Question 11 Mark
Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is $800\ m^2?$ If so, find its length and breadth.
AnswerLet the breadth of the rectangle be $x$ metres
and the length is $2x$ metres.
So, area of rectangle is $800\ sq.m$.
$(x)(2x) = 800$
$2x^2= 800$
$x^2= 400$
$x = \pm 20$
But breadth of rectangle cannot be negative, so $x = 20$
and yes, it is possible to design it.
So, Breadth is $20\ m$ and length is $40\ m.$
View full question & answer→Question 21 Mark
Find the nature of the roots of the quadratic equation $3 x ^ { 2 } - 4 \sqrt { 3 } x + 4 = 0$. If the real roots exist. Find it.
AnswerThe given quadratic equation is
$3 x ^ { 2 } - 4 \sqrt { 3 } x + 4 = 0$
Here, $a = 3,$ $b = - 4 \sqrt { 3 }$, $c = 4$
$\therefore$ discriminant $= b^2- 4ac$
$= ( - 4 \sqrt { 3 } ) ^ { 2 } - 4 ( 3 ) ( 4 )$
$= 48 - 48 = 0$
Hence, the given quadratic equation
has two equal real roots.
The roots are $= - \frac { b } { 2 a } , - \frac { b } { 2 a }$
$\text { i.e. } - \frac { ( - 4 \sqrt { 3 } ) } { 2 \times 3 } , - \frac { ( - 4 \sqrt { 3 } ) } { 2 \times 3 } , \text { i.e. } \frac { 2 } { \sqrt { 3 } } , \frac { 2 } { \sqrt { 3 } }$
View full question & answer→Question 31 Mark
Find the nature of the roots of the quadratic equation $2x^2- 3x + 5 = 0$. If the real roots exist. Find it.
AnswerThe given equation is
$2 x^2-3 x+5=0$
Here, $a=2, b=-3, c=5$
Therefore, discriminant $=b^2-4 a c$
$ =(-3)^2-4(2)(5) $
$=9-40=-31<0$
So, the given equation has no real roots.
View full question & answer→Question 41 Mark
Find two numbers whose sum is $27$ and product is $182$.
AnswerLet the two numbers be $x, 27-x$.
According to the question $x(27-x)=182 $
$\Longrightarrow 27 x-x^2-182=0$
$ \Longrightarrow x^2-27 x+182=0$
$\Longrightarrow x^2-13 x-14 x+182=0$
$ \Longrightarrow x(x-13)-14(x-13)=0 $
$\Longrightarrow(x-13)(x-14)=0$
Either $x-13=0$ or $x-14=0 $
$\Longrightarrow x=13,14$
Hence, the required numbers are $13,14$
View full question & answer→Question 51 Mark
John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with. Represent situation mathematically (quadratic equation).
AnswerLet the number of John’s marbles be $x.$
Therefore, number of Jivanti’s marble $= 45 − x$
After losing 5 marbles,
Number of John’s marbles $= x − 5$
Number of Jivanti’s marbles $= 45 − x − 5 = 40 − x$
Given that the product of their marbles is $124.$
$\therefore (x – 5) (40 – x) = 124$
$\Rightarrow x^2– 45x +324 = 0$
View full question & answer→Question 61 Mark
Find the roots of the quadratic equation $100x^2- 20x + 1 = 0$ by factorization.
AnswerWe have, $100x^2- 20x +1 = 0$
$\Rightarrow 100x^2-10x -10x +1 = 0$
$\Rightarrow 10x (10x -1) - 1 (10x -1) = 0$
$\Rightarrow (10x -1) (10x -1) = 0$
Either $10x -1 = 0$ or $10x -1 = 0$
$\Rightarrow$ x = ${1\over10},\,{1\over10}$
$\therefore x={1\over10}$ are the repeated roots.
View full question & answer→Question 71 Mark
Find the roots of the quadratic equation $2x^2-x+{1 \over8} = 0$ by factorization.
AnswerWe have, $2x^2-x+{1\over8}=0$
$\implies 2x^2-{1\over2}x-{1\over2}x+{1\over8}=0$
$\implies x(2x-{1\over2}) - {1\over4}(2x-{1\over2})=0$
$\implies (2x-{1\over2}) (x-{1\over4})=0$
$Either \,(2x-{1\over2})\, =0\,or\, (x-{1\over4})=0$
$\implies x = {1\over4},\,{1\over4}$
So, this root is repeated root.
$\therefore $ both the roots are ${1\over4}$.
View full question & answer→Question 81 Mark
Find the roots of the quadratic equation $\sqrt { 2 } x ^ { 2 } + 7 x + 5 \sqrt { 2 } = 0$ by factorization.
