We have, a2 + 5a - 24 = 0
∴ a2 + 5a = 24 .....(i)
Third term = (\(\frac 12\) coefficient of x)2
\(= (\frac 12 \times 5)^2\)
\(= \frac{25}4\)
Adding \( \frac{25}4\) to both sides of equation (i), we get
\(a^2 + 5a + \frac{25}4 = 24 + \frac{25}4\)
4a2 + 20a + 25 = 96 + \(\frac{25}4\)
(2a)2 + 20a + (5)2 = \(\frac{121}4\)
(2a + 5)2 = (\(\frac{11}2\))2
Taking square roots on both sides
2a + 5 = ± \(\frac{11}2\)
2a = -5 ± \(\frac{11}2\)
\(2a = -5 + \frac{11}2\)
\( 2a = \frac{-10 + 11}2\)
\(a = \frac 1{2 \times 2}\)
\(a=\frac 14\)
or
\(2a = -5 - \frac{11}2\)
\( 2a = \frac{-10 - 11}2\)
\(a = \frac {-21}{2 \times 2}\)
\(a = \frac{-21}4\)
Hence \(\frac 14\) and \( \frac{-21}4\) are the roots of the given quadratic equation.