If the roots of \(x^2+bx+c=0\) are \(-6\) and \(13\), what is the value of \(b+c\)?
Answer and explanation
Correct answer: -85
For the monic quadratic \(x^2+bx+c=0\), Vieta's formulas give sum of roots \(\alpha+\beta=-b\) and product \(\alpha\beta=c\). Here \(\alpha+\beta=-6+13=7\), so \(-b=7\) and hence \(b=-7\). The product is \(-6\times13=-78\), so \(c=-78\). Therefore \(b+c=-7+(-78)=-85\). A common error (leading to option C = 71) is to take the product as +78 instead of -78. Exam tip: apply Vieta’s relations directly and watch sign conventions for sum and product of roots in monic quadratics.
Frequently asked questions
What is the correct answer to this question?
-85
Why is this the correct answer?
For the monic quadratic \(x^2+bx+c=0\), Vieta's formulas give sum of roots \(\alpha+\beta=-b\) and product \(\alpha\beta=c\). Here \(\alpha+\beta=-6+13=7\), so \(-b=7\) and hence \(b=-7\). The product is \(-6\times13=-78\), so \(c=-78\). Therefore \(b+c=-7+(-78)=-85\). A common error (leading to option C = 71) is to take the product as +78 instead of -78. Exam tip: apply Vieta’s relations directly and watch sign conventions for sum and product of roots in monic quadratics.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Introduction to Quadratic Equations.
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