Question 13 Marks
In each of the following, determine whether the given values are solution of the given equation or not:
$a^2 x^2-3 a b x+2 b^2=0 ; x=\frac{a}{b}, x=\frac{b}{a}$.
Answer$ a^2 x^2-3 a b x+2 b^2=0 ; x=\frac{a}{b}, x=\frac{b}{a} \text {. } $
Now on substituting $x =\frac{a}{b}$ in L.H.S.
$\text { L.H.S. }=a^2 x^2-3 a b x+2 b^2$
$=a^2 \times\left(\frac{a}{b}\right)^2-3 a b \times \frac{a}{b}+2 b^2$
$=\frac{a^4}{b^2}-3 a^2+2 b^2$
$=\frac{a^4-3 a^2 b^2+2 b^2}{b^2} \ =a^4-3 a^2 b^2+2 b^4 \neq 0 \neq \text { R.H.S. }$
$\therefore x =\frac{a}{b}$ is not a solution of the equation
Put $x=\frac{b}{a}$ in L.H.S. of given equation
$ \text { L.H.S. }=a^2 \times\left(\frac{b}{a}\right)^2-3 a b \times \frac{b}{a}+2 b^2$
$ =b^2-3 b^2+2 b^2 $
$ =3 b^2-3 b^2$
$ =0$
$=\text { R.H.S. } $
$\therefore x=\frac{b}{a}$ is a solution of the given equation.
View full question & answer→Question 23 Marks
In each of the following, determine whether the given values are solution of the given equation or not:
$x^2-\sqrt{2}-4=0 ; x=-\sqrt{2}, x=-2 \sqrt{2}$
Answer$x^2-\sqrt{2}-4=0 ; x=-\sqrt{2}, x=-2 \sqrt{2}$
$ \text { Now } x =-\sqrt{2} $
$\Rightarrow \text { L.H.S. }=(-\sqrt{2})^2-\sqrt{2} \times(-\sqrt{2})-4=0$
$=2+2-4$
$=0$
$=\text { R.H.S. }$
$\therefore x=-\sqrt{2}$ is a solution of the given equation.
$\text { Now } x =-2 \sqrt{2}$
$\Rightarrow \text { L.H.S. }=(-2 \sqrt{2})^2-\sqrt{2}(-2 \sqrt{2})-4=0 $
$=8+4-4 \neq 0 \neq \text { R.H.S. }$
$\therefore x =-2 \sqrt{2} \text { is not a solution of the equation. }$
View full question & answer→Question 33 Marks
In each of the following, determine whether the given values are solution of the given equation or not:
$2x^2 - x + 9 = x^2 + 4x + 3; x = 2, x = 3$
Answer$2x^2 - x + 9 = x^2 + 4x + 3; x = 2, x = 3$
$2x^2 - x + 9 = x^2 + 4x + 3$
$2x^2- x^2 - x - 4x + (9 - 3) = 0$
$x^2 + 5x + 6 = 0$
Now $x = 2$
$L.H.S. = (2)^2 - 5 x 2 + 6 = 0$
$10 - 10 = 0 = R.H.S.$
$\therefore x = 2$ is a solution of the given equation.
On substituting $x = 3$ in $L.H.S$. of equation $(1)$
$\Rightarrow L.H.S. = (3)^2 - 5 x 3 + 6$
$\Rightarrow = 15 - 15$
$= 0$
$= R.H.S.$
$\therefore x = 3$ is a solution of the given equation.
View full question & answer→Question 43 Marks
In each of the following, determine whether the given values are solution of the given equation or not:
$x^2-3 \sqrt{3}+6=0 ; x=\sqrt{3}, x=-2 \sqrt{3}$
Answer$x^2-3 \sqrt{3}+6 ; x=\sqrt{3}, x=-2 \sqrt{3}$
Now substitute $x=\sqrt{3}$ in L.H.S. of given equation.
$\text { L.H.S. }=(\sqrt{3})^2-3 \sqrt{3} \times \sqrt{3}+6=0 $
$=3-9+6 $
$=0 $
$=\text { R.H.S. }$
$x=\sqrt{3}$ is a solution of the given equation.
Substitute $x=-2 \sqrt{3}$ in L.H.S. of given equation
$\Rightarrow(-2 \sqrt{3})^2-3 \sqrt{3} \times-2 \sqrt{3}+6=0 $
$ \Rightarrow \text { L.H.S. }=12+18+6 \neq 0 \neq \text { R.H.S. }$
$x=-2 \sqrt{3}$ is not a solution of the given equation.
View full question & answer→Question 53 Marks
In each of the following, determine whether the given values are solution of the given equation or not:
$x^2 - 3x + 2 = 0; x = 2, x = -1$
AnswerSubstitute $x = 2$ in $L.H.S.$ of given equation
$L.H.S. = (2)^2 - 3x 2 x2$
$= 6 - 6$
$= 0$
$\Rightarrow L.H.S. = 0$
$= R.H.S.$
Substitute x = -1 in L.H.S. of given equation.
$L.H.S. = (-1)^2 - 3 x -1 + 2 = 0$
$= 1 + 3 + 2 \neq 0 \neq R.H.S.$
$x = 2$ is a solution and $x = -1$ is not solution of the given equation.
View full question & answer→Question 63 Marks
In each of the following determine whether the given values are solutions of the equation or not.
