For the quadratic equation \((a-1)x^2+2ax+(a+3)=0\) to have equal roots, what is the value of \(a\)?
Answer and explanation
Correct answer: \(a=\frac{3}{2}\)
For the given quadratic equation, \(A=a-1\), \(B=2a\), and \(C=a+3\). Equal roots require the discriminant \(D=B^2-4AC\) to be zero. Thus, \(D=(2a)^2-4(a-1)(a+3)=4(3-2a)\). Setting \(D=0\) gives \(3-2a=0\), so \(a=\frac{3}{2}\). The value \(a=1\) is not valid because it makes \(A=0\), so the equation is no longer quadratic. Exam tip: For equal-root questions, set the discriminant to zero and then verify that the coefficient of \(x^2\) is non-zero.
Frequently asked questions
What is the correct answer to this question?
\(a=\frac{3}{2}\)
Why is this the correct answer?
For the given quadratic equation, \(A=a-1\), \(B=2a\), and \(C=a+3\). Equal roots require the discriminant \(D=B^2-4AC\) to be zero. Thus, \(D=(2a)^2-4(a-1)(a+3)=4(3-2a)\). Setting \(D=0\) gives \(3-2a=0\), so \(a=\frac{3}{2}\). The value \(a=1\) is not valid because it makes \(A=0\), so the equation is no longer quadratic. Exam tip: For equal-root questions, set the discriminant to zero and then verify that the coefficient of \(x^2\) is non-zero.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Nature of Roots.
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