Question 12 Marks
Five times a certain whole number is equal to three less than twice the square of the number. Find the number.
AnswerLet number $=x$
Now according to the condition,
$
\begin{aligned}
& 5 x =2 x ^2-3 \\
& \Rightarrow 2 x ^2-5 x -3=0 \\
& \Rightarrow 2 x ^2-6 x + x -3=0 \\
& \Rightarrow 2 x ( x -3)+1( x -3)=0 \\
& \Rightarrow( x -3)(2 x +1)=0
\end{aligned}
$
Either $x-3=0$,
then $x=3$
or
$
2 x+1=0 \text {, }
$
then $2 x=-1$
$
\Rightarrow x -\frac{1}{2}
$But it is not possible as the number is whole number.
$
\therefore \text { Number }=3 .
$
View full question & answer→Question 22 Marks
At an annual function of a school, each student gives the gift to every other student. If the number of gifts is $1980$, find the number of students.
AnswerLet the number of students $= x$
then the number of gifts given $= x – 1$
Total number of gifts $= x (x – 1)$
According to the condition,
$x (x – 1) = 1980$
$\Rightarrow x^2 - x - 1980 = 0$
$\Rightarrow x^2 - 45x + 44x - 1980 = 0$
$\Rightarrow x(x - 45) + 44(x - 45) = 0$
$\Rightarrow (x - 45)(x + 44) = 0$
Either $x - 45 = 0,$
then $x = 45$
or
$x + 44 = 0,$
then $x = -44,$
but it is not possible as it is negative.
Hence number of students $= 45.$
View full question & answer→Question 32 Marks
Find the value (s) of k for which each of the following quadratic equation has equal roots : $kx^2 – 4x – 5 = 0$
Answer$
k x^2-4 x-5=0
$
Here $a=k, b=-4, c=5$
$
\begin{aligned}
& \therefore D=b^2-4 a c \\
& =(-4)^2-4 \times k \times(-5) \\
& =16+20 k
\end{aligned}
$
$\because$ Roots are equal.
$
\begin{aligned}
& \therefore D=0 \\
& \Rightarrow b^2-4 a c=0 \\
& \therefore 16+20 k=0 \\
& \Rightarrow 20 k=-16 \\
& \Rightarrow k=\frac{-16}{20} \\
& =\frac{-4}{5}
\end{aligned}
$
Hence $k =\frac{-4}{5}$.
View full question & answer→Question 42 Marks
Without solving the following quadratic equation, find the value of ‘p’ for which the given equations have real and equal roots: $x^2 + (p – 2)x + p = 0.$
Answer$x^2 + (p – 2)x + p = 0$
Here $a = 1, b = (p - 3), c = p$
$\because $ Equation has real and equal roots
$\therefore b^2 - 4ac = 0$
$\Rightarrow (p - 3)^2- 4(1)(p) = 0$
$\Rightarrow (p - 3)^2 - 4p = 0$
$\Rightarrow p^2 + 9 - 6p - 4p = 0$
$\Rightarrow p^2 - 10p + 9 = 0$
$\Rightarrow p^2 - 9p - p + 9 = 0$
$\Rightarrow p(p - 9) -1(p - 9) = 0$
$\Rightarrow (p - 1)(p - 9) = 0$
$\therefore p = 1, 9.$
View full question & answer→Question 52 Marks
Without solving the following quadratic equation, find the value of ‘p’ for which the given equations have real and equal roots: $px^2 – 4x + 3 = 0$
Answer$\begin{aligned} & p x^2-4 x+3=0 \\ & \text { Here } a=p, b=-4, c=3 \\ & \therefore D=b^2-4 a c \\ & =(-4)^2-4 \times p \times 3 \\ & =16-12 p \\ & \because \text { The roots are equal } \\ & \therefore D=0 \\ & \Rightarrow b^2-4 a c=0 \\ & \Rightarrow 16-12 p=0 \\ & \Rightarrow 12 p=16 \\ & \Rightarrow p=\frac{16}{12}=\frac{4}{3} \\ & \therefore p=\frac{4}{3} .\end{aligned}$
View full question & answer→Question 62 Marks
Find the values of $p$ for which the equation $3x^2 – px + 5 = 0$ has real roots.
