i. m2 + 5m + 5 = 0
Comparing the above equation with am2 + bm + c = 0, we get
a = 1, b = 5, c = 5
∴ b2 – 4ac = (5)2 – 4 × 1 × 5
= 25 – 20 = 5
\(m = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(= {-5 \pm \sqrt{5} \over 2(1)}\)
∴ \(m= {-5 \pm \sqrt{5} \over 2}\)
∴ \(m= {-5 + \sqrt{5} \over 2}\) or \(m= {-5 - \sqrt{5} \over 2}\)
∴ The roots of the given quadratic equation are \( {-5 + \sqrt{5} \over 2}\) and \({-5 - \sqrt{5} \over 2}\).
ii. 5m2 + 2m+1 = 0
Comparing the above equation with am + bm + c = 0, we get
a = 5, b = 2, c = 1
∴ b2 – 4ac = (2)2 -4 × 5 × 1
= 4 – 20
= -16
∴ b2 – 4ac < 0
∴ Roots of the given quadratic equation are not real.
iii. x2 – 4x – 3 = 0
Comparing the above equation with ax + bx + c = 0, we get
a = 1, b = -4, c = -3
∴ b2 – 4ac = (-4)2 – 4 × 1 × -3
= 16 + 12
= 28
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(= {-(-4) \pm \sqrt{28} \over 2(1)}\)
\(= {4 \pm \sqrt{ 4\times7} \over 2}\)
\(= {4 \pm \sqrt{ 7} \over 2}\)
\(= {2(2 \pm \sqrt{ 7} \over 2}\)
∴ \(x = {2\pm \sqrt{ 7} }\)
∴ \(x = {2 + \sqrt{ 7} }\) or \(x = {2 - \sqrt{ 7} }\)
∴ the roots of the given quadraticd equation are \( {2 + \sqrt{ 7} }\) and \({2 - \sqrt{ 7} }\).