Let the first number be x and the second number is 27 − x.
Therefore, their product = x (27 − x)
It is given that the product of these numbers is 182.
Therefore, x(27-x) = 182
⇒ \(x^2 -27x +182 = 0\)
⇒ \(x^2 - 13x -14x +182 =0\)
⇒ \(x(x-13) -14 (x-13) =0\)
⇒ \((x-13)(x-14) = 0\)
x – 13 = 0 or x − 14 = 0
i.e., x = 13 or x = 14
If first number = 13,then
Other number = 27 − 13 = 14
If first number = 14, then
Other number = 27 − 14 = 13
Therefore, the numbers are 13 and 14.
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. 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 Rs 750. Find out the number of toys produced on that day.
Find the roots of the following quadratic equations by factorisation:
(i) \(x^2 – 3x – 10 = 0\)
(ii) \(2x^2 + x – 6 = 0\)
(iii) \(\sqrt2x^2+7x+5\sqrt2 = 0\)
(iv) \(2x^2 – x + \frac{1}8\)\( = 0\)
(v) \(100x^2 – 20x + 1 = 0\)