Answer$\begin{aligned} & 3 x^2-p x+5=0 \\ & \text { Here, } a=3, b=-p, c=5 \\ & \therefore b^2-4 a c \\ & =(-p)^2-4 \times 3 \times 5 \\ & =p 2-60 \\ & \therefore \text { Roots are real } \\ & \therefore b^2-4 a c \geq 0 \\ & \therefore p^2-60 \geq 0 \\ & \Rightarrow p^2 \geq 60 \\ & \Rightarrow p \geq \pm \sqrt{60}= \pm 2 \sqrt{15} \\ & \therefore p \leq-2 \sqrt{15} \text { or } p \geq 2 \sqrt{15} .\end{aligned}$
View full question & answer→Question 72 Marks
Find the least positive value of k for which the equation $x^2 + kx + 4 = 0$ has real roots.
Answer$x^2 + kx + 4 = 0$
Here, $a = 1, b = k, c = 4$
$b^2 - 4ac$
$= k^2 - 4 x 1 x 4$
$= k^2 - 16$
$\because $ Roots are real and positive.
$\therefore k^2 - 16 \geq 0$
$\Rightarrow k^2\geq 16$
$\Rightarrow k \geq 4$
$\Rightarrow k = 4.$
View full question & answer→Question 82 Marks
Find the value(s) of p for which the equation $2x^2 + 3x + p = 0$ has real roots.
Answer$\begin{aligned} & 2 x^2+3 x+p=0 \\ & \text { Here, } a=2, b=3, c=p \\ & b^2-4 a c \\ & =(3)^2-4 \times 2 x p \\ & =9-8 p \\ & \because \text { Roots are real } \\ & \therefore b^2-4 a c>0 \\ & \Rightarrow 9-8 p \geq 0 \\ & 9 \geq 8 p \\ & \Rightarrow 8 p \leq 9 \\ & \Rightarrow p \leq \frac{9}{8} .\end{aligned}$
View full question & answer→Question 92 Marks
Solve equation using formula : $256x^2 – 32x + 1 = 0$
Answer$
256 x^2-32 x+1=0
$
Here $a=256, b=-32, c=1$
$
\begin{aligned}
& D=b^2-4 a c \\
& =(-32)^2-4 \times 256 \times 1 \\
& =1024-1024 \\
& =0
\end{aligned}
$
$
\begin{aligned}
& \because x=\frac{-b \pm \sqrt{D}}{2 a} \\
& =\frac{-(-32) \pm \sqrt{0}}{2 \times 256} \\
& =\frac{32}{512} \\
& =\frac{1}{16} \\
& x_1=\frac{1}{16}, x_2=\frac{1}{16}
\end{aligned}
$
Hence $x=\frac{1}{16}, \frac{1}{16}$.
View full question & answer→Question 102 Marks
Solve the equation $5x^2 – 3x – 4 = 0$ and give your answer correct to $3$ significant figures:
AnswerWe have $5 x^2-3 x-4=0$
Here $a=5, b=-3, c=-4$
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$=\frac{3 \pm \sqrt{9+4 \times 5 \times 4}}{2 \times 5}$
$=\frac{3 \pm \sqrt{89}}{10}$
$x=\frac{3^{10}+9.43}{10}$ or $x=\frac{3-9.43}{10}$
$\Rightarrow x=\frac{12.43}{10}$ or $x=\frac{-6.43}{10}$
$\Rightarrow x =1.24$ or $x =-0.643$.
View full question & answer→Question 112 Marks
Solve the following equation: $x-\frac{18}{x}=6$ Give your answer correct to two x significant figures.
