What is the repeated (double) root of the equation \(x^2+2x+1=0\)?
Answer and explanation
Correct answer: -1
Factor the quadratic: \(x^2+2x+1=(x+1)^2\). Thus \((x+1)^2=0\) gives the double root \(x=-1\). A common mistake is choosing \(1\), which would correspond to \((x-1)^2\) and the polynomial \(x^2-2x+1\). Exam tip: check either factorisation or compute the discriminant \(b^2-4ac\); if it equals 0, the quadratic has a repeated root.
Frequently asked questions
What is the correct answer to this question?
-1
Why is this the correct answer?
Factor the quadratic: \(x^2+2x+1=(x+1)^2\). Thus \((x+1)^2=0\) gives the double root \(x=-1\). A common mistake is choosing \(1\), which would correspond to \((x-1)^2\) and the polynomial \(x^2-2x+1\). Exam tip: check either factorisation or compute the discriminant \(b^2-4ac\); if it equals 0, the quadratic has a repeated root.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Roots of a Quadratic Equation.
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