AnswerWe have, $\sqrt { 2 } x ^ { 2 } + 7 x + 5 \sqrt { 2 } = 0$
$\Rightarrow \quad \sqrt { 2 } x ^ { 2 } + 2 x + 5 x + 5 \sqrt { 2 } = 0$
$\Rightarrow \sqrt { 2 } x ( x + \sqrt { 2 } ) + 5 ( x + \sqrt { 2 } ) = 0$
$\Rightarrow \quad ( x + \sqrt { 2 } ) ( \sqrt { 2 } x + 5 ) = 0$
$\Rightarrow \quad x = - \sqrt { 2 }$ and $\frac{-5}{\sqrt{2}}$
View full question & answer→Question 91 Mark
Find the roots of the quadratic equation $2x^2+ x – 6 = 0$ by factorization.
AnswerWe have,$ 2x^2+x - 6 = 0$
$\Rightarrow 2x^2+4x -3x - 6 = 0$
$\Rightarrow 2x (x + 2) - 3(x + 2) = 0$
$\Rightarrow (2x - 3) (x + 2) = 0$
Either $2x - 3 = 0 or x +2 = 0x = {3 \over 2},\, -2$
$\therefore x = {3 \over 2},\, -2$ are the required roots.
View full question & answer→Question 101 Mark
Find the roots of the quadratic equation $ x^2-3 x-10=0 $ by factorization.
Answer$ x^2-3 x-10=0 $
$ \Rightarrow x^2-5 x+2 x-10=0 $
$ \Rightarrow x(x-5)+2(x-5)=0 $
$ \Rightarrow(x-5)(x+2)=0 $
$ \Rightarrow x=5,-2$
View full question & answer→Question 111 Mark
Represent the situation in the form of the quadratic equation:
Rohan's mother is $26$ years older than him. The product of their ages $3$ years from now will be $360$. We would like to find Rohan's present age.
AnswerLet Rohan's present age be $x$ years.
Then, his mother's age is $(x + 26)$ years.
Rohan's age after $3$ years $= (x + 3)$ years.
After $3$ years the age of Rohan's mother $= (x + 26 + 3)$ years $= (x + 29)$ years.
According to the question,
$ (x + 3)(x + 29) = 360.$
$\Rightarrow x^2+ 32x - 273 = 0.$
This is the required quadratic equation.
View full question & answer→Question 121 Mark
Represent the situation in the form of the quadratic equation:
The area of a rectangular plot is $528\ m^2$. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
AnswerLet the breadth of the plot be $'x'$ m
$\therefore$ Length $= (2x + 1) m$
Now, Area of the plot $= 528\ m^2$
$\Rightarrow L \times B = 528\ m^2$
$\Rightarrow$ $(2x+1) \times x=528$
$\Rightarrow 2x^2+x-528=0$
This is the required quadratic equation.
View full question & answer→Question 131 Mark
Check whether the given equation is quadratic equation or not: $x^3-4 x^2-x+1=(x-2)^3$
AnswerWe have given that, $x^3-4 x^2-x+1=(x-2)^3$
Applying identity on $R.H.S.$ we get,
$(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}$
$\Rightarrow x^{3}-4 x^{2}-x+1=x^{3}-8-6 x^{2}+12 x$
$\Rightarrow x^{3}-4 x^{2}-x+1-x^{3}+8+6 x^{2}-12 x=0$
$\Rightarrow 2 x^{2}-13 x+9=0$
Degree of the equation is $2$ and
It is of the form $a x^{2}+b x+c=0,$ with $a \neq 0$ , therefore, the given equation is quadratic.
View full question & answer→Question 141 Mark
Check the equation is quadratic equation or not: $ (x+2)^3=2 x\left(x^2-1\right) $
AnswerWe have the following equation,
$ (x+2)^3=2 x\left(x^2-1\right) $
$ \Rightarrow x^3+8+6 x(x+2)=2 x^3-2 x $
$ \Rightarrow x^3+6 x^2+12 x+8=2 x^3-2 x $
$ \Rightarrow x^3-6 x^2-14 x-8=0$
$\text { Cleary, It is not in the form of } a x^2+b x+c=0 \text {. }$
Therefore, it is not a quadratic equation.
View full question & answer→Question 151 Mark
Check the equation is quadratic equation: $x^2+3 x+1=(x-2)^2$
AnswerWe have $x^2+3 x+1=(x-2)^2$
$\Rightarrow x^2+3 x+1=x^2-4 x+4$
$\Rightarrow 7x - 3 = 0$
It is not of the form $a x^2+b x+c=0, a \neq 0$
$\therefore$ the given equation is not a quadratic equation.
View full question & answer→Question 161 Mark
Check the equation is quadratic equation or not: $(2x - 1) (x - 3) = (x + 5) (x - 1)$
AnswerWe have $(2x -1) (x - 3) = (x + 5) (x - 1)$
$\Rightarrow 2 x^2-6 x-x+3=x^2-x+5 x-5 $
$ \Rightarrow 2 x^2-7 x+3=x^2+4 x-5$
$ \Rightarrow x^2-11 x+8=0$
$\text { which is of the form } a x^2+b x+c=0, a \neq 0$
$\therefore $ the given equation is a quadratic equation.