Answer$\begin{aligned} & x-\frac{18}{x}=6 \\ & \Rightarrow x ^2-6 x -18=0 \\ & a =1, b =-6, c =-18 \\ & x =\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\ & =\frac{6 \pm \sqrt{36+72}}{2} \\ & =\frac{6 \pm \sqrt{108}}{2} \\ & =\frac{6 \pm 6 \sqrt{3}}{2} \text { or } \frac{6(1-1.73)}{2} \\ & =3 \times 2.73 \text { or } 3 \times-0.73 \\ & =8.19 \text { or }-2.19 .\end{aligned}$
View full question & answer→Question 122 Marks
Solve equation by factorization :$ x^2– 4x – 12 = 0$,when x\in $N$
Answer$x^2 – 4x – 12 = 0$
$\Rightarrow x^2 – 6x + 2x – 12 = 0$
$\Rightarrow x (x – 6) + 2 (x – 6) = 0$
$\Rightarrow (x – 6) (x + 2) = 0$
Either $x – 6 = 0,$ then $x = 6$
or $x + 2 = 0$, then $x = -2$
But -2 is not a natural number
$\therefore x = 6.$
View full question & answer→Question 132 Marks
Solve equation by factorization : $3(x – 2)^2 = 147$
Answer$3(x – 2)^2 = 147$
$3(x^2 - 4x + 4) = 147$
$\Rightarrow 3x^2 - 12x + 12 - 147 = 0$
$\Rightarrow 3x^2 - 12x - 135 = 0$
$\Rightarrow x^2 - 4x - 45 = 0 ...$ (dividing by $3$)
$\Rightarrow x^2 - 9x + 5x - 45 = 0$
$\Rightarrow x(x - 9) + 5(x - 9) = 0$
$\Rightarrow (x - 9) (x + 5) = 0$
EIther $x - 9 = 0,$
then $x = 9$
or
$x + 5 = 0,$
then $x = -5$
Hence $x = 9, -5.$
View full question & answer→Question 142 Marks
Solve equation by factorization : $(x – 4)^2 + 5^2 = 13^2$
Answer$(x – 4)^2 + 5^2 = 13^2$
$x^2 – 8x + 16 + 25 = 169$
$x^2- 8x + 41 - 169 = 0$
$\Rightarrow x^2- 8x - 128 = 0$
$\Rightarrow x^2 - 16x + 8x - 128 = 0$
$\Rightarrow x (x - 16) + 8(x - 16) = 0$
$\Rightarrow (x - 16) (x + 8) = 0$
Either $x - 16 = 0,$
then $x = 16$
or
$x + 8 = 0,$
then $x = -8$
Hence $x = 16, -8.$
View full question & answer→Question 152 Marks
Solve equation by factorization : $\frac{2}{3} x^2-\frac{1}{3} x=1$
Answer$
\begin{aligned}
& \frac{2}{3} x^2-\frac{1}{3} x=1 \\
& \Rightarrow 2 x^2-x=3 \\
& \Rightarrow 2 x^2-x-3=0 \\
& \Rightarrow 2 x^2-3 x+2 x-3=0 \\
& \Rightarrow x(2 x-3)+1(2 x-3)=0 \\
& \Rightarrow(2 x-3)(x+1)=0
\end{aligned}
$
Either $2 x-3=0$,
then $2 x =3$
$
\Rightarrow x =\frac{3}{2}
$
or
$
x+1=0 \text {, }
$
then $x=-1$
Hence $x=\frac{3}{2},-1$.
View full question & answer→Question 162 Marks
Solve equation by factorization : $6p^2+ 11p – 10 = 0$
Answer$
\begin{aligned}
& 6 p^2+11 p-10=0 \\
& \Rightarrow 6 p^2+15 p-4 p-10=0 \\
& \Rightarrow 3 p(2 p+5)-2(2 p+5)=0 \\
& (2 p+5)(3 p-2)=0
\end{aligned}
$
Either $2 p+5=0$, then $2 p=-5 \Rightarrow p=\frac{-5}{2}$
or $3 p-2=0$, then $3 p=2 \Rightarrow p=\frac{2}{3}$
Hence $p=\frac{-5}{2}, \frac{2}{3}$.
View full question & answer→Question 172 Marks
Solve equation by factorization : x(6x – 1) = 35
Answer$
\begin{aligned}
& x(6 x-1)=35 \\
& \Rightarrow 6 x^2-x-35=0 \\
& \Rightarrow 6 x^2-15 x+14 x-35=0 \\
& \Rightarrow 3 x(2 x-5)+7(2 x-5)=0 \\
& \Rightarrow(2 x-5)(3 x+7)=0
\end{aligned}
$
Either $2 x-5=0$,
then $2 x=5$
$
\Rightarrow x =\frac{5}{2}
$
or
$
3 x+7=0
$
then $3 x=-7$
$
\Rightarrow x =\frac{-7}{3}
$
Hence $x=\frac{5}{2}, \frac{-7}{3}$.
View full question & answer→Question 182 Marks
Solve equation by factorization : $3x^2= x + 4$
Answer$
\begin{aligned}
& 3 x^2=x+4 \\
& \Rightarrow 3 x^2-x-4=0 \\
& \Rightarrow 3 x^2-4 x+3 x-4=0 \\
& \Rightarrow x(3 x-4)+1(3 x-4)=0 \\
& \Rightarrow(3 x-4)(x+1)=0
\end{aligned}
$
Either $3 x-4=0$,
then $3 x =4$
$
\Rightarrow x =\frac{4}{3}
$
or
$
x+1=0 \text {, }
$
then $x=-1$
Hence $x=\frac{4}{3},-1$.
