If \((\alpha,\beta)\) are the roots of \(x^2-19x+84=0\), what is the value of \((\alpha+\beta)^2\)?

361

For a quadratic \(ax^2+bx+c=0\), the sum of roots is \(\alpha+\beta=-\dfrac{b}{a}\) and the product is \(\alpha\beta=\dfrac{c}{a}\). In this equation \(a=1,\; b=-19,\; c=84\), so \(\alpha+\beta=-\dfrac{-19}{1}=19\). Therefore \((\alpha+\beta)^2=19^2=361\). Option B (84) is the product \(\alpha\beta\), not the square of the sum. Options C (256) and D (324) are \(16^2\) and \(18^2\) respectively and are incorrect because the sum is 19, not 16 or 18. Exam tip: identify \(a,b,c\) quickly and use \(\alpha+\beta=-b/a\); watch the sign of \(b\).

This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Introduction to Quadratic Equations.