Examples - Chapter 4 Quadratic Equations class 10 ncert solutions Maths - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Examples (Revised) : Chapter 4 - Quadratic Equations - Ncert Solutions class 10 - Maths

Example 1 :

Represent the following situations mathematically:
(i) 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.
(ii) 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 toys produced in a day. On a particular day, the total cost of production was $₹ 750$. We would like to find out the number of toys produced on that day.

Solution :
(i) Let the number of marbles John had be $x$.
Then the number of marbles Jivanti had $=45-x$ (Why?).
The number of marbles left with John, when he lost 5 marbles $=x-5$
The number of marbles left with Jivanti, when she lost 5 marbles $=45-x-5$
$
=40-x
$

$
\begin{aligned}
\text { Therefore, their product } & =(x-5)(40-x) \\
& =40 x-x^2-200+5 x \\
& =-x^2+45 x-200
\end{aligned}
$

So, $\quad-x^2+45 x-200=124$
(Given that product $=124$ )
i.e., $\quad-x^2+45 x-324=0$
i.e. $\quad x^2-45 x+324=0$

Therefore, the number of marbles John had, satisfies the quadratic equation
$
x^2-45 x+324=0
$
which is the required representation of the problem mathematically.
(ii) Let the number of toys produced on that day be $x$.
Therefore, the cost of production (in rupees) of each toy that day $=55-x$
So, the total cost of production (in rupees) that day $=x(55-x)$
Therefore,
$
x(55-x)=750
$
i.e.,
$
55 x-x^2=750
$
i.e.,
$
-x^2+55 x-750=0
$
i.e.,
$
x^2-55 x+750=0
$

Therefore, the number of toys produced that day satisfies the quadratic equation
$
x^2-55 x+750=0
$
which is the required representation of the problem mathematically.

Example 2 :

Check whether the following are quadratic equations:
(i) $(x-2)^2+1=2 x-3$
(ii) $x(x+1)+8=(x+2)(x-2)$
(iii) $x(2 x+3)=x^2+1$
(iv) $(x+2)^3=x^3-4$

Solution :
(i) LHS $=(x-2)^2+1=x^2-4 x+4+1=x^2-4 x+5$
Therefore, $(x-2)^2+1=2 x-3$ can be rewritten as
$
\begin{aligned}
& x^2-4 x+5=2 x-3 \\
& x^2-6 x+8=0 \\
&
\end{aligned}
$
i.e.,

It is of the form $a x^2+b x+c=0$.
Therefore, the given equation is a quadratic equation.

(ii) 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.
(iii) Here,
$
\text { LHS }=x(2 x+3)=2 x^2+3 x
$

So,
$
\begin{aligned}
x(2 x+3) & =x^2+1 \text { can be rewritten as } \\
2 x^2+3 x & =x^2+1
\end{aligned}
$

Therefore, we get $x^2+3 x-1=0$
It is of the form $a x^2+b x+c=0$.
So, the given equation is a quadratic equation.
(iv) Here,
$
\text { 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.
Remark : Be careful! In (ii) above, the given equation appears to be a quadratic equation, but it is not a quadratic equation.

In (iv) above, the given equation appears to be a cubic equation (an equation of degree 3) and not a quadratic equation. But it turns out to be a quadratic equation. As you can see, often we need to simplify the given equation before deciding whether it is quadratic or not.

Example 3 :

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

Let us first split the middle term $-5 x$ as $-2 x-3 x$ [because $(-2 x) \times(-3 x)=$ $\left.6 x^2=\left(2 x^2\right) \times 3\right]$.
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 solutions of the equation.
In other words, 1 and $\frac{3}{2}$ are the roots of the equation $2 x^2-5 x+3=0$.
Verify that these are the roots of the given equation.

Note that we have found the roots of $2 x^2-5 x+3=0$ by factorising $2 x^2-5 x+3$ into two linear factors and equating each factor to zero.

Example 4 :

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

We have
$
\begin{aligned}
6 x^2-x-2 & =6 x^2+3 x-4 x-2 \\
& =3 x(2 x+1)-2(2 x+1) \\
& =(3 x-2)(2 x+1)
\end{aligned}
$

The roots of $6 x^2-x-2=0$ are the values of $x$ for which $(3 x-2)(2 x+1)=0$
Therefore, $3 x-2=0$ or $2 x+1=0$,
i.e.,
$
x=\frac{2}{3} \quad \text { or } \quad x=-\frac{1}{2}
$

Therefore, the roots of $6 x^2-x-2=0$ are $\frac{2}{3}$ and $-\frac{1}{2}$.
We verify the roots, by checking that $\frac{2}{3}$ and $-\frac{1}{2}$ satisfy $6 x^2-x-2=0$.