$x^2 + x + 1 = 0; x = 1, x = -1.$
AnswerGiven equation is
$x^2 + x + 1 = 0; x = 1, x = -1$
Substitute $x = 1$ in $L.H.S.$
$L.H.S. = (1)^2 + (1) + 1$
$= 3 \neq R.H.S. \neq 0$
Hence, x = 1 is not a solution of the given equation.
Now substitute x = -1 in $L.H.S.$
$L.H.S. = (-1)^2+ (-1) + 1$
$= 1 - 1 + 1$
$= 1 \neq R.H.S. \neq 0$
Hence, $x = -1$ is not a solution of the given equation.
View full question & answer→Question 73 Marks
In each of the following determine whether the given values are solutions of the equation or not.
$9 x^2-3 x-2=0 ; x=-\frac{1}{3}, x=\frac{2}{3}$
AnswerGiven equation is
$9 x^2-3 x-2=0 ; x=-\frac{1}{3}, x=\frac{2}{3}$
Substitute $x=-\frac{1}{3}$ in the L.H.S.
L.H.S. $=9\left(-\frac{1}{3}\right)^2-3 \times\left(-\frac{1}{3}\right)-2$
$=9 \times \frac{1}{9}+1-2 $
$ =2-2$
$ =0 $
$ =\text { R.H.S. }$
Hence, $x=-\frac{1}{3}$ is a solution of the equation.
Again put $x=\frac{2}{3}$
L.H.S. $=9\left(\frac{2}{3}\right)^2-3\left(\frac{2}{3}\right)-2$
$=9 \times \frac{4}{9}-2-2 $
$ =4-4 $
$ =0$
$ =\text { R.H.S. }$
Hence, $x=\frac{2}{3}$ is a solution of the equation.
View full question & answer→Question 83 Marks
In each of the following determine whether the given values are solutions of the equation or not.
$x^2 + 6x + 5 = 0; x = -1, x = -5$
AnswerGiven equation is
$x^2 + 6x + 5 = 0; x = -1, x = -5$
Substitute $x = -1$ in $L.H.S.$
$L.H.S. = (-1)^2+ 6 x (-1) + 5$
$= 1 - 6 + 5$
$= 6 - 6$
$= 0$
Hence, $x = -1$ is a solution of the given equation.
Again put $x = -5$ in $L.H.S.$
$L.H.S. = (-5)^2 + 6 x (-5) + 5$
$= 25 - 30 + 5$
$= 30 - 30$
$= 0$
Hence, $x = -5$ is also a solution of the given equation.
View full question & answer→Question 93 Marks
If the ratio of the roots of the equation
$1 x^2+n x+n=0$ is $p: q$, Prove that
$\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}=0$
AnswerLet $\alpha, \beta$ be the roots of
$1 x ^2+ nx + n =0, \alpha+\beta=\frac{n}{l}$ and $\alpha \beta=\frac{n}{l}$.
$\frac{\alpha}{\beta}=\frac{p}{q} \ldots \text { (given) }$
Now L.H.S.
$=\sqrt{\frac{p}{q}}+\sqrt{\frac{q}{p}}+\sqrt{\frac{n}{l}}$
$ =\sqrt{\frac{\alpha}{\beta}}+\sqrt{\frac{\beta}{\alpha}}+\sqrt{\frac{n}{l}} $
$ =\frac{\alpha+\beta}{\sqrt{\alpha \beta}}+\sqrt{\frac{n}{l}}$
$=\frac{\frac{-n}{l}}{\sqrt{\frac{n}{l}}}+\sqrt{\frac{n}{l}},\left[\because \alpha+\beta=\frac{-n}{l} \alpha \beta=\frac{n}{l}\right]$
$=-\sqrt{\frac{n}{l}}+\sqrt{\frac{n}{l}}$
$= 0$
$= R.H.S.$
Hence proved.
View full question & answer→Question 103 Marks
Given that one root of the quadratic equation $ax^2 + bx + c = 0$ is three times the other, show that $3b^2 – 16ac$.
AnswerThe given quadratic equation is
$ax^2 + bx + c = 0$
Let α be the one root
Then other root $= 3\alpha$
Now, Sum of the root $=\frac{-b}{a}$
$\Rightarrow \alpha+3 \alpha=\frac{-b}{a}$
$\Rightarrow 4 \alpha=\frac{-b}{a}$
$\Rightarrow \alpha=\frac{-b}{4 a}$.....(i)
Also product of the root $\alpha \times 3\alpha$
$=\frac{c}{a} $
$=3 \alpha^2=\frac{c}{a}$
From equation (i) $3\left(-\frac{b}{4 a}\right)^2=\frac{c}{a}$
$\Rightarrow 3 \times \frac{b^2}{16 a^2}=\frac{c}{a}$
$\Rightarrow 3 b^2=16 \text { ac Proved }$
View full question & answer→Question 113 Marks
$\sqrt{3} x^2+11 x+6 \sqrt{3}=0$
Answer$\sqrt{3} x^2+11 x+6 \sqrt{3}=0$
$ \Rightarrow \sqrt{3} x^2+9 x+2 x+6 \sqrt{3}=0 $
$ \Rightarrow \sqrt{3} x(x+3 \sqrt{3})+2(x+3 \sqrt{3})=0$
$ \Rightarrow(x+3 \sqrt{3})(\sqrt{3} x+2)=0 $
$ x +3 \sqrt{3}=0 \text { or } \sqrt{3} x+2=0$
$\Rightarrow x =-3 \sqrt{3}$ or $x=-\frac{2}{\sqrt{3}}$ are two roots of the equation.
