1 Marks Question

Question 11 Mark

Find the nature of the roots of the quadratic equation $2 x^2-3 x+5=0$. If the real roots exist. Find it.

Answer

The 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 qiven equation has no real roots.

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Question 21 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).

Answer

Let 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.
$\therefore(x-5)(40-x)=124$
$\Rightarrow x^2-45 x+324=0$

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Question 31 Mark

Find the roots of the quadratic equation $100x^2- 20x + 1 = 0$ by factorization.

Answer

We 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.

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Question 41 Mark

Find the roots of the quadratic equation $2x^2-x+{1 \over8} = 0$ by factorization.

Answer

We 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 ${1\over4}$.

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Question 51 Mark

Find the roots of the quadratic equation $\sqrt { 2 } x ^ { 2 } + 7 x + 5 \sqrt { 2 } = 0$ by factorization.

Answer

We 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 $\frac{-5}{\sqrt{2}}$

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Question 61 Mark

Find the roots of the quadratic equation $2x^2 + x – 6 = 0$ by factorization.

Answer

We have, $2 x^2+x-6=0$
$\Rightarrow 2 x^2+4 x-3 x-6=0$
$\Rightarrow 2 x(x+2)-3(x+2)=0$
$\Rightarrow(2 x-3)(x+2)=0$
Either $2 x -3=0$ or $x +2=0 \quad x=\frac{3}{2},-2$ $\therefore x=\frac{3}{2},-2$ are the required roots.

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Question 71 Mark

Find the roots of the quadratic equation $x^2-3 x-10=0$ by factorization.

Answer

$x^2−3x - 10=0$
$\Rightarrow x^2−5x +2x - 10=0$
$\Rightarrow x (x−5)+2(x−5)=0$
$\Rightarrow (x−5)(x +2)=0$
$\Rightarrow x =5,−2$

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Question 81 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.

Answer

Let 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.

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Question 91 Mark

Check whether the given equation is quadratic equation or not: $x^3 - 4x^2 - x + 1 = (x - 2)^3$

Answer

We have given that, $x^3 - 4x^2 - x + 1 = (x - 2)^3$
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.

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Question 101 Mark

Check the equation is quadratic equation or not:$ (x + 2)^3 = 2x(x^2- 1)$

Answer

We have the following equation,
$(x+2)^3=2 x\left(x^2-1\right)$
$\Rightarrow x^3+8+6 x(x+2)=2 x^3-2 x$
$\Rightarrow x^3+6 x^2+12 x+8=2 x^3-2 x$
$\Rightarrow x^3-6 x^2-14 x-8=0$
Cleary, It is not in the form of $ax ^2+ bx + c =0$.
Therefore, it is not a quadratic equation.

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Question 111 Mark

Check the equation is quadratic equation: $x^2+3 x+1=(x-2)^2$

Answer

We have $x^2+3 x+1=(x-2)^2$
$\Rightarrow x^2+3 x+1=x^2-4 x+4$
$\Rightarrow 7 x-3=0$
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.

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Question 121 Mark

Check the equation is quadratic equation or not: (2x - 1) (x - 3) = (x + 5) (x - 1)

Answer

$\text { We have }(2 x-1)(x-3)=(x+5)(x-1)$
$\Rightarrow 2 x^2-6 x-x+3=x^2-x+5 x-5$
$\Rightarrow 2 x^2-7 x+3=x^2+4 x-5$
$\Rightarrow x^2-11 x+8=0$
which is of the form $a x^2+b x+c=0, a \neq 0$
$\therefore$ the given equation is a quadratic equation.

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Question 131 Mark

Check the equation is quadratic equation or not: (x - 3) (2x + 1) = x(x + 5)

Answer

The given equation is (x - 3) (2x +1) = x (x+5)
$\Rightarrow$$ 2x^2 + x - 6x - 3 = x^2 + 5x$
$\Rightarrow$ $2x^2 - 5x - 3 = x^2 + 5x$
$\Rightarrow$$ x^2 - 10x - 3 = 0$
It is in the form of $ax^2 + bx + c = 0$, $a\ne 0$
$\therefore$ the given equation is a quadratic equation.