View full question & answer→Question 192 Marks
Solve equation by factorization : $21x^2 – 8x – 4 = 0$
Answer$
\begin{aligned}
& 21 x^2-8 x-4=0 \\
& \Rightarrow 21 x^2-14 x+6 x-4=0 \\
& \Rightarrow 7 x(3 x-2)+2(3 x-2)=0 \\
& \Rightarrow(3 x-2)(7 x+2)=0
\end{aligned}
$
Either $3 x-2=0$,
then $3 x =2$
$
\Rightarrow x =\frac{2}{3}
$
or
$
7 x+2=0 \text {, }
$
then $7 x =-2$
$
\Rightarrow x=\frac{-2}{7}
$
Hence $x=\frac{2}{3}, \frac{-2}{7}$.
View full question & answer→Question 202 Marks
Solve equation by factorization : $3x^2 – 5x – 12 = 0$
Answer$ 3 x^2-5 x-12=0$
$\Rightarrow 3 x^2-9 x+4 x-12=0$
$\Rightarrow 3 x(x-3)+4(x-3)=0$
$\Rightarrow(x-3)(3 x+4)=0$
Elther $x-3=0$,
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 212 Marks
Solve equation by factorization : x(2x + 5) = 3
Answer$\begin{aligned} & x(2 x+5)=3 \\ & \Rightarrow 2 x^2+5 x-3=0 \\ & \Rightarrow 2 x^2+6 x-x-3=0 \\ & \Rightarrow 2 x(x+3)-1(x+3)=0 \\ & \Rightarrow(x+3)(2 x-1)=0 \\ & \text { Either } x+3=0, \\ & \text { then } x=-3 \\ & \text { or } \\ & 2 x-1=0 \\ & \text { then } 2 x=1 \\ & \Rightarrow x=\frac{1}{2} \\ & \therefore x=-3, \frac{1}{2} .\end{aligned}$
View full question & answer→Question 222 Marks
Solve equation by factorization : $x^2 – 3x – 10 = 0$
Answer$x^2 – 3x – 10 = 0$
$\Rightarrow x^2 – 5x + 2x – 10 = 0$
$\Rightarrow x(x – 5) + 2(x – 5) = 0$
$\Rightarrow (x - 5) (x + 2) = 0$
Either $x - 5 = 0,$
then $x = 5$
or
$x + 2 = 0,$
then $x = -2$
Hence $x = 5, -2.$
View full question & answer→Question 232 Marks
Find the values of $x$ if $p + 7 = 0, q – 12 = 0$ and $x^2 + px + q = 0,$
Answer$p + 7 = 0$, then $p = – 7$
and $q – 12 = 0$, then $q = 12$
Substituting the values of p and q in the given quadratic equation,
$x^2 – 7x + 12 = 0$
$\Rightarrow x^2 – 3x – 4x + 12 = 0$
$\Rightarrow x(x – 3) – 4(x – 3) = 0$
$\Rightarrow (x – 3) (x – 4) = 0$
Either $x – 3 = 0,$
then $x = 3$
or
$x – 4 = 0,$
then $x = 4$
Hence $x = 3, 4.$
View full question & answer→Question 242 Marks
Find the values of $x$ if $p + 1 =0$ and $x^2 + px – 6 = 0$
Answer$p + 1 = 0$, then $p = – 1$
Substituting the value of p in the given quadratic equation
$x^2 + ( – 1)x – 6 = 0$
$\Rightarrow x^2 – x – 6 = 0$
$\Rightarrow x^2 – 3x + 2x – 6 = 0$
$\Rightarrow x(x – 3) + 2(x – 3) = 0$
$\Rightarrow (x – 3) (x + 2) = 0$
Either $x – 3 = 0,$
then $x = 3$
or
$x + 2 = 0,$
then $x = – 2$
Hence $x = 3, -2.$
View full question & answer→Question 252 Marks
Solve equation by factorization : x (2x + 1) = 6
Answer$
\begin{aligned}
& x(2 x+1)=6 \\
& 2 x^2+x-6=0 \\
& 2 x^2+4 x-3 x-6=0 \\
& \Rightarrow 2 x(x+2)-3(x+2)=0 \\
& \Rightarrow(x+2)(2 x-3)=0
\end{aligned}
$
Either $x+2=0$,
then $x=-2$
or
$
2 x-3=0 \text {, }
$
then $2 x=3$
$
\Rightarrow x =\frac{3}{2}
$
Hence $x=-2, \frac{3}{2}$.