View full question & answer→Question 123 Marks
$\frac{2}{x^2}-\frac{5}{x}+2=0$
Answer$\frac{2}{x^2}-\frac{5}{x}+2=0 $
$ \Rightarrow \frac{2-5 x+2 x^2}{x^2}=0 $
$ \Rightarrow 2 x^2-5 x +2=0 $
$ \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$
$ \Rightarrow x -2=0 \text { or } 2 x -1=0$
$\Rightarrow x=2$ or $x=\frac{1}{2}$ are two root of the equation.
View full question & answer→Question 133 Marks
$10 x-\frac{1}{x}=3$
Answer$10 x-\frac{1}{x}=3 $
$\Rightarrow \frac{10 x^2-1}{x}=3 $
$ \Rightarrow 10 x^2-3 x-1=0 $
$ \Rightarrow 10 x^2-5 x+2 x-1=0$
$ \Rightarrow 5 x(2 x-1)+1(2 x-1)=0$
$ \Rightarrow(2 x-1)(5 x+1)=0$
$ \Rightarrow 2 x-1=0 \text { or } 5 x+1=0$
$x=\frac{1}{2}$ or $x=-\frac{1}{5}$ are two roots of the equation.
View full question & answer→Question 143 Marks
Find the values of k so that the sum of tire roots of the quadratic equation is equal to the product of the roots in each of the following:
$2x^2 - (3k + 1)x - k + 7 = 0.$
Answer$2 x^2-(3 k+1) x-k+7=0$
Here,
$a= 2,$
$b = -(3k + 1)$
$c = -k + 7$
Sum of roots = $\frac{-b}{a}$
$=\frac{3 k+1}{2}$
Product of roots $=\frac{c}{a}$
$=\frac{-k+7}{2}$
Sum of roots = Product of roots
$\frac{3 k+1}{2}=\frac{-k+7}{2}$
$6 k +2=-2 k +14$
$8 k =12$
$\Rightarrow k =\frac{12}{8}$
$\therefore k =\frac{3}{2} .$
View full question & answer→Question 153 Marks
Find the value of k so that sum of the roots of the quadratic equation is equal to the product of the roots:
$(k + 1)x^2 + (2k + 1)x - 9 = 0, k + 1 \neq 0.$
AnswerThe given equation is
$(k+1) x^2+(2 k+1) x-9=0$
Here, $a = k + 1, b = (2k + 1)$ and $c = -9.$
Sum of the roots $\alpha+\beta=\frac{-(2 k+1)}{k+1}$
and $\alpha \beta=\frac{c}{a}=\frac{-9}{k+1}$
Since, Sum of the roots = Product of the roots
Then, $\left(\frac{2 k+1}{k+1}\right)=\frac{9}{k+1}$
$\Rightarrow 2 k +1=9$
$\Rightarrow 2 k =9-1$
$\Rightarrow 2 k =8$
$\Rightarrow k =\frac{8}{2}$
$=4$
$\Rightarrow k =4 .$
View full question & answer→Question 163 Marks
Find the value of $k$ so that sum of the roots of the quadratic equation is equal to the product of the roots:
$kx^2+ 6x - 3k = 0, k \neq 0$
AnswerThe given quadratic equation is
$k x^2+6 x-3 k=0$
Here, $a=k, b=6$ and $c=-3 k$
Sum of the roots $\alpha+\beta=\frac{-b}{a}=\frac{-6}{k}$
and product of the roots $\alpha \beta=\frac{c}{a}=\frac{-3 k}{k}$
$\Rightarrow \frac{-6}{k}=-3 $
$ \Rightarrow k =\frac{+6}{+3} $
$\Rightarrow k =2 .$
View full question & answer→Question 173 Marks
Determine whether the given quadratic equations have equal roots and if so, find the roots:
$\frac{4}{3} x^2-2 x+\frac{3}{4}=0$
AnswerWe have
$\frac{4}{3} x^2-2 x+\frac{3}{4}=0$
Here, $a=\frac{4}{3}, b=-2$ and $c=\frac{3}{4}$
Discriminant
$=b^2-4 a c$
$ =(-2)^2-4 \times \frac{4}{3} \times \frac{3}{4} $
$ =4-4 $
$ =0$
So, the given equation has two real and equal roots given by
$a=\frac{-b+\sqrt{b^2-4 a c}}{2 a}$
$=\frac{+2+0}{4}$
$=\frac{3}{4}$
and $\beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}$
$=\frac{+2-0}{2 \times \frac{4}{3}} $$=\frac{3}{4} .$
View full question & answer→Question 183 Marks
Determine whether the given quadratic equations have equal roots and if so, find the roots:
$x^2+ 2x + 4 = 0$
AnswerThe given quadratic equation is
$x^2+ 2x + 4 = 0$
Here, $a = 1, b = 2$ and $c = 4$
Descriminant
$= b^2 - 4ac$
$= (2)^2 - 4 x 1 x 4$
$= 4 - 16$
$= -12 < 0$
Hence, the given equation has no real roots.
