Difference between revisions of "Double root"
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<math>3(x^2+2x+1)=0 \Longrightarrow x^2+2x+1=0</math> | <math>3(x^2+2x+1)=0 \Longrightarrow x^2+2x+1=0</math> | ||
− | and now we plug the [[ | + | and now we plug the [[coefficients]] 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> | <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>. | so again, the quadratic has a double root of <math>-1</math>. |
Latest revision as of 14:33, 20 August 2020
The double root instance occurs when a quadratic equation is the square of a binomial, or when its discriminant is equal to .
Example
Find all solutions to the quadratic equation .
Plugging the values of and into the quadratic formula, we get
so this quadratic has a double root of .
We also could have solved this problem by factoring a out of the left side and dividing:
and now we plug the coefficients into the quadratic formula:
so again, the quadratic has a double root of .