View full question & answer→Question 262 Marks
Solve equation by factorization : $3 x-\frac{8}{x}=2$
Answer$
\begin{aligned}
& 3 x-\frac{8}{x}=2 \\
& \frac{3 x^2-8}{x}=2 \\
& \Rightarrow 3 x^2-8=2 x \\
& \Rightarrow 3 x^2-2 x-8=0 \\
& \Rightarrow 3 x^2-6 x+4 x-8=0 \\
& \Rightarrow 3 x(x-2)+4(x-2)=0 \\
& \Rightarrow(x-2)(3 x+4)=0
\end{aligned}
$
Either $x-2=0$,
then $x=2$
or
$
3 x+4=0 \text {, }
$
then $3 x=-4$
$
\Rightarrow x =\frac{-4}{3}
$
Hence $x=2, \frac{-4}{3}$.
View full question & answer→Question 272 Marks
Solve equation by factorization $:\frac{x^2}{15}-\frac{x}{3}-10=0$
Answer$\frac{x^2}{15}-\frac{x}{3}-10=0$
$\Rightarrow x^2-5 x-150=0 \ldots\left\{\begin{array}{l}\because-150=-15 \times 10 \\ -5=-15+10\end{array}\right\}$
$\Rightarrow x^2-15 x+10 x-150=0$
$ \Rightarrow x(x-15)+10(x-15)=0$
$ \Rightarrow(x-15)(x+10)=0$
Elther $x-15,$
then $x=15$
or
$x+10=0$
then $x=-10$
$\therefore x=15,-10 \text {. }$
View full question & answer→Question 282 Marks
Solve equation by factorization : $\frac{2}{x^2}-\frac{5}{x}+2=0, x \neq 0$
Answer$ \frac{2}{x^2}-\frac{5}{x}+2=0, x \neq 0$
$ \Rightarrow 2-5 x+2 x^2=0$
$ \Rightarrow 2 x^2-5 x+2=0 \ldots$
$\because 2 \times 2=4$
$4=-4 \times(-1)-5=-4-1$
$ \Rightarrow 2 x^2-4 x-x+2=0$
$ \Rightarrow 2 x(x-2)-1(x-2)=0$
$ \Rightarrow(x-2)(2 x-1)=0$
$ \text { Either } x-2=0$
$ \text { then } x=2$
$ \text { or }$
$ 2 x =1=0$
$ \text { then } 2 x=1$
$ \Rightarrow x=\frac{1}{2}$
$ \therefore x=2, \frac{1}{2} .$
View full question & answer→Question 292 Marks
Solve equation by factorization : $x+\frac{1}{x}=2 \frac{1}{20}$
Answer$
\begin{aligned}
& x+\frac{1}{x}=2 \frac{1}{20} \\
& \Rightarrow 20 x^2+20=41 x \\
& \Rightarrow 20 x^2-41 x+20=0 \\
& \Rightarrow 20 x^2-16 x-25 x+20=0 \\
& \Rightarrow 4 x(5 x-4)-5(5 x-4)=0 \\
& \Rightarrow(5 x-4)(4 x-5)=0
\end{aligned}
$
Either $5 x-4=0$,
then $5 x =4$
$
\Rightarrow x =\frac{4}{5}
$
or
$
4 x-5=0 \text {, }
$
then $4 x=5$
$
\Rightarrow x =\frac{5}{4}
$
Hence $x=\frac{4}{5}, \frac{5}{4}$.
View full question & answer→Question 302 Marks
Solve equation by factorization : $a^2x^2 + (a^2+ b^2)x + b^2 = 0, a \neq 0$
Answer$a^2 x^2+\left(a^2+b^2\right) x+b^2=0$
$ \Rightarrow a^2 x(x+1)+b^2(x+1)=0$
$ \Rightarrow(x+1)\left(a^2 x+b^2\right)=0$
$ \Rightarrow(x+1)=0, \text { then } x=-1$
or
$a^2 x+b^2=0 \text {, then } a^2 x=-b^2 \text {. }$
$\Rightarrow x =\frac{-b^2}{a^2}$
Hence $x =-1, \frac{-b^2}{a^2}$.
