Question

Solve using the quadratic formula  t^2 + 53 = (3 - t)(2t + 1)

Please explain how you solved this problem, thank you.

Answer 1

Given :

t ^2 + 53 = (3 - t )(2t + 1)

Apply  property : (a + b )(c + d ) = (a + b)c + (a + b)d.

t ^2 + 53 = (3 - t )(2t ) + (3 - t )(1)

Apply distributive property : a (b - c) = ab - ac.

t ^2 + 53 = 6t - 2t ^2 + 3 - t

t ^2 + 53 = - 2t ^2 + 5t + 3

Add 2t ^2 to each side.

t ^2 + 53 + 2t ^2 = 5t + 3

3t ^2 + 53 = 5t + 3

Subtract 5t from each side.

3t ^2 - 5t + 53 = 3

Subtract 3 from each side.

3t ^2 - 5t + 53 - 3 = 0

3t ^2 - 5t + 50 = 0.

3t ^2 - 5t + 50 , is quadratic, use the quadratic formula to find the rooots of the related quadratic equation.

t = [- b ± sqrt(b ^2 - 4ac)] / 2a.

Substitute b = - 5, a = 3, and c = 50.

t = [- (- 5) ± √ [(- 5) ^2 - 4 * 3 * 50)] / 2 * 3

  = [5 ± √ (25 - 600)] / 6

 = [5 ± √ - 575] / 6

  = (5/6) [1 ± √23i ].

Solution of the equation is (5/6) [1 ± √23i ].