Solution:

Find the roots of the following quadratic equations by factorisation:

(i) $x^2 – 3x – 10 = 0$

Factorise: $x^2 – 5x + 2x – 10 = 0$

$(x – 5)(x + 2) = 0$

Roots: $x = 5$ or $x = -2$

(ii) $2x^2 + x – 6 = 0$

Factorise: $2x^2 + 4x – 3x – 6 = 0$

$(2x – 3)(x + 2) = 0$

Roots: $x = \dfrac{3}{2}$ or $x = -2$

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

Factorise: $\sqrt{2}x^2 + 2\sqrt{2}x + 5x + 5\sqrt{2} = 0$

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

Roots: $x = -\dfrac{5}{\sqrt{2}}$ or $x = -\sqrt{2}$

(iv) $2x^2 – x + \dfrac{1}{8} = 0$

Multiply through by 8 to clear the fraction: $16x^2 – 8x + 1 = 0$

Factorise: $(4x – 1)(4x – 1) = 0$

Roots: $x = \dfrac{1}{4}$

(v) $100x^2 – 20x + 1 = 0$

Factorise: $(10x – 1)(10x – 1) = 0$

Roots: $x = \dfrac{1}{10}$

Solution:

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.

Let the number of marbles John had initially be $x$. Then, the number of marbles Jivanti had initially will be $45 - x$.

After both lose 5 marbles each, John has $x - 5$ marbles, and Jivanti has $(45 - x) - 5 = 40 - x$ marbles.

The product of the number of marbles they now have is given as 124, so:

$(x - 5)(40 - x) = 124$

Expanding the left-hand side:

$x(40 - x) - 5(40 - x) = 124$

$40x - x^2 - 200 + 5x = 124$

Combining like terms:

$40x + 5x - x^2 = 124 + 200$

$45x - x^2 = 324$

Rearranging the equation:

$x^2 - 45x + 324 = 0$

Now, we solve this quadratic equation using the quadratic formula:

$x = \frac{-(-45) \pm \sqrt{(-45)^2 - 4(1)(324)}}{2(1)}$

$x = \frac{45 \pm \sqrt{2025 - 1296}}{2}$

$x = \frac{45 \pm \sqrt{729}}{2}$

$x = \frac{45 \pm 27}{2}$

So, $x = \frac{45 + 27}{2} = 36$ or $x = \frac{45 - 27}{2} = 9$.

If $x = 36$, then John had 36 marbles and Jivanti had $45 - 36 = 9$ marbles initially. If $x = 9$, then John had 9 marbles and Jivanti had $45 - 9 = 36$ marbles initially.

Thus, the possible initial number of marbles are:

John had 36 marbles and Jivanti had 9 marbles, or John had 9 marbles and Jivanti had 36 marbles.

(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 Rs.750. We would like to find out the number of toys produced on that day.

Let the number of toys produced be $x$.

The cost of production of each toy is $55 - x$.

The total cost of production is given as Rs.750, so:

$x(55 - x) = 750$

Expanding the left-hand side:

$55x - x^2 = 750$

Rearranging the equation:

$x^2 - 55x + 750 = 0$

Now, we solve this quadratic equation using the quadratic formula:

$x = \frac{-(-55) \pm \sqrt{(-55)^2 - 4(1)(750)}}{2(1)}$

$x = \frac{55 \pm \sqrt{3025 - 3000}}{2}$

$x = \frac{55 \pm \sqrt{25}}{2}$

$x = \frac{55 \pm 5}{2}$

So, $x = \frac{55 + 5}{2} = 30$ or $x = \frac{55 - 5}{2} = 25$.

Thus, the number of toys produced on that day is either 30 or 25.

Solution:

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

Let the two numbers be $x$ and $y$.

According to the given conditions:

• Sum: $x + y = 27$
• Product: $xy = 182$

We have the system of equations:

• $x + y = 27$
• $xy = 182$

We can solve for $x$ and $y$ using the quadratic equation. First, express $y$ in terms of $x$ from the first equation:

$y = 27 - x$

Substitute this into the second equation:

$x(27 - x) = 182$

Expand the equation:

$27x - x^2 = 182$

Rearrange it into standard quadratic form:

$x^2 - 27x + 182 = 0$

Now, solve this quadratic equation using the quadratic formula:

$x = \frac{-(-27) \pm \sqrt{(-27)^2 - 4(1)(182)}}{2(1)}$

$x = \frac{27 \pm \sqrt{729 - 728}}{2}$

$x = \frac{27 \pm \sqrt{1}}{2}$

$x = \frac{27 \pm 1}{2}$

This gives two possible solutions for $x$:

• $x = \frac{27 + 1}{2} = 14$
• $x = \frac{27 - 1}{2} = 13$

Thus, the two numbers are $14$ and $13$.

