Quadratic Equations - Solved Examples

Exercise 4.3

Exercise 4.1

(i) $x^2 - 3x - 10 = 0$

$⇒ x^2 - 5x + 2x - 10 = 0$
$⇒ x(x - 5) + 2(x - 5) = 0$
$⇒ (x - 5)(x + 2) = 0$

i.e. either $x - 5 = 0$ or $x + 2 = 0$
⇒ $x = 5$ or $x = -2$

Hence the roots of given quadratic equation are $5$ and $-2$

(ii) $2x^2 + x - 6 = 0$

$⇒ 2x^2 + 4x - 3x - 6 = 0$
$⇒ 2x(x + 2) - 3(x + 2) = 0$
$⇒ (x + 2)(2x - 3) = 0$

i.e. either $x + 2 = 0$ or $2x - 3 = 0$
⇒ $x = -2$ or $x = \dfrac{3}{2}$

Hence the roots of given quadratic equation are $-2$ and $\dfrac{3}{2}$

(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$

$⇒ \sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0$
$⇒ x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0$
$⇒ (x + \sqrt{2})(\sqrt{2}x + 5) = 0$

i.e. either $x + \sqrt{2} = 0$ or $\sqrt{2}x + 5 = 0$
⇒ $x = -\sqrt{2}$ or $x = -\dfrac{5}{\sqrt{2}}$

Hence the roots of given quadratic equation are $-\sqrt{2}$ and $-\dfrac{5}{\sqrt{2}}$

(iv) $2x^2 - x + \dfrac{1}{8} = 0$

$⇒ \dfrac{1}{8}(16x^2 - 8x + 1) = 0$

$⇒ \dfrac{1}{8}(16x^2 - 4x - 4x + 1) = 0$

$⇒ \dfrac{1}{8}× (4x(4x - 1) - (4x - 1)) = 0$

$⇒ \dfrac{1}{8}× (4x - 1)(4x - 1)) = 0$

$⇒ \dfrac{1}{8}× (4x - 1)^2 = 0$

i.e. $4x - 1 = 0$
⇒ $x = \dfrac{1}{4}$

Hence the root of given quadratic equation is $\dfrac{1}{4}$

(v) $100x^2 - 20x + 1 = 0$

$⇒ 100x^2 - 20x + 1 = 0$
$⇒ 100x^2 - 10x - 10x + 1 = 0$
$⇒ 10x(10x - 1) - (10x - 1) = 0$
$⇒ (10x - 1)(10x - 1) = 0$

i.e. $10x - 1 = 0$
⇒ $x = \dfrac{1}{10}$

Hence the root of given quadratic equation is $\dfrac{1}{10}$

Solve the problems given in Example 1

(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 that John had be $x$.
∴ The number of marbles Jivanti had will be = $45 − x$

After losing $5$ marbles each,
John will have $x - 5$ marbles and
Jivanti will have $45 - x - 5$ marbles = $40 - x$ marbles

Product of marbles now = $(x - 5)(40 - x) = 124$
⇒ $-x^2 + 40x + 5x - 200 = 124$
⇒ $-x^2 + 45x - 324 = 0$
⇒ $x^2 - 45x + 324 = 0$
⇒ $x^2 - 36x - 9x + 324 = 0$
⇒ $x(x - 36) - 9(x - 36) = 0$
⇒ $(x - 36)(x - 9) = 0$

i.e. either $x - 36 = 0$ or $x - 9 = 0$
⇒ $x = 36$ or $x = 9$

Hence John and Jivanti started with $36$ and $9$ marbles

(ii)
Let the number of toys produced in a day be $x$.
∴ The cost of production of each toy (in rupees) that day = $55 - x$
Hence, the total cost of production (in rupees) on that day = $x(55 - x)$
∴ $x(55 - x) = 750$
⇒ $55x - x^2 = 750$
⇒ $x^2 - 55x + 750 = 0$
⇒ $x^2 - 25x - 30x + 750 = 0$
⇒ $x(x - 25) - 30(x - 25) = 0$
⇒ $(x - 25)(x - 30) = 0$
i.e. either $x - 25 = 0$ or $x - 30 = 0$
⇒ $x = 25$ or $x = 30$

Hence number of toys produced on that day is $25$ or $30$

Find two numbers whose sum is $27$ and product is $182$.

