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Class 10MathematicsQuadratic EquationsRoots of a Quadratic EquationHardLevel 32
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If \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(x^2-3x-28=0\), what is the value of \((\alpha+1)(\beta+1)\)?

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Answer and explanation

Correct answer: -24

By Vieta’s formulas, \(\alpha+\beta=3\) and \(\alpha\beta=-28\). Therefore, \((\alpha+1)(\beta+1)=\alpha\beta+\alpha+\beta+1=-28+3+1=-24\). Hence, option A is correct. Remember to include the final \(+1\) term when expanding the product; omitting it leads to an incorrect result.

Related tags

Quadratic EquationsRoots Of Quadratic EquationVietas FormulasSum And Product Of RootsAlgebraic Expressions

Frequently asked questions

What is the correct answer to this question?

-24

Why is this the correct answer?

By Vieta’s formulas, \(\alpha+\beta=3\) and \(\alpha\beta=-28\). Therefore, \((\alpha+1)(\beta+1)=\alpha\beta+\alpha+\beta+1=-28+3+1=-24\). Hence, option A is correct. Remember to include the final \(+1\) term when expanding the product; omitting it leads to an incorrect result.

Which subject and chapter does this question cover?

This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Roots of a Quadratic Equation.

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