neq-2$
Answer$\left(\frac{x}{x+2}\right)^2-7\left(\frac{x}{x+2}\right)+12=0 ; x \neq-2$
Let $\frac{x}{x+2}=y$
then $y^2-7 y+12=0$
$\Rightarrow y^2-4 y-3 y+12=0 $
$\Rightarrow y(y-4)-3(y-4)=0$
then $y = 4$ and $y = 3$
$\Rightarrow \frac{x}{x+2}=4 \text { and } \frac{x}{x+2}=3$
$ \Rightarrow \frac{x}{x+2}=4 \text { and } \frac{x}{x+2}=3$
$\Rightarrow4 x +8= x \text { and } 3 x +6= x $
$\Rightarrow x=\frac{-8}{3} \text { and } x=-3$
View full question & answer→Question 223 Marks
Solve of the following equations, giving answer upto two decimal places. $3x^2 – x – 7 =0$
Answer$3 x^2-x-7=0$
Here $a = 3, b = -1$ and $c = -7$
$x=\frac{-b \pm \sqrt{(-1)^2-4(3)(-7)}}{2(3)}$
$\Rightarrow x=\frac{1 \pm \sqrt{85}}{6}=\frac{1 \pm 9.22}{6}$
$\Rightarrow x=\frac{10.22}{6} \text { and } \frac{-8.22}{6}=1.70 \text { and }-1.37$
View full question & answer→Question 233 Marks
Solve the following equation and give your answer correct to $3$ significant figures: $5x^2 – 3x – 4 = 0$
AnswerConsider the given equation
Using quadratic formula, we have
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$\Rightarrow x=\frac{-(-3) \pm \sqrt{(-3)^2-4 \times 5 \times(-4)}}{2 \times 5}$
$\Rightarrow x=\frac{3 \pm \sqrt{9+80}}{2 \times 5}$
$\Rightarrow x=\frac{3 \pm \sqrt{89}}{10}$
$\Rightarrow x=1.243 \text { or } x=-0.643$
View full question & answer→Question 243 Marks
Solve : $x^4 - 10x^2 +9 =0$
Answer$x^4 - 10x^2 +9 =0$
$\Rightarrow x^4 - 9x^2 - x^2 + 9 = 0$
$\Rightarrow x^2(x^2 - 9) -1(x^2 - 9) = 0$
$\Rightarrow (x^2 - 9)(x^2 - 1) = 0$
if $x^2 - 9 = 0$ or$ x^2 - 1 = 0$
$\Rightarrow x^2 = 9$ or $x^2 = 1$
$\Rightarrow x = ± 3$ or $x = ± 1$
View full question & answer→Question 253 Marks
Solve: $x^4-2 x^2-3=0$
Answer$x^4-2 x^2-3=0$
$\Rightarrow x ^4-3 x ^2+ x 2-3=0$
$\Rightarrow x^2\left(x^2-3\right)+1\left(x^2-3\right)=0$
$\Rightarrow\left(x^2-3\right)\left(x^2+1\right)=0$
If $x^2-3=0$ or $x^2+1=0$
$\Rightarrow x^2=3 \text { or } x^2+1=0$
$\Rightarrow x= \pm \sqrt{3}$
View full question & answer→Question 263 Marks
Solve the following equations for x and give, in each case, your answer correct to 3 decimal places
$2x^2 + 11x + 4= 0$
Answer$2 x^2+11 x+4=0$
Here a= 2, b = 11 and c = 4
$\therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(11) \pm \sqrt{(11)^2-4(2)(4)}}{2(2)}$
$=\frac{-11 \pm \sqrt{(11)^2-4(2)(4)}}{2(2)}$
$=\frac{-11 \pm \sqrt{89}}{4}$
$=\frac{-11 \pm 9.433}{4}$
$=\frac{-11+9.433}{4}$ and $\frac{-11-9.433}{4}$
$=-0.392$ and -5.108
View full question & answer→Question 273 Marks
Solve the following equations for x and give, in each case, your answer correct to 3 decimal places