View full question & answer→Question 193 Marks
Determine whether the given quadratic equations have equal roots and if so, find the roots:
$x^2 + 5x + 5 = 0$
AnswerThe given quadratic equation is
$x^2+5 x+5=0$
Here, $a = 1, b = 5$ and $c = 5$
Discriminant
$=b^2-4 a c$
$=(5)^2-4 \times 1 \times 5$
$ =25-20$
$ =5>0$
so the given equation has real roots given by
$a=\frac{-b+\sqrt{b^2-4 a c}}{2 a}$
$=\frac{-5+\sqrt{25-4 \times 1 \times 5}}{2 \times 1}$
$ =\frac{-5+\sqrt{5}}{2}$
$\text { and } \beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}$
$ =\frac{-5-\sqrt{25-4 \times 1 \times 5}}{2 \times 1}$
$ =\frac{-5-\sqrt{5}}{2}$
View full question & answer→Question 203 Marks
Solve the following quadratic equation:
$4x^2- 4ax + (a^2 - b^2) = 0$ where $a , b ∈ R$.
Answer$4 x^2-4 a x+\left(a^2-b^2\right)=0 \text { where } a, b \in R$
$ \Rightarrow 4 x^2-\{2(a+b) x+2(a-b) x\}+a^2-b^2=0 $
$\Rightarrow\left\{4 x^2-2(a+b) x\right\}-\left\{2(a-b) x-\left(a^2-b^2\right)\right\}=0 $
$\Rightarrow 2 x\{2 x-(a+b)\}-(a-b)\{2 x-(a+b)\}=0$
$ \Rightarrow\{2 x-(a+b)\}\{2 x-(a-b)\}=0 $
$ \Rightarrow 2 x-(a+b)=0$
or
$2 x -( a - b )=0$
$\Rightarrow x =\frac{a+b}{2} \text { or } x =\frac{a-b}{2} .$
View full question & answer→Question 213 Marks
Solve the following quadratic equation by factorisation:
$x^2 + 3x - 18 = 0$
AnswerThe given quadratic equation is
$x^2 + 3x - 18 = 0$
$\Rightarrow x^2 + 6x - 3x - 18 = 0$
$\Rightarrow x(x + 6) -3(x + 6) = 0$
$\Rightarrow (x + 6) (x - 3) = 0$
when $x + 6 = 0$
$x = -6$
when $x - 3 = 0$
$x = 3$
$\therefore x = {-6, 3}.$
View full question & answer→Question 223 Marks
Two pipes running together can $1$ fill a cistern in $11\ 1/9$ minutes. If one pipe takes $5$ minutes more than the other to fill the cistern find the time when each pipe would fill the cistern.
AnswerLet x minutes be time taken by the larger pipe to fill the cistern then the smaller pipe taken $(x + 5)$ minutes. These two pipes would fil $\frac{1}{x}$ and $\frac{1}{x+5}$ of the cistern in a minute, respectively.
$\frac{1}{x}+\frac{1}{x+5}=\frac{9}{100}$
$\Rightarrow 9 x^2-155 x-500=0$
$\Rightarrow 9 x^2+25 x-180 x-500=0$
$\Rightarrow x(9 x+25)-20(9 x+25)=0$
$\Rightarrow(9 x+25)(x-20)=0$
$\Rightarrow x-20=0$
$\text { and } 9 x+25=0$
$x=20$
$\text { and } x=-\frac{25}{9} \ldots \text { (negligible) }$
Hence the time taken by the pipes to fill the cistern in $20$ minutes and $25$ minutes.
View full question & answer→Question 233 Marks
Car A travels x km for every litre of petrol, while car B travels $(x + 5)$ km for every litre of petrol.
Write down the number of litres of petrol used by car A and car B in covering a distance of $40\ km.$
AnswerGiven Distance $= 400\ km$
Car A travels $x$ km/litre.
Car B travels $(x + 5)$ km/litre.
No, of litre used by car
$A=\frac{\text { Distance }}{\text { Speed of car A }} $
$ =\frac{400}{x} \text { litre. }$
No. of litre used by car B $=\frac{\text { Distance }}{\text { Speed of car B }}$
$=\frac{400}{x+5}$ liter.
View full question & answer→Question 243 Marks
An aeroplane travelled a distance of $400$ km at an average speed of x km/hr. On the return journey the speed was increased by $40$ km/hr. Write down the expression for the time taken for
the return Journey. If the return journey took $30$ minutes less than the onward journey write down an equation in x and find its value.
AnswerTIme taken for the onward journey = $=\frac{400}{x+40}$ hours.
According to the question,
$\frac{400}{x+40}=\frac{400}{x}-\frac{1}{2} $
$\Rightarrow 800 x =800( x +40)- x ( x +40)$
$ \Rightarrow x ^2+40 x -32,000=0$
$ \Rightarrow( x +200)( x -1600)=0$
$ \Rightarrow x =-200 \text { (inadmissible) or } 160$
Hence, the required value of x is $160.$
View full question & answer→Question 253 Marks
Two pipes flowing together can fill a cistern in $6$ minutes. If one pipe takes $5$ minutes more than the other to fill the cistern, find the time in which each pipe would fill the cistern.
AnswerLet the time taken by the two pipes to fill the cistern be x and $x + 5$ min. respectively.
In $1$ min., the first pipe can fill $\frac{1}{x}$ of the cistern. In $1$ min., $\frac{1}{x+5}$ of the cistern then
$\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6} $
$\Rightarrow \frac{x+5+x}{x(x+5)}=\frac{1}{6} $
$ \Rightarrow \frac{2 x+5}{x^2+5 x}=\frac{1}{6}$
$\Rightarrow x^2+5 x=12 x+30$
$ \Rightarrow x^2-7 x-30=0 $
$ \Rightarrow x^2-10 x+3 x-30=0 $
$\Rightarrow x(x-10)+3(x-10)=0 $
$ \Rightarrow(x-10)(x+3)=0 $
$ \Rightarrow x-10=0 \text { or } x=-3$
$ \Rightarrow x=10 \text { or } x=-3$
Since, time cannot be negative.