View full question & answer→Question 312 Marks
Solve equation by factorization : $x^2 – (p + q)x + pq = 0$
Answer$x^2 – (p + q)x + pq = 0$
$x^2 - px - qx + pq = 0$
$x(x - p) -q)x - p) = 0$
$\Rightarrow (x - p) (x - q) = 0$
Either $x - p = 0,$
then $x = p$
or
$x - q = 0,$
then$ x = q$
Hence$ x = p, q.$
View full question & answer→Question 322 Marks
Solve equation by factorization :$ a^2x^2 + 2ax + 1 = 0, a \neq 0$
Answer$ a^2 x^2+2 a x+1=0$
$\Rightarrow a^2 x^2+a x+a x+1=0$
$\Rightarrow a x(a x+1)+1(a x+1)=0$
$\Rightarrow a x(a x+1)(a x+1)=0$
$\Rightarrow(a x+1)^2=0$
$\therefore a x+1=0$
$\Rightarrow a x=-1$
$\Rightarrow x=-\frac{1}{a}$
Hence $x =-\frac{1}{a}, \frac{1}{a}$.
View full question & answer→Question 332 Marks
Solve equation by factorization : $2x^2 – 9x + 10 = 0$,when $x\in Q$
Answer$ 2 x^2-9 x+10=0 $
$ \Rightarrow 2 x^2-4 x-5 x+10=0 $
$ \Rightarrow 2 x(x-2)-5(x-2)=0 $
$ \Rightarrow(x-2)(2 x-5)=0$
Either $x-2=0$,
then $x=2$,
or
$2 x-5=0 \text {, }$
then $2 x=5$
$\Rightarrow x =\frac{5}{2}$
When $x \in Q$ then $x=2, \frac{5}{2}$.
View full question & answer→Question 342 Marks
Solve equation by factorization : $2x^2 – 9x + 10 = 0$,when $x\in N$
Answer$2 x^2-9 x+10=0$
$\Rightarrow 2 x^2-4 x-5 x+10=0$
$\Rightarrow 2 x(x-2)-5(x-2)=0$
$\Rightarrow(x-2)(2 x-5)=0$
Either $x-2=0$,
then $x =2$,
or
$2 x-5=0 \text {, }$
then $2 x=5$
$\Rightarrow x =\frac{5}{2}$
When $x \in N$, then $x=2$.
View full question & answer→Question 352 Marks
Solve equation by factorization : $5x^2– 8x – 4 = 0$ when $x\in Q$
Answer$5 x^2-8 x-4=0$
$\because 5 \times(-4)=-20$
$-20=-10+2$
$-8=-10+2$
$\Rightarrow 5 x^2-10 x+2 x-4=0$
$\Rightarrow 5 x(x-2)+2(x-2)=0$
$\Rightarrow(x-2)(5 x+2)=0 \ldots(\text { Zero Product Rule })$
$\text { Either } x-2=0$
$\text { then } x=2$
$\text { or }$
$5 x+2=0$
$\text { then } 5 x=-2$
$\Rightarrow x=-\frac{2}{5}$
$\therefore x=2,-\frac{2}{5} .$
View full question & answer→Question 362 Marks
Solve equation by factorization : $2x^2 – 8x – 24 = 0$ when x∈I
Answer$2x^2 – 8x – 24 = 0$
$\Rightarrow x^2– 4x – 12 = 0 ...$(Dividing by $2)$
$\Rightarrow x^2 – 6x + 2x – 12 = 0$
$\Rightarrow x (x – 6) + 2 (x – 6) = 0$
$\Rightarrow (x – 6) (x + 2) = 0$
Either$ x – 6 = 0$, then, $x = 6$
or $x + 2 = 0$, then$ x = – 2$
Hence$ x = 6, – 2.$
View full question & answer→Question 372 Marks
If $-\frac{1}{2}$ is a solution of the equation $3 x^2+2 k x-3=0$, find the value of $k$.
Answer$x=-\frac{1}{2}$ is a solution of the
$
3 x^2+2 k x-3=0 \text {, }
$
Substituting the value of $x$ in the given equation
$
\begin{aligned}
& 3\left(\frac{-1}{2}\right)^2+2 k\left(\frac{-1}{2}\right)-3=0 \\
& 3 \times \frac{1}{4}-k-3=0 \\
& \frac{3}{4}-k-3=0 \\
& \Rightarrow k =\frac{3}{4}-3 \\
& =-\frac{9}{4}
\end{aligned}
$
Hence $k =-\frac{9}{4}$.