Solution:

Let one of the numbers be $x$.
Then the other number will be $27 - x$

Product of the two numbers = $182$
∴ $x(27 - x) = 182$
⇒ $-x^2 + 27x = 182$
⇒ $x^2 - 27x + 182 = 0$
⇒ $x^2 - 14x - 13x + 182 = 0$
⇒ $x(x - 14) - 13(x - 14) = 0$
⇒ $(x - 14)(x - 13) = 0$
i.e. either $x - 14 = 0$ or $x - 13 = 0$
⇒ $x = 14$ or $x = 13$

Hence the numbers are $13$ and $14$

Find two consecutive positive integers, sum of whose squares is $365$.

Solution:

Let the first positive integer be $x$
Hence the next consecutive integer will be $x + 1$

Sum of squares of these numbers is $365$.
∴ $x^2 + (x + 1)^2 = 365$
⇒ $x^2 + x^2 + 2x + 1 = 365$
⇒ $2x^2 + 2x + 1 = 365$
⇒ $2x^2 + 2x + 1 - 365 = 0$
⇒ $2x^2 + 2x - 364 = 0$
⇒ $x^2 + x - 182 = 0$
⇒ $x^2 + 14x - 13x - 182 = 0$
⇒ $x(x + 14) - 13(x + 14) = 0$
⇒ $(x + 14)(x - 13) = 0$
i.e. either $x + 14 = 0$ or $x - 13 = 0$
⇒ $x = -14$ or $x = 13$

As the numbers are positive integers, value of $x$ cannot be $-14$
∴ $x = 13$ and $x + 1 = 13 + 1 = 14$

Hence the numbers are $13$ and $14$

The altitude of right triangle is $7$ cm less than its base. If the hyptenuse is $13$ cm, find the other two sides.

Solution:

Let the base of right triangle be $x$ cm
Hence, its altitude will be $x - 7$ cm

As per Pythagoras theorem,
$Hyptenuse^2$ = $Base^2$ + $Altitude^2$
∴ $13^2 = x^2 + (x - 7)^2$
⇒ $169 = x^2 + x^2 - 14x + 49$
⇒ $x^2 + x^2 - 14x + 49 - 169 = 0$
⇒ $2x^2 - 14x - 120 = 0$
⇒ $x^2 - 7x - 60 = 0$
⇒ $x^2 - 12x + 5x - 60 = 0$
⇒ $x(x - 12) + 5(x - 12) = 0$
⇒ $(x - 12)(x + 5) = 0$
i.e. either $x - 12 = 0$ or $x + 5 = 0$
⇒ $x = 12$ or $x = -5$

As the value of base cannot be negative, $x = 12$
As altitude is $7$ cm less than base, altitude = $x - 7 = $ $ 12 - 7 = 5$

Hence base is $12$ cm and altitude is $5$ cm

A cottage industry produces certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the cost of the production on that day was ₹ 90, find the number of articles produced and the cost of each article.

Solution:

Let the number of articles produced on that day be $x$
Therefore, the cost (in rupees) of each article will be $(3 + 2x)$

Total cost of production = Cost of each article × Total number of articles
i.e. $90 = x(3 + 2x)$
$⇒ 2x^2 + 3x - 90 = 0$
$⇒ 2x^2 + 15x - 12x - 90 = 0$
$⇒ 2x(x - 6) + 15(x - 6) = 0$
$⇒ (2x + 15)(x - 6) = 0$
i.e. either $2x + 15 = 0$ or $x - 6 = 0$
⇒ $x = -\dfrac{15}{2}$ or $x = 6$

As number of articles cannot be negative, $x = 6$
∴ cost of each article = $3 + 2x =$ $ 3 + 12 = 15$

Hence cost of each article is ₹ $15$ and number of articles is $6$