So, $x = 10$ and $x + 5 = 10 + 5 = 15.$
View full question & answer→Question 263 Marks
Some students planned a picnic. The budget for the food was $Rs. 480.$ As eight of them failed to join the party, the cost of the food for each member increased by $Rs. 10.$ Find how many students went for the picnic.
AnswerLet the total no. of student be $x.$
Cost of food for each = ₹ $\frac{480}{x}$
When $8$ students failed to join, then cost of food for each = $\frac{480}{x-8}$
According to question
$\frac{480}{x-8}-\frac{480}{x}=10 $
$\frac{480 x-480(x-8)}{x(x-8)}=10 $
$\frac{480(x-x+8)}{x(x-8)}=10$
$x^2-8 x-384=0$
$ x^2-24 x+16 x-384=0 $
$(x-24)(x+16)=0 $
$ x=24$
The no. of students went for picnic
$= 24 - 8$
$= 16.$
View full question & answer→Question 273 Marks
The speed of an express train is $x\ km/hr$ arid the speed of an ordinary train is $12\ km/hr$ less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of $240\ km,$ find the speed of the express train.
AnswerLet the speed of express train is $x\ Km/hr.$
Speed of ordinary train is $(x - 12)\ km/hr.$
Time require to cover for each train is $\frac{240}{x}$ and $\frac{240}{x-12}$ respectively.
According to question
$\frac{240}{x-12}-\frac{240}{x}=1$
$ \frac{240 x-240(x-12)}{(x-12)(x)}=1 $
$ 240 x-240(x-12)=x(x-12) $
$ x^2-12 x-2880=0$
$ (x-60)(x+48)=0 $
$\therefore x=60 km / hr .$
Speed of the express train is $60\ km/hr.$
View full question & answer→Question 283 Marks
Solve the following equation by reducing it to quadratic equation:
$\sqrt{3 x^2-2}+1=2 x$.
Answer$\sqrt{3 x^2-2}+1=2 x$
$\Rightarrow \sqrt{3 x^2-2} $
$ =2 x-1$
On squaring both sides, we get
$3 x ^2-2=4 x ^2+1-4 x $
$\Rightarrow- x ^2+4 x -3=0 $
$\Rightarrow x ^2-4 x +3=0$
$\Rightarrow x ^2-3 x - x +3=0 $
$\Rightarrow x ( x -3)-1( x -3)=0$
$ \Rightarrow( x -3)( x -1)=0$
$ \Rightarrow x =3 \text { or } x =1$
Hence, the solutions are $[3, 1].$
View full question & answer→Question 293 Marks
Solve for x : $9^{x+2}-6.3^{x+1}+1=0$
AnswerGiven equation $9^{x+2}-6.3^{x+1}+1=0$
$\Rightarrow 9^x \cdot 9^2-6 \cdot 3^x \cdot 3^1+1=0$
$\Rightarrow 81 \cdot\left(3^2\right)^x-18 \cdot 3^x+1=0$
$\Rightarrow 81 \cdot 3^{2 x}-18 \cdot 3^x+1=0$
Putting $3^x=y$, then it becomes $81 y^2-18 y+1=0$
$\Rightarrow 81 y^2-9 y-9 y+1=0$
$\Rightarrow 9 y(9 y-1)-1(9 y-1)=0$
$\Rightarrow(9 y-1)(9 y-1)=0$
$\Rightarrow 9 y=1$
$\Rightarrow y =\frac{1}{9}$
$\text { But } 3^{ x }=\frac{1}{9}=\frac{1}{3^2}=3^{-2}$
$\therefore x =-2$
Hence, the required root is $-2.$
View full question & answer→Question 303 Marks
Solve the following by reducing them to quadratic equations:
$z^4 - 10z^2 + 9 = 0.$
AnswerGiven equation $z^4 - 10z^2 + 9 = 0$
Putting $z^2 = x$, then given equation reduces to the form $x^2 - 10x + 9 = 0$
$\Rightarrow x^2 - 9x - x + 9 = 0$
$\Rightarrow x(x - 9) -1(x - 9) = 0$
$\Rightarrow (x - 9) (x - 1) = 0$
$\Rightarrow x - 9 = 0$ or $x - 1 = 1$
$\Rightarrow x = 9$ or $x = 1$
But $z^2 = x$
$\therefore z^2 = 9$
$\Rightarrow z = ±3$
or
$z^2 = 1$
$z = ±1$
Hence, the required roots are $±3, ±1.$
View full question & answer→Question 313 Marks
Solve the following by reducing them to quadratic equations:
$x^4 - 26x2 + 25 = 0$
AnswerGiven $x^4-26 x^2+25=0$
Putting $x^2=y$, the given equation reduces to the form $y^2-26 y+25=0$
$\Rightarrow y^2 - 25y - y + 25 = 0$
$\Rightarrow y(y - 25) -1(y- 25) = 0$
$\Rightarrow (y - 25) (y - 1) = 0$
$\Rightarrow y - 25 = 0$ or $y - 1 = 0$
$\Rightarrow y = 25$ or $y = 1$
$\therefore x^2 = 25$
$\Rightarrow x = ± 5$
or
$x^2 = 1$
$x = ±1$
Hence, the required roots are $\pm 5, \pm 1$.