View full question & answer→Question 382 Marks
In each of the following, determine whether the given numbers are solutions of the given equation or not: $x^2-\sqrt{2} x-4=0, x=-\sqrt{2}, 2 \sqrt{2}$
Answer$x^2-\sqrt{2} x-4=0, x=-\sqrt{2}, 2 \sqrt{2}$
(a) $x=-\sqrt{2}$
Substituting $x =-\sqrt{2}$
L.H.S. $=x^2-\sqrt{2} x-4$
$=(-\sqrt{2})^2-\sqrt{2}(-\sqrt{2})-4$
$=2+2-4$
$=0$
$=$ R.H.S.
$\therefore x=-\sqrt{2}$ is its solution.
(b) $x=-2 \sqrt{2}$
Substituting $x =-2 \sqrt{2}$
$
\begin{aligned}
& \text { L.H.S. }=x^2-\sqrt{2} x-4 \\
& =(-2 \sqrt{2})^2-\sqrt{2}(-2 \sqrt{2})-4 \\
& =8-4-4 \\
& =8-8 \\
& =0 \\
& =\text { R.H.S. }
\end{aligned}
$
$\therefore x=-2 \sqrt{2}$ is its solution.
View full question & answer→Question 392 Marks
In each of the following, determine whether the given numbers are solutions of the given equation or not: $x^2-3 \sqrt{3} x+6=0 ; \sqrt{3},-2 \sqrt{3}$
Answer$
x^2-3 \sqrt{3} x+6=0 ; \sqrt{3},-2 \sqrt{3}
$
(a) Substituting the value of $x=\sqrt{3}$
$
\begin{aligned}
& \text { L.H.S. }=x^2-3 \sqrt{3} x+6 \\
& =(\sqrt{3})^2-3 \sqrt{3} \times \sqrt{3}+6 \\
& =3-9+6 \\
& =0 \\
& =\text { R.H.S. }
\end{aligned}
$
$\therefore x=\sqrt{3}$ is its solution.
(b) $x =2 \sqrt{3}$
Substituting $x =2 \sqrt{3}$
$
\begin{aligned}
& \text { L.H.S. }=x^2-3 \sqrt{3} x+6 \\
& =(-2 \sqrt{3})^2-3 \sqrt{3}(-2 \sqrt{3})+6 \\
& =12+18+6 \\
& =36 \neq 0 \\
& \because x=-2 \sqrt{3} \text { is not its solution. }
\end{aligned}
$
View full question & answer→Question 402 Marks
In each of the following, determine whether the given numbers are roots of the given equations or not; $x^2 – 5x + 6 = 0; 2, – 3$
Answer$x^2 – 5x + 6 = 0; 2, – 3$
When $x = 2$, then
$(2)^2 - 5 x 2 + 6$
$= 4 - 10 + 6$
$= 10 - 10$
$= 0$
When $x = -3$, then
$(-3)^2 -5(-3) + 6$
$= 9 + 15 + 6$
$= 30 \neq 0$
$\therefore x = -3$ is not its solution
$\therefore 2$ is root of the equation by $-3$ is not a root.
View full question & answer→Question 412 Marks
Solve equation by using formula : $\frac{4}{3}-3=\frac{5}{2 x+3}, x \neq 0,-\frac{3}{2}$
Answer$
\begin{aligned}
& \frac{4}{3}-3=\frac{5}{2 x+3}, x \neq 0,-\frac{3}{2} \\
& \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=0 \\
& \Rightarrow-6 x^2-6 x+12=0 \\
& \Rightarrow x^2+x-2=0 \\
& \Rightarrow x^2+2 x-x-2=0 \\
& \Rightarrow x(x+2)-1(x+2)=0 \\
& \Rightarrow(x+2)(x-1)=0
\end{aligned}
$
Either $x+2=0$,
then $x=-2$
or
$
x-1=0 \text {, }
$
then $x=1$
$
\therefore x=1,-2 \text {. }
$
View full question & answer→Question 422 Marks
Solve equation using formula : $2x^2 – 3x – 1 = 0$
Answer$\begin{aligned} & 2 x^2-3 x-1=0 \\ & \text { Here } a=2, b=-3, c=-1 \\ & D=b^2-4 a c \\ & =(-3)^2-4 \times 2 \times(-1) \\ & =9+8 \\ & =17 \\ & \because x=\frac{-b \pm \sqrt{D}}{2 a} \\ & =\frac{-(-3) \pm \sqrt{17}}{2 \times 2} \\ & =\frac{3 \pm \sqrt{17}}{4} \\ & \therefore x=\frac{3+\sqrt{17}}{4}, \frac{3-\sqrt{17}}{4} .\end{aligned}$
View full question & answer→Question 432 Marks
Solve equation by factorisation : $x(x + 1) + (x + 2)(x + 3) = 42$
Answer$x(x + 1) + (x + 2)(x + 3) = 42$
$\Rightarrow 2x^2 + 6x + 6 – 42 = 0$
$\Rightarrow x2 + 3x – 18 = 0$
$\Rightarrow x^2+ 3x – 18 = 0 ..$(Dividing by $2)$
$\Rightarrow x^2 + 6x – 3x – 18 = 0$
$\Rightarrow x(x + 6) 3(x + 6) = 0$
$\Rightarrow (x + 6)(x – 3) = 0$
Either $x+6=0$,
then $x = –6$
or
$x-3=0$
then $x=3$
Hence $x=-6,3$.
