Quadratic Formula Math Example 5

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Example 5

hard
Solve x2+4x+5=0x^2 + 4x + 5 = 0 and describe the solutions.

Solution

  1. 1
    Apply the quadratic formula: x=−4±16−202=−4±−42x = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2}.
  2. 2
    The discriminant is −4<0-4 < 0, so there are no real solutions.
  3. 3
    The equation has two complex solutions: x=−2±ix = -2 \pm i.

Answer

No real solutions (complex: x=−2±ix = -2 \pm i)
When the discriminant b2−4acb^2 - 4ac is negative, the quadratic has no real roots. The square root of a negative number involves the imaginary unit ii.

About Quadratic Formula

A formula giving the exact solutions to any quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 directly from its three coefficients.

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More Quadratic Formula Examples

Example 1 easy

Solve [formula] by factoring.

Example 2 medium

Solve [formula] using the quadratic formula.

Example 3 hard

Solve [formula] using the quadratic formula.

Example 4 easy

Solve [formula].

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