$3x^2 – 12x – 1 =0$
Answer$3 x^2-12 x-1=0$
Here $a= 3, b = -12$ and $c = -1$
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-12) \pm \sqrt{(-12)^2-4(3)(-1)}}{2(3)}$
$=\frac{12 \pm \sqrt{156}}{6}$
$=\frac{12 \pm 12.489}{6}$
$=4.082$ and -0.082
View full question & answer→Question 283 Marks
Solve the following equation by using quadratic formula and give your answer correct to 2 decimal places : $4x^2 – 5x – 3 = 0$
AnswerGiven equation $4 x^2-5 x-3=0$
Comparing with $a x^2+b x+c=0$, we have $a = 4, b = -5, c = -3$
$\because x =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-5) \pm \sqrt{(-5)^2-4 \times 4 \times(-3)}}{2 \times 4}$
$=\frac{5 \pm \sqrt{25+48}}{8}$
$=\frac{5 \pm \sqrt{73}}{8}$
$=\frac{5 \pm 8.544}{8}$
$=\frac{5+8.544}{8}$ or $\frac{5-8.544}{8}$
$=\frac{13.544}{8}$ or $\frac{-3.544}{8}$
$= 1.693$ or $-0.443$
$= 1.69$ or $-0.44. ...($correct to $2$ demical places$)$
View full question & answer→Question 293 Marks
Solve each of the following equations for $x$ and give, in each case, your answer correct to two decimal places :
$4 x+\frac{6}{x}+13=0$
Answer$4 x+\frac{6}{x}+13=0$
$\Rightarrow 4 x^2+6+13 x=0$
$\Rightarrow 4 x^2+13 x+6=0$
Here $a = 4, b = 13$ and $c = 6$
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(13) \pm \sqrt{(13)^2-4(4)(6)}}{2(4)}$
$=\frac{-13 \pm \sqrt{73}}{8}$
$=\frac{-13 \pm 8.54}{8}$
$=\frac{-13+8.54}{8}$ and $\frac{-13-8.54}{8}$
$= -0.56$ and $-2.69$
View full question & answer→Question 303 Marks
Solve each of the following equations for x and give, in each case, your answer correct to two decimal places :
$2x^2 – 10x +5=0$
Answer$2 x^2-10 x+5=0$
Here $a= 2, b = -10$ and $c = 5$
$\therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-10) \pm \sqrt{(-10)^2-4(2)(5)}}{2(2)}$
$=\frac{10 \pm \sqrt{60}}{4}$
$=\frac{10 \pm 7.75}{4}$
$=\frac{10+7.75}{4}$ and $\frac{10-7.75}{4}$
$=4.44$ and $0.56$
View full question & answer→Question 313 Marks
Solve the following equations for x and give, in given case, your answer correct to one decimal place :
$5x^2 +10x – 3 =0$
Answer$5 x^2+10 x-3=0$
Here $a = 5, b = 10$ and $c = -3$
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(10) \pm \sqrt{(10)^2-4(5)(-3)}}{2(5)}$
$=\frac{-10 \pm \sqrt{160}}{10}$
$=\frac{-10 \pm 12.6}{10}$
$=\frac{-10+12.6}{10}$and $\frac{-10-12.6}{10}$
$= 0.26$ and $-2.26 = 0.3$ and $-2.3$
View full question & answer→Question 323 Marks
Solve the following equations for $x$ and give, in each case, your answer correct to one decimal place :
$x^2 – 8x+5=0$
Answer$x^2-8 x+5=0$
Here $a = 1, b = -8$ and $c = 5$
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-8) \pm \sqrt{44}}{2}$
