Find the solutions to the equations?

Question

1) 28x^4+179x^2+25=0

2) x^3+4x^2+13x=0?

Answer 1

(1).

The equation is 28x⁴ + 179x² + 25 = 0.

Let x² = t then equation is 28t² + 179t + 25 = 0.

28t² + 179t + 25 = 0

28t² + 175t + 4t + 25 = 0

7t(4t + 25) + 1(4t + 25) = 0

(7t + 1)(4t + 25) = 0

7t + 1 = 0 and 4t + 25 = 0

t = - 1/7 and t = - 25/4

x² = - 1/7 and x² = - 25/4         [ Since x² = t ]

x = ± i√(1/7) and x = ± 5i/2

The solutions are x = ± i/√7 and x = ± 5i/2.

Answer 2

(2).

The equation is x³ + 4x² + 13x = 0.

x(x² + 4x + 13) = 0

x = 0 and x² + 4x + 13 = 0

Compare the equation x² + 4x + 13 = 0 with general form of quadratic equation ax² + bx + c = 0.

a = 1, b = 4 and c = 13.

Substitute a = 1, b = 4 and c = 13 in the quadratic formula: x = [-b±√(b² - 4ac)]/2a.

x = [-4±√(4² - 4*1*13)]/2*1

x = [-4±√(16 - 52)]/2

x = [-4±√(-36)]/2

x = [-4±6i]/2

x = - 2 ± 3i.

The solutions are x = 0 and x = - 2 ± 3i.