A geometric situation leads to the quadratic equation \(L^2-2(a+4)L+(a^2+6a+13)=0\). What condition on \(a\) is necessary for \(L\) to have real values?
Answer and explanation
Correct answer: \(a\geq-\frac{3}{2}\)
For this quadratic, the discriminant is \(D=[-2(a+4)]^2-4(a^2+6a+13)=4(2a+3)\). For \(L\) to be real, \(D\geq0\) is required. Thus, \(4(2a+3)\geq0\), giving \(a\geq-\frac{3}{2}\). In option B, the discriminant is negative, so the roots are not real. Exam tip: use \(D\geq0\) for real roots; use \(D>0\) when two distinct real roots are required.
Frequently asked questions
What is the correct answer to this question?
\(a\geq-\frac{3}{2}\)
Why is this the correct answer?
For this quadratic, the discriminant is \(D=[-2(a+4)]^2-4(a^2+6a+13)=4(2a+3)\). For \(L\) to be real, \(D\geq0\) is required. Thus, \(4(2a+3)\geq0\), giving \(a\geq-\frac{3}{2}\). In option B, the discriminant is negative, so the roots are not real. Exam tip: use \(D\geq0\) for real roots; use \(D>0\) when two distinct real roots are required.
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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