View full question & answer→Question 323 Marks
Solve the equation $3x^2 – x – 7 = 0$ and give your answer correct to two decimal places.
Answer$3 x^2-x-7=0 $
$ a=3, b=-1, c=-7$
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} $
$ =-\frac{(-1) \pm \sqrt{(-1)^2-4.3 \cdot(-7)}}{2 \times 3} $
$ =\frac{1 \pm \sqrt{1+84}}{6} $
$ =\frac{1 \pm \sqrt{85}}{6}$
$ =\frac{1 \pm 9 \cdot 216}{6}$
$x=\frac{1+9.216}{6} \text { and } \frac{1-9.216}{6} $
$=\frac{10.216}{6} \text { and }-\frac{8.216}{6}$
$= 1.703$ and $-1.37$
View full question & answer→Question 333 Marks
Form the quadratic equation whose root are:
$2+\sqrt{5}$ and $2-\sqrt{5}$
AnswerLet $a, \beta$ be the given roots.
$\text { Then } a=2+\sqrt{5} \text { and } \beta=2-\sqrt{5} $
$a+\beta=2+\sqrt{5}+2-\sqrt{5}=4$
$ \text { and } a \beta=(2+\sqrt{5})(2-\sqrt{5})$
$\Rightarrow a+\beta=4 \text { and } a \beta=(2)^2-(\sqrt{5})^2$
$ \Rightarrow a+\beta=4 \text { and } a \beta=4-5$
$ \Rightarrow a+\beta=4 \text { and } a \beta=-1$
Required quadratic equation
$x^2-(a+\beta) x+a \beta=0$
$ \Rightarrow x^2-4 x-1=0$
View full question & answer→Question 343 Marks
Solve the quadratic equation:
$4 \sqrt{5} x^2+7 x-3 \sqrt{5}=0$
AnswerThe given equation is
$4 \sqrt{5} x^2+7 x-3 \sqrt{5}=0 . $
$\Rightarrow 4 \sqrt{5} x^2+12 x-5 x-3 \sqrt{5}=0 $
$\Rightarrow 4 x(\sqrt{5} x+3)-\sqrt{5}(\sqrt{5} x+3)=0 $
$\Rightarrow(\sqrt{5} x+3)(4 x-\sqrt{5})=0 $
$\Rightarrow \sqrt{5} x+3=0 \text { or } 4 x-\sqrt{5}=0 $
$\Rightarrow \sqrt{5} x=-3 \text { and } 4 x=\sqrt{5}$
$\Rightarrow x =-\frac{3}{\sqrt{5}} \text { and } x=\frac{\sqrt{5}}{4} $
$ \text { so } x =-\frac{3}{\sqrt{5}}, \frac{\sqrt{5}}{4}$
View full question & answer→Question 353 Marks
Find whether the value $x=\frac{1}{a^2}$ and $x=\frac{1}{b^2}$ are the solution of the equation:
$a^2 b^2 x^2-\left(a^2+b^2\right) x+1=0, a \neq 0, b \neq 0$
Answer$
a^2 b^2 x^2-\left(a^2+b^2\right) x+1=0 ; x=\frac{1}{a^2}, x=\frac{1}{b^2}
$
By putting $x=\frac{1}{2}$ in L.H.S. of equation
L.H.S. $=a^2 b^2 \times\left(\frac{1}{a^2}\right)^2-\left(a^2+b^2\right) \times \frac{1}{a^2}+1$
$=\frac{b^2}{a^2}-1-\frac{b^2}{a^2}+1=0=$ R.H.S.
By Putting $x =\frac{1}{b^2}$, in L.H.S. of equation
L.H.S. $=a^2 b^2 \times\left(\frac{1}{b^2}\right)^2-\left(a^2+b^2\right) \times \frac{1}{b^2}+1$
$=\frac{a^2}{b^2}-\frac{a^2}{b^2}-1+1=0=$ R..H.S.
Hence, $x=\frac{1}{a^2}, \frac{1}{b^2}$ are the solution of the equation.
View full question & answer→Question 363 Marks
Determine whether the given values of x is the solution of the given quadratic equation below:
$6 x^2-x-2=0 ; x=\frac{2}{3},-1$
Answer$6 x^2-x-2=0 ; x=\frac{2}{3},-1$
Now put $x = -1$ in $\text{L.H.S.}$ of equation.
$\text { L.H.S. }=6 \times(-1)^2-(-1)-2$
$ =6+1-2 $
$ =7-2=5 \neq 0 \neq \text { R.H.S. }$
Hence, $x = -1$ is not a root of the equation.
Put $x=\frac{2}{3}$ in $\text{L.H.S.}$ of equation.
$\text { L.H.S. }=6 \times\left(\frac{2}{3}\right)^2-\frac{2}{3}-2$
$=\frac{24}{9}-\frac{2}{3}-2 $
$ =\frac{8}{3}-\frac{2}{3}-2=0$
$ =8-8=0 $
$ =\text { R.H.S. }$
Hence, $x=\frac{2}{3}$ is a solution of given equation.
View full question & answer→Question 373 Marks
If x = 2 and x = 3 are roots of the equation 3x² – 2kx + 2m = 0. Find the values of k and m.
Answerx = 2 is a root of given equation
substitute x = 2 in L.H.S.
$\text { L.H.S. }=3(2)^2-2 k \times 2+2 m=0$
$12-4 k+2 m=0$
$4 k-2 m=12 \quad \ldots \text { (i) }$
Similarly when x = 2 is root of given equation
Substitute x = 3 in L.H.S.
