If the two roots of the quadratic 3x^2+5x+k are \(\frac{-5\pm i\sqrt{11}}{6}\), what is k?
+118659 If the two roots of the quadratic 3x^2+5x+k are \(\frac{-5\pm i\sqrt{11}}{6}\) , what is k?
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \frac{-5\pm i\sqrt{11}}{6} = {-5 \pm \sqrt{5^2-4*3*k} \over 6}\\ \frac{-5\pm \sqrt{-11}}{6} = {-5 \pm \sqrt{25-12k} \over 6}\\ -11=25-12k\\ -36=-12k\\ k=3 \)
+118659 If the two roots of the quadratic 3x^2+5x+k are \(\frac{-5\pm i\sqrt{11}}{6}\) , what is k?
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \frac{-5\pm i\sqrt{11}}{6} = {-5 \pm \sqrt{5^2-4*3*k} \over 6}\\ \frac{-5\pm \sqrt{-11}}{6} = {-5 \pm \sqrt{25-12k} \over 6}\\ -11=25-12k\\ -36=-12k\\ k=3 \)
+129841 Here's another way to find this:
The product of the roots = c / a ....or in this case, k /a
So....using Melody's answer.......the product of the roots =
[ 5 + i√11] / 6 * [ 5 - i√11] / 6 = [25 - 11* i^2] / 36 = [25 + 11] / 36 = 36 /36 = 1
So.....
1 = k / a
1 = k / 3
3 = k
