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what are the roots of the equation $x^2 + 4x - 16 = 0$? 1. $2\\pm2\\sqr…

Question

what are the roots of the equation $x^2 + 4x - 16 = 0$?

  1. $2\pm2\sqrt{5}$
  2. $-2\pm2\sqrt{5}$
  3. $2\pm4\sqrt{5}$
  4. $-2\pm4\sqrt{5}$

Explanation:

Step1: Recall quadratic formula

For \(ax^2 + bx + c = 0\), roots are \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\).
Here, \(a = 1\), \(b = 4\), \(c=-16\).

Step2: Calculate discriminant

Discriminant \(D = b^2 - 4ac = 4^2 - 4(1)(-16)=16 + 64 = 80\).

Step3: Simplify square root of discriminant

\(\sqrt{D}=\sqrt{80}=\sqrt{16\times5}=4\sqrt{5}\).

Step4: Apply quadratic formula

\(x=\frac{-4\pm4\sqrt{5}}{2(1)}=\frac{-4\pm4\sqrt{5}}{2}=-2\pm2\sqrt{5}\).

Answer:

  1. \(-2\pm2\sqrt{5}\)