If the roots of the equations ax^{2} + 2bx + c = 0 and are simultaneously real then prove that b^{2} = ac.

Given the roots of the equations ax^{2} + 2bx + c = 0 are real.

Comparing with standard quadratic equation Ax^{2} + Bx + C = 0

A = a B = 2b C = c

Discriminant D_{1} = B^{2} – 4AC ≥ 0

= (2b)^{2} – 4.a.c ≥ 0

= 4(b^{2} –ac) ≥ 0

= (b^{2} –ac) ≥ 0 – – – – – (1)

For the equation

Discriminant D_{2} = b^{2} – 4ac ≥ 0

=

= 4(ac – b^{2}) ≥0

= – 4(b^{2}–ac) ≥0

= (b^{2} –ac) ≥0 – – – – – (2)

The roots of the are simultaneously real if (1) and (2) are true together

b^{2} –ac = 0

b^{2} = ac

Hence proved.

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