Example 5 :

Find the roots of the quadratic equation $3 x^2-2 \sqrt{6} x+2=0$.
Solution
$
\begin{aligned}
: 3 x^2-2 \sqrt{6} x+2 & =3 x^2-\sqrt{6} x-\sqrt{6} x+2 \\
& =\sqrt{3} x(\sqrt{3} x-\sqrt{2})-\sqrt{2}(\sqrt{3} x-\sqrt{2}) \\
& =(\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})
\end{aligned}
$

So, the roots of the equation are the values of $x$ for which
$
(\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})=0
$

Now, $\sqrt{3} x-\sqrt{2}=0$ for $x=\sqrt{\frac{2}{3}}$.
So, this root is repeated twice, one for each repeated factor $\sqrt{3} x-\sqrt{2}$.
Therefore, the roots of $3 x^2-2 \sqrt{6} x+2=0$ are $\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}$.

Example 6 :

Find the dimensions of the prayer hall discussed in Section 4.1.
Solution :

In Section 4.1, we found that if the breadth of the hall is $x \mathrm{~m}$, then $x$ satisfies the equation $2 x^2+x-300=0$. Applying the factorisation method, we write this equation as
$
\begin{aligned}
2 x^2-24 x+25 x-300 & =0 \\
\text { i.e., } \quad 2 x(x-12)+25(x-12) & =0 \\
(x-12)(2 x+25) & =0
\end{aligned}
$

So, the roots of the given equation are $x=12$ or $x=-12.5$. Since $x$ is the breadth of the hall, it cannot be negative.
Thus, the breadth of the hall is $12 \mathrm{~m}$. Its length $=2 x+1=25 \mathrm{~m}$.

Example 7:

Find the discriminant of the quadratic equation $2 x^2-4 x+3=0$, and hence find the nature of its roots.
Solution:

The given equation is of the form $a x^2+b x+c=0$, where $a=2, b=-4$ and $c=3$. Therefore, the discriminant
$
b^2-4 a c=(-4)^2-(4 \times 2 \times 3)=16-24=-8<0
$

So, the given equation has no real roots.
Example 8 :

A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

Solution :

Let us first draw the diagram (see Fig. 4.2).

Let $\mathrm{P}$ be the required location of the pole. Let the distance of the pole from the gate $\mathrm{B}$ be $x \mathrm{~m}$, i.e., $\mathrm{BP}=x \mathrm{~m}$. Now the difference of the distances of the pole from the two gates $=\mathrm{AP}-\mathrm{BP}($ or, $\mathrm{BP}-\mathrm{AP})=$ $7 \mathrm{~m}$. Therefore, $\mathrm{AP}=(x+7) \mathrm{m}$.

Now, $\mathrm{AB}=13 \mathrm{~m}$, and since $\mathrm{AB}$ is a diameter,
$
\angle \mathrm{APB}=90^{\circ} \quad \text { (Why?) }
$

Therefore, $\mathrm{AP}^2+\mathrm{PB}^2=\mathrm{AB}^2$
(By Pythagoras theorem)
i.e.,
$
(x+7)^2+x^2=13^2
$

i.e.,
$
\begin{aligned}
x^2+14 x+49+x^2 & =169 \\
2 x^2+14 x-120 & =0
\end{aligned}
$

So, the distance ' $x$ ' of the pole from gate $\mathrm{B}$ satisfies the equation
$
x^2+7 x-60=0
$

So, it would be possible to place the pole if this equation has real roots. To see if this is so or not, let us consider its discriminant. The discriminant is
$
b^2-4 a c=7^2-4 \times 1 \times(-60)=289>0 \text {. }
$

So, the given quadratic equation has two real roots, and it is possible to erect the pole on the boundary of the park.
Solving the quadratic equation $x^2+7 x-60=0$, by the quadratic formula, we get

$
x=\frac{-7 \pm \sqrt{289}}{2}=\frac{-7 \pm 17}{2}
$

Therefore, $x=5$ or -12 .
Since $x$ is the distance between the pole and the gate B, it must be positive. Therefore, $x=-12$ will have to be ignored. So, $x=5$.

Thus, the pole has to be erected on the boundary of the park at a distance of $5 \mathrm{~m}$ from the gate $B$ and $12 \mathrm{~m}$ from the gate $A$

Example 9 :

Find the discriminant of the equation $3 x^2-2 x+\frac{1}{3}=0$ and hence find the nature of its roots. Find them, if they are real.
Solution :

Here $a=3, b=-2$ and $c=\frac{1}{3}$.
Therefore, discriminant $b^2-4 a c=(-2)^2-4 \times 3 \times \frac{1}{3}=4-4=0$.
Hence, the given quadratic equation has two equal real roots.
The roots are $\frac{-b}{2 a}, \frac{-b}{2 a}$, i.e., $\frac{2}{6}, \frac{2}{6}$, i.e., $\frac{1}{3}, \frac{1}{3}$.