Let the age of one friend be x years.
Age of the other friend will be (20 − x) years.
4 years ago,
age of 1st friend = (x − 4) years
And, age of 2nd friend = (20 − x − 4) = (16 − x) years
Given that, \((x − 4) (16 − x) = 48\)
\(16x − 64 − x^2 + 4x = 48 \)
\(− x^2 + 20x − 112 = 0 \)
\(x^2 − 20x + 112 = 0 \)
Comparing this equation with \(ax^2 + bx + c = 0\), we obtain
a = 1, b = −20, c = 112
Discriminant = \(b^2 − 4ac\)
= \((− 20)^2 − 4 (1) (112)\)
= \(400 − 448\)
= \(−48 \)
As \(b^2 − 4ac < 0\), Therefore, no real root is possible for this equation and hence, this situation is not possible.
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) \(2x^2 + kx + 3 = 0\) (ii) \(kx (x – 2) + 6 = 0\)