Solution:

Question:

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

Solution:

Let the two consecutive positive integers be $x$ and $x+1$.

The sum of their squares is given by:

$x^2 + (x+1)^2 = 365$

Expand the equation:

$x^2 + (x^2 + 2x + 1) = 365$

Combine like terms:

$2x^2 + 2x + 1 = 365$

Subtract 365 from both sides:

$2x^2 + 2x + 1 - 365 = 0$

$2x^2 + 2x - 364 = 0$

Divide the entire equation by 2:

$x^2 + x - 182 = 0$

Now, solve the quadratic equation $x^2 + x - 182 = 0$ using the quadratic formula:

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 1$, and $c = -182$.

Substitute the values of $a$, $b$, and $c$ into the formula:

$x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-182)}}{2(1)}$

$x = \frac{-1 \pm \sqrt{1 + 728}}{2}$

$x = \frac{-1 \pm \sqrt{729}}{2}$

$x = \frac{-1 \pm 27}{2}$

Now, consider both possible values for $x$:

1. $x = \frac{-1 + 27}{2} = \frac{26}{2} = 13$

2. $x = \frac{-1 - 27}{2} = \frac{-28}{2} = -14$ (not a positive integer)

Thus, the only valid solution is $x = 13$.

The two consecutive positive integers are 13 and 14.

Solution:

Given:

The hypotenuse of the right triangle is 13 cm.

The altitude is 7 cm less than the base.

Let the base be $x$ cm.

Then, the altitude will be $(x - 7)$ cm.

According to the Pythagorean theorem:

$x^2 + (x - 7)^2 = 13^2$

Expanding the equation:

$x^2 + (x^2 - 14x + 49) = 169$

Simplifying the equation:

$2x^2 - 14x + 49 = 169$

Bringing all terms to one side:

$2x^2 - 14x + 49 - 169 = 0$

$2x^2 - 14x - 120 = 0$

Dividing the entire equation by 2:

$x^2 - 7x - 60 = 0$

Using the quadratic formula:

$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(1)(-60)}}{2(1)}$

Solving for $x$:

$x = \frac{7 \pm \sqrt{49 + 240}}{2}$

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

$x = \frac{7 \pm 17}{2}$

Thus, $x = \frac{7 + 17}{2} = 12$ or $x = \frac{7 - 17}{2} = -5$.

Since $x$ represents the base of the triangle, it cannot be negative. Therefore, $x = 12$ cm.

Now, the altitude is:

Altitude = $x - 7 = 12 - 7 = 5$ cm.

Therefore, the base is 12 cm, and the altitude is 5 cm.

Solution:

Given:

The cost of production of each article is 3 more than twice the number of articles produced.

Let the number of articles produced be $x$.

The cost of production of each article is $2x + 3$.

The total cost of production is Rs. 90.

Thus, the total cost of production can be expressed as:

$(2x + 3) \times x = 90$.

Solving the equation:

Expanding the equation:

$2x^2 + 3x = 90$.

Rearranging the terms:

$2x^2 + 3x - 90 = 0$.

Solving this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 2$, $b = 3$, and $c = -90$.

Substituting the values:

$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-90)}}{2(2)}$

$x = \frac{-3 \pm \sqrt{9 + 720}}{4}$

$x = \frac{-3 \pm \sqrt{729}}{4}$

$x = \frac{-3 \pm 27}{4}$.

Now, we have two possible values for $x$:

$x = \frac{-3 + 27}{4} = \frac{24}{4} = 6$

or

$x = \frac{-3 - 27}{4} = \frac{-30}{4}$ (which is negative and not possible).

Therefore, the number of articles produced is $x = 6$.

Cost of each article:

The cost of each article is $2x + 3 = 2(6) + 3 = 12 + 3 = 15$.