How many real roots does x4 + 12x − 5 have?

Answer 1

Let x^4 + 12x - 5 = 0.

Factor x^4 + 12x - 5 = 0 by factor by grouping.

(x^2 − 2x + 5)(x^2 + 2x - 1) = 0.

Apply zero product property.

x^2 − 2x + 5 = 0 and x^2 + 2x + 1 = 0

• x^2 − 2x + 5 = 0 is a quadratic,use quadratic formula to find roots of the related quadratic equation.

x = [ - b ± √ (b^2 - 4ac) ]/2a.

Compare the above equation with standard form of the quadratic equation ax^2 + bx + c = 0.

a = 1, b = - 2, and c = 5

x = [ 2 ± √ ((- 2)^2 - 4 * 1 * 5) ]/2 * 1

  = [ 2 ± √ (4 - 20) ]/2

  = [ 2 ± √ (- 16) ]/2

  = 1 ± 2i.

The solutions of x^2 − 2x + 5 = 0 are imaginary values i.e, 1 ± 2i.

• x^2 + 2x - 1 = 0 is a quadratic,use quadratic formula to find roots of the related quadratic equation.

x = [ - b ± √ (b^2 - 4ac) ]/2a.

Compare the above equation with standard form of the quadratic equation ax^2 + bx + c = 0.

a = 1, b =  2, and c = - 1

x = [ - 2 ± √ ((2)^2 - 4 * 1 * (- 1)) ]/2 * 1

  = [ - 2 ± √ (4 + 4) ]/2

  = [ - 2 ± √ 8 ]/2

  = - 1 ± √ 2

The solutions of x^2 + 2x - 1 = 0 are real values i.e, - 1 ± √ 2.

Therefore, x^4 + 12x - 5 has two real roots.