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Solve the following quadratic equations.

i. m2 + 5m = 5 = 0

ii. 5m2 + 2m + 1 = 0

iii x2 - 4x - 3 = 0

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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 \({2 - \sqrt{ 7} }\).

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