If \(\alpha\) and \(\beta\) are the two roots of \(2x^2+3x+1=0\), the value of \(\alpha^2+\beta^2\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\frac{5}{4}\)
For \(ax^2+bx+c=0\), the sum of the roots is \(-\frac{b}{a}\) and the product is \(\frac{c}{a}\). The sum of the squares follows from \((\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta\).
\(\alpha+\beta=-\frac{3}{2}\), \(\alpha\beta=\frac{1}{2}\)
\(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\frac{9}{4}-1=\frac{5}{4}\)