The equation \(x^2-2(3t+1)x+(5t^2+2t+4)=0\) has no real roots. What is the correct interval for \(t\)?
Answer and explanation
Correct answer: \(-\frac{3}{2}<t<\frac{1}{2}\)
A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=1\), \(b=-2(3t+1)\), and \(c=5t^2+2t+4\). Thus, \(D=4(3t+1)^2-4(5t^2+2t+4)=4(2t-1)(2t+3)\). Therefore, \((2t-1)(2t+3)<0\), which gives \(-\frac{3}{2}<t<\frac{1}{2}\). At the endpoints, \(D=0\), so the equation has equal real roots rather than no real roots. Exam tip: for a positive leading coefficient, a factored quadratic is negative between its two distinct roots.
Frequently asked questions
What is the correct answer to this question?
\(-\frac{3}{2}<t<\frac{1}{2}\)
Why is this the correct answer?
A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=1\), \(b=-2(3t+1)\), and \(c=5t^2+2t+4\). Thus, \(D=4(3t+1)^2-4(5t^2+2t+4)=4(2t-1)(2t+3)\). Therefore, \((2t-1)(2t+3)<0\), which gives \(-\frac{3}{2}<t<\frac{1}{2}\). At the endpoints, \(D=0\), so the equation has equal real roots rather than no real roots. Exam tip: for a positive leading coefficient, a factored quadratic is negative between its two distinct roots.
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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