For the equation \\((q+2)x^2-2(q-1)x+q=0\\), which value of \\(q\\) gives equal roots, given that \\(q\\ne-2\\)?
Answer and explanation
Correct answer: \\(q=\\frac{1}{4}\\)
For a quadratic equation \\(ax^2+bx+c=0\\), equal roots occur when the discriminant \\(D=b^2-4ac\\) is zero. Here, \\(a=q+2\\), \\(b=-2(q-1)\\), and \\(c=q\\). Thus, \\(D=4(q-1)^2-4q(q+2)=4(1-4q)\\). Setting \\(D=0\\) gives \\(1-4q=0\\), so \\(q=\\frac14\\). At \\(q=1\\), the discriminant is not zero, while \\(q=-2\\) removes the quadratic term and is also excluded by the condition. Exam tip: For equal roots, begin by setting the discriminant equal to zero.
Frequently asked questions
What is the correct answer to this question?
\\(q=\\frac{1}{4}\\)
Why is this the correct answer?
For a quadratic equation \\(ax^2+bx+c=0\\), equal roots occur when the discriminant \\(D=b^2-4ac\\) is zero. Here, \\(a=q+2\\), \\(b=-2(q-1)\\), and \\(c=q\\). Thus, \\(D=4(q-1)^2-4q(q+2)=4(1-4q)\\). Setting \\(D=0\\) gives \\(1-4q=0\\), so \\(q=\\frac14\\). At \\(q=1\\), the discriminant is not zero, while \\(q=-2\\) removes the quadratic term and is also excluded by the condition. Exam tip: For equal roots, begin by setting the discriminant equal to 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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