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Question 141 Mark

Check the equation is quadratic equation or not: (x - 2) (x + 1) = (x - 1) (x + 3)

Answer

Let P be the position of the pole and A & B be the opposite fixed gates. Let, BP = x metres.
$\therefore$ AP = x + 7
In right triangle APB,

$A P^2+B P^2=A B^2$
$\Rightarrow(x+7)^2+x^2=13^2$
$\Rightarrow x^2+49+14 x+x^2=169$
$\Rightarrow 2 x^2+14 x+49-169=0$
$\Rightarrow 2 x^2+14 x-120=0$
$\Rightarrow 2\left(x^2+7 x-60\right)=0$
$\Rightarrow x^2+7 x-60=0$
$\Rightarrow x^2+12 x-5 x-60=0$
$\Rightarrow x(x+12)-5(x+12)=0$
$\Rightarrow(x+12)(x-5)=0$
Either x+12 = 0, then x = -12 which is not possible being negative or x - 5 = 0, then x = 5.
Thus P is at a distance of 5m from B and 5+7 = 12m from A.

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Question 151 Mark

Check the equation is quadratic equation or not: $x^2 - 2x = (-2)(3 - x)$

Answer

$x^2−2x =(−2)(3−x )$
$\Rightarrow x^2−2x =−6+2x$
$\Rightarrow x^2−2x−2x +6=0$
$\Rightarrow x^2−4x +6=0$
Here, degree of equation is 2.
Therefore, it is a Quadratic Equation.

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Question 161 Mark

Check the equation is quadratic equation or not: $(x + 1)^2= 2(x – 3)$

Answer

The given equation is $(x+1)^2=2(x-3) \Longrightarrow x^2+2 x+1-2 x+6=0$
$\Longrightarrow x^2+7=0$
$\Longrightarrow x^2+0 \cdot x+7=0$
Which is of the form $a x^2+b x+c=0$
Hence, the given equation is a quadratic equation.

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Question 171 Mark

Find the roots of the quadratic equation $6x^2 - x - 2 = 0.$

Answer

$6x^2- x - 2 = 0$
$or, 6x^2 + 3x - 4x - 2 = 0$
$or, 3x(2x + 1) -2(2x + 1) = 0$
$or, (2x + 1)(3x - 2) = 0$
$\Rightarrow$ 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 $\frac { 2 } { 3 } \text { and } - \frac { 1 } { 2 }$.

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Question 181 Mark

Find the roots of the quadratic equation $2 x^2-5 x+3=0$ by factorisation.

Answer

Let us first split the middle term $-5 x$ as $-2 x,-3 x$ [because $(-2 x) \times(-3 x)=6 x^2=\left(2 x^2\right) \times 3$ ] So, $2 x^2-5 x+3=2 x^2-2 x-3 x+3=2 x(x-1)-3(x-1)=(2 x-3)(x-1)$
Now, $2 x^2-5 x+3=0$ can be rewritten as $(2 x-3)(x-1)=0$
So, the values of $x$ for which $2 x^2-5 x+3=0$ are the same for which $(2 x-3)(x-1)=0$
i.e., either $2 x-3=0$ or $x-1=0$.
Now, $2 x-3=0$ gives $x=\frac{3}{2}$ and $x-1=0$ gives $x=1$.
So, $x=\frac{3}{2}$ and $x=1$ are the roots of the given quadratic equation.

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Question 191 Mark

Check the equation is quadratic equation or not: $(x+2)^3=x^3-4$

Answer

Here, 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.

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Question 201 Mark

Check the equation is quadratic equation or not: $x(2 x+3)=x^2+1$

Answer

Taking, 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 $a^2+b x+c=0$. Also degree is 2 . So, the given equation is a quadratic equation.

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Question 211 Mark

Check the equation is quadratic equation or not: x(x + 1) + 8 = (x + 2) (x – 2)

Answer

Since $x(x+1)+8=x^2+x+8$ 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.

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Question 221 Mark

Check the equation is quadratic equation or not: $(x-2)^2+1=2 x-3$

Answer

Taking 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 $a x^2+b x+c=0$. Also degree of equation is 2
Therefore, the given equation is a quadratic equation.

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Question 231 Mark

Find the dimensions of the prayer hall given in the below figure.

Answer

Let 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
$\Rightarrow x^2 – 55x + 750 = 0$

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