$=\frac{8+2 \sqrt{11}}{2}$
$=4 \pm \sqrt{11}$
$=4 \pm 3.3=7.3$ and $0.7$
View full question & answer→Question 333 Marks
Solve each of the following equations using the formula:
$\frac{4}{x}-3=\frac{5}{2 x+3}$
Answer$\frac{4}{x}-3=\frac{5}{2 x+3}$
$\Rightarrow \frac{4-3 x}{x}=\frac{5}{2 x+3} $
$ \Rightarrow(4-3 x)(2 x+3)=5 x $
$ \Rightarrow 8 x+12-6 x^2-9 x=5 x $
$ \Rightarrow 6 x^2+6 x-12=0$
Here $a=1, b=1$ and $c=-2$
Then $x =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(1) \pm \sqrt{-1}^2-4(1)(-2)}{2(1)}$
$=\frac{1 \pm \sqrt{9}}{2}=\frac{-1 \pm 3}{2}$
$=\frac{-1+3}{2}$ and $\frac{-1-3}{2}=1$and -2
View full question & answer→Question 343 Marks
Solve each of the following equations using the formula:
$\sqrt{6} x^2-4 x-2 \sqrt{6}=0$
Answer$\sqrt{6} x^2-4 x-2 \sqrt{6}=0$
Here $a =\sqrt{6}, b=-4$ and $c=-2 \sqrt{6}$
Then $x =\frac{-b \pm \sqrt{b}^2-4 a c}{2 a}$
$=\frac{-4(4) \pm \sqrt{\left((-4)^2-4 \sqrt{6}\right)(-2 \sqrt{6})}}{2(\sqrt{6})}$
$=\frac{4 \pm \sqrt{64}}{2 \sqrt{6}}=\frac{4 \pm 8}{2 \sqrt{6}}$
$=\frac{4+8}{2 \sqrt{6}}$ and $\frac{4-8}{2 \sqrt{6}}$
$=\frac{6}{\sqrt{6}}$ and $-\frac{2}{\sqrt{6}}=\sqrt{6}$ and $-\frac{\sqrt{6}}{3}$
View full question & answer→Question 353 Marks
Solve each of the following equations using the formula :
$\frac{2 x+3}{x+3}=\frac{x+4}{x+2}$
Answer$\frac{2 x+3}{x+3}=\frac{x+4}{x+2}$
$\Rightarrow(2 x+3)(x+2)=(x+3)(x+4) $
$ \Rightarrow 2 x^2+4 x+3 x+6=x^2+4 x+3 x+12 $
$\Rightarrow x^2-6=0$
Then $x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(0) \pm \sqrt{(0)^2-4(1)(-6)}}{2(1)}$
$=\frac{0 \pm \sqrt{24}}{2}=\frac{0 \pm 2 \sqrt{6}}{2}$
$=-\sqrt{6}$ and $\sqrt{6}$
View full question & answer→Question 363 Marks
Solve for x using the quadratic formula. Write your answer correct to two significant figures.
$(x – 1)^2 – 3x + 4 = 0$
Answer$(x-1)^2-3 x+4=0 $
$ \Rightarrow x^2-2 x+1-3 x+4=0 $
$ \Rightarrow x^2-5 x+5=0$
Here $a = 1,b = -5$ and $c= 5$
$\therefore x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-5) \pm \sqrt{-5}^2-4 \times 1 \times 5}{2 \times 1}$
$=\frac{5 \pm \sqrt{25-20}}{2}$
$\therefore x=\frac{5+2.24}{2} \text { or } x=\frac{5-2.24}{2}$
$\Rightarrow x=\frac{7.24}{2} \text { or } x=\frac{2.76}{2}$
$\Rightarrow x=3.6 \text { or } x=1.4$
View full question & answer→Question 373 Marks
Solve each of the following equations using the formula:
$x^2-6=2 \sqrt{2} x$
Answer$x^2-6=2 \sqrt{2} x$
$\Rightarrow x^2-2 \sqrt{2} x-6=0$
Here $a=1, b=-2 \sqrt{2}$ and $c=-6$
Then $x =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=-\frac{(-2 \sqrt{2}) \pm \sqrt{-2 \sqrt{(2)}^2-4(1)(-6)}}{2(1)}$