$\text { L.H.S. }=3(3)^2-2 k \times 3+2 m=0$
$27-6 k+2 m=0$
$6 k-2 m=27 \quad \text {...(ii) }$
On solving equation (i) and (ii), we get
$k =\frac{15}{2}$ and $m =9$
View full question & answer→Question 383 Marks
A two digit number is such that the product of its digit is $14.$ When $45$ is added to the number, then the digit interchange their places. Find the number.
AnswerLet the ten's digit be x, then unit's digi$=\left(\frac{14}{x}\right)$
Then, the number is $\left(10 x+\frac{14}{x}\right)$
Where $45$ is added to the number, the digits get interchanged.
$\therefore 10 x+\frac{14}{x}+45=10 \times \frac{14}{x}+x $
$ \Rightarrow x^2+5 x-14=0 $
$ \Rightarrow(x+7)(x+2)=0$
$ \Rightarrow x=2$
and $x = -7$ (inadmissible)
Hence, the number is $\left(10 x+\frac{14}{x}\right)$
$=\left(10 \times 2+\frac{14}{2}\right)$
$= 27$
View full question & answer→Question 393 Marks
Five years ago, a woman’s age was the square of her son’s age. Ten years later her age will be twice that of her son’s age. Find:
The present age of the woman.
AnswerLet the age of son be $x$ years five years ago.
$\therefore$ Mother's age be $x ^2$ years five years ago.
After ten years son's age be $(x+15)$ years and woman's age $\left(x^2+15\right)$
Given $x^2+15=2(x+15)$
$x^2 + 15 = 2x + 30$
$x^2 - 2x - 15 = 0$
$(x - 5) (x + 3) = 0$
$x = 5$
Or $x = -3$ (not possible)
$\therefore$ Woman's present age
$= 25 + 5$
$= 30 years.$
View full question & answer→Question 403 Marks
Five years ago, a woman’s age was the square of her son’s age. Ten years later her age will be twice that of her son’s age. Find:
The age of the son five years ago.
AnswerLet the age of son be x years five years ago.
$\therefore$ Mother's age be $x^2$ years five years ago.
After ten years son's age be $(x + 15)$ years and woman's age $(x^2 + 15)$
Given $x^2 + 15 = 2(x + 15)$
$x^2 + 15 = 2x + 30$
$x^2 - 2x - 15 = 0$
$(x - 5) (x + 3) = 0$
$x = 5$
Or $x = -3$ (not possible)
$\therefore$ Son's age five years ago $= 5$years.
View full question & answer→Question 413 Marks
A two digit positive number is such that the product of its digits is $6.$ If $9$ is added to the number, the digits interchange their places. Find the number.
AnswerLet the unit's digit be x then tens digit will be $\frac{6}{x}$, then two digit number is $\frac{60}{x}+x$.
From question,
$\frac{60}{x}+x+9=10 x+\frac{6}{x}$
$60+x^2+9 x=10 x^2+6$
$9 x^2-9 x-54=0$
$x^2-x-6=0$
$x^2-3 x+2 x-6=0$
$x(x-3)+2(x-3)=0$
$(x-3)(x+2)=0$
$\Rightarrow x=-2 \text { or } 3$
As x can't be-ve
So, required two digit number
$=\frac{60}{3}+3 $
$ =23$
View full question & answer→Question 423 Marks
Divide $29$ into two parts so that the sum of the square of the parts is $425.$
AnswerLet the parts be $x$ and $29-x$
According to the problem
$x^2 + (29 - x)^2 = 425$
$\Rightarrow x^2 + 841 + x^2 - 58x - 425 = 0$
$\Rightarrow 2x^2 - 58x + 416 = 0$
$\Rightarrow x^2- 29x + 208 = 0$
$\Rightarrow x^2 - 16x - 13x + 208 = 0$
$\Rightarrow x(x - 16) - 13(x - 16) = 0$
$\Rightarrow (x - 16) (x - 13) = 0$
$\Rightarrow x - 16 = 0$ or $x - 13 = 0$
$\Rightarrow x = 16$ or $x = 13$
When $x=16$ When $x=13$
Then $29-x=13$ Then $29-x=16$
Hence, the parts are $16$ and $13.$
View full question & answer→Question 433 Marks
Find two consecutive positive even integers whose squares have the sum $340.$
AnswerLet two consecutive positive even integers be $2x, 2x + 2$
$\therefore (2x)^2 + (2x + 2)^2 = 340$
$\Rightarrow 4x^2+ 4x^2 + 4 + 8x = 340$
$\Rightarrow 8x^2 + 8x - 336 = 0$
$\Rightarrow x^2+ x - 42 = 0$
$\Rightarrow x^2 + 7x - 6x - 42 = 0$
$\Rightarrow x(x + 7) -6(x + 7) = 0$
$\Rightarrow (x + 7) (x + 6) = 0$
$\Rightarrow x + 7 = 0$ or $x - 6 = 0$
$\Rightarrow x = -7$ or $x = 6$
Negative integer is not required, therefore,$ x = 6.$
Hence, integers are $6 x 2, (6 x 2)+ 2.$
i.e., $12$ and $14.$
View full question & answer→Question 443 Marks
Determine, if $3$ is a root of the given equation
$\sqrt{x^2-4 x+3}+\sqrt{x^2-9}=\sqrt{4 x^2-14 x+16}$.