View full question & answer→Question 442 Marks
Solve equation by factorisation : $
\sqrt{3} x^2+10 x+7 \sqrt{3}=0
$
Answer$
\begin{aligned}
& \sqrt{3} x^2+10 x+7 \sqrt{3}=0 \\
& \Rightarrow \sqrt{3} x^2+3 x+7 x+7 \sqrt{3}=0 \\
& \Rightarrow \sqrt{3} x(x+\sqrt{3})+7(x+\sqrt{3})=0 \\
& \Rightarrow(x+\sqrt{3})(\sqrt{3}+7)=0
\end{aligned}
$
Either $x+\sqrt{3}=0$,
then $x =-\sqrt{3}$
or
$
\sqrt{3} x+7=0 \text {, }
$
then $\sqrt{3} x=-7$
$
\begin{aligned}
& \Rightarrow x =\frac{-7}{\sqrt{3}} \\
& 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 452 Marks
Two natural numbers are in the ratio 3 : 4. Find the numbers if the difference between their squares is 175.
AnswerRatio in two natural numbers $=3: 4$
Let the numbers be $3 x$ and $4 x$
According to the condition,
$
\begin{aligned}
& (4 x)^2-(3 x)^2=175 \\
& \Rightarrow 16 x^2-9 x^2=175 \\
& \Rightarrow 7 x^2=175 \\
& \Rightarrow x^2=\frac{175}{7}=25=( \pm 5)^2 \\
& \therefore x=5,-5
\end{aligned}
$
But $x-5$ is not a natural number
$
\therefore x =5
$
$\therefore$ Natural numbers are $3 x_2 4 x$
$
\begin{aligned}
& =3 \times 5,4 \times 5 \\
& =15,20 .
\end{aligned}
$
View full question & answer→Question 462 Marks
Find two natural numbers which differ by $3$ and whose squares have the sum of $117.$
AnswerLet the numbers be $x $and $x-3$. Then,
$x^2 + (x-3)2 = 117,$
$\Rightarrow x^2 + x^2 + 9 - 6x =117$
$\Rightarrow 2 x^2 -6x - 108 = 0$
$\Rightarrow x^2 - 3x - 54 = 0$
$\Rightarrow x^2 - 9x + 6x - 54 = 0$
$\Rightarrow x (x - 9) + 6(x - 9) = 0$
$\Rightarrow (x-9) (x+6 ) = 0$
$\Rightarrow x = 9$ (As the number have to be natural number)
Hence the numbers are $6$ and $9.$
View full question & answer→Question 472 Marks
Find the values of m so that the quadratic equation $3x^2 – 5x – 2m = 0$ has two distinct real roots.
Answer$\begin{aligned} & 3 x^2-5 x-2 m=0 \\ & \text { Here } a=3, b=-5, c=-2 m \\ & \therefore D=b^2-4 a c \\ & =(5)^2-4 \times 3 \times(-2 m) \\ & =25+24 m \\ & \because \text { Roots are real and distinct } \\ & \therefore D>0 \\ & 25+24 m>0 \\ & 24 m>-25 \\ & m>-\frac{25}{24} .\end{aligned}$
View full question & answer→Question 482 Marks
Discuss the nature of the roots of the following equation:$x^2-\frac{1}{2} x-4=0$
Answer$
x^2-\frac{1}{2} x-4=0
$
Here $a=1, b=\frac{1}{2}, c=-4$
$
\begin{aligned}
& \therefore D=b^2-4 a c \\
& =\left(\frac{1}{2}\right)^2-4 \times 1 \times(-4) \\
& =\frac{1}{4}+16 \\
& =\frac{65}{4} \\
& \because D>0
\end{aligned}
$
$\therefore$ Roots are real and distinct.
View full question & answer→