$=\frac{2 \sqrt{2} \pm \sqrt{32}}{2}$
$=\frac{2 \sqrt{2} \pm 4 \sqrt{2}}{2}$
$=\frac{2 \sqrt{2}+4 \sqrt{2}}{2}$ and $\frac{2 \sqrt{2}-4 \sqrt{2}}{2}$
$=6 \frac{\sqrt{2}}{2}$ and $-2 \frac{\sqrt{2}}{2}$
$=3 \sqrt{2}$ and $-\sqrt{2}$
View full question & answer→Question 383 Marks
Solve each of the following equations using the formula:
$\frac{1}{15} x^2+\frac{5}{3}=\frac{2}{3} x$
Answer$\frac{1}{15} x^2+\frac{5}{3}=\frac{2}{3} $
$ \Rightarrow x^2++25=10 x $
$\ \Rightarrow x^2-10 x+25=0$
Here $a = 1, b = − 10$ and $c = 25$
Then $x =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-10) \pm \sqrt{-10}^2-4(1)(25)}{2(1)}$
$=\frac{10 \pm \sqrt{0}}{2}=5$
View full question & answer→Question 393 Marks
Solve each of the following equations using the formula:
$\frac{2}{3} x=-\frac{1}{6} x^2-\frac{1}{3}$
Answer$\frac{2}{3} x=-\frac{1}{6} x^2-\frac{1}{3}$
$\Rightarrow 4 x=-x^2-2 $
$\Rightarrow x^2+4 x+2=0$
Here $a = 1, b = 4$ and $c = 2$
Then $x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(-4) \pm \sqrt{(4)^2-4(1)(2)}}{2(1)}$
$=\frac{4 \pm \sqrt{8}}{2}=\frac{-4 \pm 2 \sqrt{2}}{2}$
$=-2 \pm \sqrt{2}$
View full question & answer→Question 403 Marks
Solve the following equations using the formula
$2 x^2+7 x+5=0$
Answer$2 x^2+7 x+5=0$
Here a = 2, b = 7 and c = 5
Then $x=\frac{-b \pm \sqrt{b^2-4 a c}}{2(a)}$
$=\frac{-(7) \pm \sqrt{(7)^2-4(2)(5)}}{2(2)}$
$=\frac{-7 \pm 3}{4}=\frac{-7+3}{4}$
and $\frac{-7-3}{4}$
$=-1$ and $-\frac{5}{2}$
View full question & answer→Question 413 Marks
Solve each of the following equations using the formula:
$3 x^2+2 x-1=0$
Answer$3 x^2+2 x-1=0$
Here a = 3, b = 2 and c = −1
Then $x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{-(2) \pm \sqrt{(2)^2-4(3)(1)}}{2(3)}$
$=\frac{-2 \pm 4}{6}=\frac{-2 \pm 4}{6}$
and $\frac{-2-4}{6}=\frac{1}{3}$
and -1
View full question & answer→Question 423 Marks
Solve each of the following equations using the formula:
$x^2+2 x-6=0$
Answerx^2 + 2x – 6 = 0
Here a = 1, b = 2 and c = − 6
Then $x =\frac{-b \pm \sqrt{b}^2-4 a c}{2 a}$
$=-\frac{(2) \pm \sqrt{(2)^2-4(1)(-6)}}{2(1)}$
$=\frac{-2 \pm \sqrt{28}}{2}$
$=\frac{-2 \pm 2 \sqrt{7}}{2}=-1 \pm \sqrt{7}$
View full question & answer→Question 433 Marks
Solve each of the following equations using the formula:
$x^2+6 x-10=0$
Answer$x^2+6 x-10=0$
Here a = 1, b = 6 and c = – 10
Then $x =-\frac{b \pm \sqrt{b}^2-4 a c}{2 a}$
$=\frac{-(6) \pm \sqrt{(-6)^2-4(1)(-10)}}{2(1)}$
$=\frac{-6 \pm \sqrt{76}}{2}$
$=\frac{-6 \pm 2 \sqrt{19}}{2}$
and $\frac{-6-2 \sqrt{19}}{2}=-3+\sqrt{19}$ and $-3-\sqrt{19}$
View full question & answer→Question 443 Marks
Solve the following equation using the formula.
