Difference between revisions of "Double root"

Revision as of 10:54, 15 May 2014

The double root instance occurs when a quadratic equation is the square of a binomial, or when its discriminant is equal to $0$ .

Find all solutions to the quadratic equation $3x^2+6x+3=0$ .

Plugging the values of $a, b$ and $c$ into the quadratic formula, we get

$\frac{-6\pm\sqrt{6^2-4(3)(3)}}{2(3)}=\frac{-6\pm\sqrt{36-36}}{6}=\frac{-6\pm\sqrt{0}}{6}=\frac{-6}{6}=-1$

so this quadratic has a double root of $-1$ .

We also could have solved this problem by factoring a $3$ out of the left side and dividing:

$3(x^2+2x+1)=0 \Longrightarrow x^2+2x+1=0$

and now we plug the coeficcients into the quadratic formula:

$\frac{-2\pm\sqrt{2^2-4(1)(1)}}{2(1)}=\frac{-2\pm\sqrt{4-4}}{2}=\frac{-2}{2}=-1$

so again, the quadratic has a double root of $-1$ .

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 The '''double root instance''' occurs when a [[quadratic equation]] is the [[square]] of a [[binomial]], or when its [[discriminant]] is equal to <math>0</math>.
+==Example==
+Find all solutions to the quadratic equation <math>3x^2+6x+3=0</math>.
+Plugging the values of <math>a, b</math> and <math>c</math> into the [[quadratic formula]], we get
+<math>\frac{-6\pm\sqrt{6^2-4(3)(3)}}{2(3)}=\frac{-6\pm\sqrt{36-36}}{6}=\frac{-6\pm\sqrt{0}}{6}=\frac{-6}{6}=-1</math>
+so this quadratic has a double root of <math>-1</math>.
+We also could have solved this problem by [[factoring]] a <math>3</math> out of the left side and dividing:
+<math>3(x^2+2x+1)=0 \Longrightarrow x^2+2x+1=0</math>
+and now we plug the [[coeficcients]] into the quadratic formula:
+<math>\frac{-2\pm\sqrt{2^2-4(1)(1)}}{2(1)}=\frac{-2\pm\sqrt{4-4}}{2}=\frac{-2}{2}=-1</math>
+so again, the quadratic has a double root of <math>-1</math>.