AnswerSubstituting $x = 3$ in the given eqution
$\text { L.H.S. }=\sqrt{(3)^2-4 \times 3+3}+\sqrt{(3)^2-9} $
$ =\sqrt{9-12+3}+\sqrt{9-9} $
$ =0+0 $
$ =0$
$\text { R.H.S. }=\sqrt{4(3)^2-14 \times 3+16} $
$ =\sqrt{36-42+16}$
$=\sqrt{52-42} $
$ =\sqrt{10}$
Since, $L.H.S. \neq R.H.S.$
Therefore, $x = 3$ is not a root of the given equation.
View full question & answer→Question 453 Marks
If an integer is added to its square the sum is $90$. Find the integer with the help of a quadratic equation.
AnswerLet the required integer be $x.$
Then according to the given condition
$\Rightarrow x + x^2 = 90$
$\Rightarrow x^2+ x - 90 = 0$
$\Rightarrow x^2 + 10x - 9x - 90 = 0$
$\Rightarrow x(x + 10) - 9(x +10) = 0$
$\Rightarrow (x +10) (x - 9) = 0$
$\Rightarrow x + 10 = 0$ or $x - 9 = 0$
$\Rightarrow x = -10$ or $x = 9$
Hence, the required integer is $9$ or $-10.$
View full question & answer→Question 463 Marks
Find two consecutive natural numbers whose squares have the sum $221.$
AnswerLet the number be $x_1 x+1$
Then $x^2+(x+1)^2=221$
$\Rightarrow x^2 + x^2 + 1 + 2x - 221 = 0$
$\Rightarrow 2x^2 + 2x - 220 = 0$
$\Rightarrow x^2 + x - 110 = 0$
$\Rightarrow x^2 + 11x - 10x - 110 = 0$
$\Rightarrow x(x + 11) - 10(x + 11) = 0$
$\Rightarrow (x = 11) (x - 10) = 0$
$\Rightarrow x = -11 or x = 10$
But $x=-11$ is rejected ...[ $\because$ It cannot been as its is a natural number]
$\therefore x=10$
Hence, required numbers are 10, $10+1$.
i.e., $10$ and $11.$
View full question & answer→Question 473 Marks
Solve $x^{2/3}+ x^{1/3} - 2 = 0.$
AnswerGiven equation is $x^{2/3}+ x^{1/3} - 2 = 0$
Putting $x^{1/3} = y,$ the given equation becomes
$y^2 + y - 2 = 0$
$\Rightarrow y^2 + 2y - y - 2 = 0$
$\Rightarrow y(y + 2) - 1(y + 2) = 0$
$\Rightarrow (y + 2) (y - 1) = 0$
$\Rightarrow y + 2 = 0$ or $y - 1 = 0$
$\Rightarrow y = -2$ or $y = 1$
But $x^{1/3} = y$
$\therefore x^{1/3} = -2 or x^{1/3} = 1$
$\Rightarrow x =(-2)^3$ or $x = (1)^3$
$\Rightarrow x = -8$ or $x = 1$
Hence, roots are$ -8, 1.$
View full question & answer→Question 483 Marks
If one root of the equation $2x^2– px + 4 = 0$ is $2,$ find the other root. Also find the value of $p.$
AnswerThe given quadratic equation is
$2x^2 – px + 4 = 0$
one root $= 2$
Let the other root be a
The sum of the roots $2 + a = \frac{-(-p)}{2}=\frac{p}{2}$
$a =\frac{p}{2}-2 \quad \ldots (i)$
The product of the roots
$a \times 2=\frac{4}{2}=2$
$ \Rightarrow a=1$
$\text { Now } 2+a=\frac{p}{2}$
$ \Rightarrow 2+1=\frac{p}{2}$
$ \Rightarrow p=6$
View full question & answer→Question 493 Marks
Find the value of k for which the following equation has equal roots:
$(k - 12)x^2+ 2(k - 12)x + 2 = 0.$
AnswerThe given equation is
$(k-12) x^2+2(k-12) x+2=0$
Here, $a=k-12, b=2(k-12)$ and $c =2$
Since, the given equation has two equal real roots
then we must have $b^2-4 a c=0$
$\Rightarrow [2(k - 12)]^2 - 4(k - 12) x 2 = 0$
$\Rightarrow 4(k - 12)^2 - 8(k - 12) = 0$
$\Rightarrow 4(k - 12) {k - 12 - 2} = 0$
$\Rightarrow (k - 12) (k - 14) = 0$
$\Rightarrow k - 12 = 0$ or $k - 14 = 0$
$\Rightarrow k = 12$ or $k = 14.$
Note: But at $k =12$, terms of $x ^2$ and x in the equation vanish hence only $k =14$ is acceptable.
View full question & answer→Question 503 Marks
Without solving the following quadratic equation, find the value of ‘p’ for which the given equation has real and equal roots:
$x^2 + (p – 3) x + p = 0$
Answer$x^2 + (p – 3) x + p = 0$
Here $a = 1, b = p - 3, c = p$
For real and equal roots
$D = b^2 - 4ac = 0$
$(p - 3)^2 - 4 \times 1 \times p = 0$
$p^2 - 6p + 9 - 4p = 0$
$p^2 - 10p + 9 = 0$
$\Rightarrow p^2 - p - 9 (p - 1) = 0$
$p(p - 1) - 9 (p - 1) = 0$
$\Rightarrow (p - 1) (p - 9) = 0$
$\Rightarrow p = 1$ or $p = + 9$
View full question & answer→