$x^2-10 x+21=0$
Answer$x^2-10 x+21=0$
Here $a = 1, b = - 10$ and $c = 21$
Then $x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$x=\frac{-(-10) \pm \sqrt{(-10)^2-4(1)(21)}}{2(1)}$
$x =\frac{10 \pm 4}{2} $
$x =\frac{10+4}{2} \text { or } x=\frac{10-4}{2}$
$ =7 \text { and } 3$
View full question & answer→Question 453 Marks
Solve the following equation using the formula
$x^2-6 x=27$
Answer$x^2-6 x=27$
$\Rightarrow x^2-6 x-27=0$
Here a= 1, b= -6 and c = -27
Then $x=\frac{-b \pm \sqrt{(-6)^2-4(1)(-27)}}{2(1)}$
$=\frac{6 \mp 12}{2}=\frac{6+12}{2}$ and $\frac{6-12}{2}=9$ and -3
View full question & answer→Question 463 Marks
Solve equation using factorisation method:
$x+\frac{1}{x}=2.5$
Answer$x+\frac{1}{x}=2.5$
$\Rightarrow \frac{x^2+1}{x}=\frac{5}{2}$
$\Rightarrow 2 x^2+2=5 x$
$\Rightarrow 2 x^2-5 x+2=0$
$\Rightarrow 2 x(x-2)-1(x-2)=0$
$\Rightarrow(x-2)(2 x-1)=0$
Since $x - 2 = 0$ or $2x - 1 = 0$
$\therefore x - 2 = 0$
$\Rightarrow x = 2$
$\therefore 2x - 1 = 0$
$\Rightarrow 2x = 1$
$\Rightarrow x=\frac{1}{2}$
then $x =2$ or $x =\frac{1}{2}$
View full question & answer→Question 473 Marks
Solve equation using factorisation method:
$\frac{9}{2} x=5+x^2$
Answer$\frac{9}{2} x=5+x^2$
$\Rightarrow 9x = 10 + 2x^2$
$\Rightarrow 2x^2 - 9x + 10 = 0$
$\Rightarrow 2x^2 - 5x - 4x + 10 = 0$
$\Rightarrow x(2x - 5) - 2(2x -5) = 0$
$\Rightarrow (2x - 5) (x - 2) = 0$
since $2x - 5 = 0$ or $x - 2 = 0$
$\therefore 2x - 5 = 0$
$\therefore 2x = 5$
$\therefore x =\frac{5}{2}$
$\therefore x - 2 = 0$
$\therefore x = 2$
then $x=\frac{5}{2}$ or $x=2$
View full question & answer→Question 483 Marks
Solve equation using factorisation method:
$x (x - 5) = 24$
Answer$x (x - 5) = 24$
$\Rightarrow x^2 - 5x - 24 = 0$
$\Rightarrow x^2 - 8x + 3x - 24 = 0$
$\Rightarrow x (x - 8) + 3 (x - 8) = 0$
$\Rightarrow (x - 8) (x + 3) = 0$
since $x - 8 = 0$ or $x + 3 = 0$
then $x = 8$ or $x = -3$
View full question & answer→Question 493 Marks
Solve : $\frac{ x }{ a }-\frac{ a + b }{ x }=\frac{ b ( a + b )}{ ax }$
Answer$\frac{x}{ a }-\frac{ a + b }{x}=\frac{ b ( a + b )}{ a x} $
$ \Rightarrow \frac{x^2- a ^2- ab }{a x}=\frac{ ab + b ^2}{a x}$
$\Rightarrow x^2-a^2-a b=a b+b^2 $
$ \Rightarrow x^2=a^2+b^2+2 a b$
$ \Rightarrow x^2=(a+b)^2 $
$ \Rightarrow x^2=(a+b)(a+b)$
$\Rightarrow x=\sqrt{(a+b)(a+b)}$
$x= \pm(a+b)$
View full question & answer→Question 503 Marks
If $x=-3$ and $x=\frac{2}{3}$ are solutions of quadratic equation $m x^2+7 x+n=0$,
find the values of m and n.
Answer$m x^2+7 x+n=0$
Put $x = -3$ in given equation
$m(-3)^2 + 7 (-3) + n = 0$
$\Rightarrow 9m - 21 + n = 0$
$\Rightarrow 9m + n = 21.............(1)$
Put $x=\frac{2}{3}$ in given equation
$m \left(\frac{2}{3}\right)^2+7\left(\frac{2}{3}\right)+n=0$
$\Rightarrow `(4"m")/9 + 14/3 + n = 0$
$\Rightarrow 4m + 9n = -42.........(2)$
solving thses equations we get
$m = 3$ and $n = -6$
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