For a real parameter \(t\), which value of \(t\) will make the equation \((t-1)x^2-2(t+2)x+(t+5)=0\) have equal roots?
Answer and explanation
Correct answer: No real value
For equal roots, a quadratic equation must satisfy the condition \(D=b^2-4ac=0\). Here, \(a=t-1\), \(b=-2(t+2)\), and \(c=t+5\). Thus, \(D=4(t+2)^2-4(t-1)(t+5)=4[(t+2)^2-(t^2+4t-5)]=36\), which is never zero for any real \(t\). Therefore, no real value of \(t\) produces equal roots. Also, when \(t=1\), the coefficient of \(x^2\) becomes zero, so the equation is linear rather than quadratic. Exam tip: check \(a\neq0\) first, and then apply the equal-roots condition \(D=0\).
Frequently asked questions
What is the correct answer to this question?
No real value
Why is this the correct answer?
For equal roots, a quadratic equation must satisfy the condition \(D=b^2-4ac=0\). Here, \(a=t-1\), \(b=-2(t+2)\), and \(c=t+5\). Thus, \(D=4(t+2)^2-4(t-1)(t+5)=4[(t+2)^2-(t^2+4t-5)]=36\), which is never zero for any real \(t\). Therefore, no real value of \(t\) produces equal roots. Also, when \(t=1\), the coefficient of \(x^2\) becomes zero, so the equation is linear rather than quadratic. Exam tip: check \(a\neq0\) first, and then apply the equal-roots condition \(D=0\).
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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