View full question & answer→Question 171 Mark
Check the equation is quadratic equation or not:$(x - 3) (2x + 1) = x(x + 5)$
AnswerThe given equation is $(x - 3) (2x +1) = x (x+5)$
$ \Rightarrow 2 x^2+x-6 x-3=x^2+5 x $
$ \Rightarrow 2 x^2-5 x-3=x^2+5 x$
$ \Rightarrow x^2-10 x-3=0$
$\text { It is in the form of } a x^2+b x+c=0, a \neq 0$
$\therefore$ the given equation is a quadratic equation.
View full question & answer→Question 181 Mark
Check the equation is quadratic equation or not: $(x - 2) (x + 1) = (x - 1) (x + 3)$
AnswerThe given equation is $(x-2) (x+1) = (x-1) (x+3)$
$ \Rightarrow x^2-2 x+x-2=x^2+3 x-x-3 $
$ \Rightarrow 3 x-1=0$
$ \text { Which is not of the form } a x^2+b x+c=0, a \neq 0$
Hence, the given equation is not a quadratic equation.
View full question & answer→Question 191 Mark
Check the equation is quadratic equation or not: $ x^2-2 x=(-2)(3-x)$
Answer$ x^2-2 x=(-2)(3-x)$
$ \Rightarrow x^2-2 x=-6+2 x $
$ \Rightarrow x^2-2 x-2 x+6=0$
$ \Rightarrow x^2-4 x+6=0$
Here, degree of equation is $2$.
Therefore, it is a Quadratic Equation.
View full question & answer→Question 201 Mark
Check the equation is quadratic equation or not: $(x + 1)^2= 2(x – 3)$
AnswerThe given equation is $(x+1)^2= 2 (x-3)$
$\implies x^2+ 2x + 1 - 2x + 6 = 0$
$\implies x^2+ 7 = 0$
$\implies x^2+ 0.x + 7 = 0$
Which is of the form $ax^2+ bx + c = 0$
Hence, the given equation is a quadratic equation.
View full question & answer→Question 211 Mark
Find the roots of the quadratic equation $6 x^2-x-2=0$.
Answer$6 x^2-x-2=0$
or, $6 x^2+3 x-4 x-2=0$
or, $3 x(2 x+1)-2(2 x+1)=0$
or, $(2 x+1)(3 x-2)=0$
$\Rightarrow$ either $3x - 2 = 0$ or $2x + 1 = 0$
$\therefore \quad x = \frac { 2 } { 3 } \text { or } x = - \frac { 1 } { 2 }$
Therefore, Roots of equation are $\frac { 2 } { 3 } \text { and } - \frac { 1 } { 2 }$.
View full question & answer→Question 221 Mark
Check the equation is quadratic equation or not: $(x + 2)^3= x^3– 4$
AnswerHere, $LHS$ $=(x+2)^3=x^3+6 x^2+12 x+8$
Therefore, $(x+2)^3=x^3-4$ can be rewritten as
$x^3+6 x^2+12 x+8=x^3-4$
i.e., $6 x^2+12 x+12=0$ or, $x^2+2 x+2=0$
It is of the form $a x^2+b x+c=0$.
So, the given equation is a quadratic equation.
View full question & answer→Question 231 Mark
Check the equation is quadratic equation or not: $x(2x + 3) = x^2+ 1$
AnswerTaking, L.H.S. $=x(2 x+3)=2 x^2+3 x$
So, $x(2 x+3)=x^2+1$ can be rewritten as
$2 x^2+3 x=x^2+1$
Therefore, we get $x^2+3 x-1=0$
It is of the form $a x^2+b x+c=0$. Also degree is $2$ .
So, the given equation is a quadratic equation.
View full question & answer→Question 241 Mark
Check the equation is quadratic equation or not: $x(x + 1) + 8 = (x + 2) (x – 2)$
AnswerSince $x(x+1)+8=x^2+x+8$ and $(x+2)(x-2)=x^2-4$
Therefore, $x^2+x+8=x^2-4$
i.e., $x+12=0$
It is not of the form $a x^2+b x+c=0$
Therefore, the given equation is not a quadratic equation.
View full question & answer→Question 251 Mark
Check the equation is quadratic equation or not: $(x - 2)^2+ 1 = 2x - 3$
AnswerTaking L.H.S. $=(x-2)^2+1=x^2-4 x+4+1=x^2-4 x+5$
Therefore, $(x-2)^2+1=2 x-3$ can be rewritten as
$x^2-4 x+5=2 x-3$
i.e., $x^2-6 x+8=0$
It is of the form $\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0$. Also degree of equation is $2$
Therefore, the given equation is a quadratic equation.
View full question & answer→Question 261 Mark
A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be $55$ minus the number of articles produced in a day. On a particular day, the total cost of production was $Rs.\ 750$. We would like to find out the number of toys produced on that day. Represent situation mathematically (quadratic equation)
AnswerLet the number of toys produced be $x.$
$\therefore$ Cost of production of each toy $= Rs\ (55 − x)$
It is given that, total production of the toys $= Rs\ 750$
$\therefore x(55 – x) = 750$
$⇒ x^2 – 55x + 750 = 0$
View full question & answer→