Question

If alpha and beta
are the roots of 5x²-px+1=0
and alpha-beta=1, then find p
​

Answer 1

Step-by-step explanation:

$5 {x}^{2} - px + 1 = 0 \\ \alpha - \beta = \: 1 \: \: ...........(1) \\ \\ \alpha + \beta = \frac{ - b}{a} \\ = > \alpha + \beta = - \frac{ - p}{5} \\ = > \alpha + \beta = \frac{p}{5} \\ \\ \alpha \beta = \frac{c}{a} \\ \alpha \beta = \frac{1}{5} \: \: \: \: \: \: ........(2) \\ \\ putting \: the \: values \: of \: \: \: \ \: beta in \: \: equation \: (1) \\ \alpha - \beta = 1 \\ = > \alpha - \frac{1}{5 \alpha } = 1 \\ = > 5 { \alpha }^{2} - 1 = 5 \alpha \\ = > 5 { \alpha }^{2} - 5 \alpha - 1 = 0 \\ = > \alpha = \frac{ 5 + - \sqrt{ {5}^{2} + 4 \times 5 \times 1} }{2 \times 5} \\ = > \alpha = \frac{5 + - \sqrt{25 + 20} }{10} \\ = > \alpha = \frac{5 + - \sqrt{45} }{10} \\ = > \alpha = \frac{5 + - 3 \sqrt{5} }{10} \\ = > \alpha = \frac{5 + 3 \sqrt{5} }{10} \: \: or \: x = \frac{5 - 3 \sqrt{5} }{10} \\ \\ \\ \\ \beta = \frac{1}{5 \alpha } \\ = \frac{1}{5 \times 5 + 3 \sqrt{5} } \times 10 \\ = \frac{2}{5 + 3 \sqrt{5} } \\ \\ \alpha + \beta = \frac{p}{5} \\ = > \frac{5 + 3 \sqrt{5} }{10} + \frac{2}{5 + 3 \sqrt{5} } = \frac{p}{5} \\ = > \frac{( {5 + 3 \sqrt{5} )}^{2} + 2 \times 10}{10(5 + 3 \sqrt{5} )} = \frac{p}{5 } \\ = > \frac{ {5}^{2} + 30 \sqrt{5} + 45 + 20}{2(5 + 3 \sqrt{5} )} = p \\ = > \frac{25 + 65 + 30 \sqrt{5} }{10 + 6 \sqrt{5} } = p \\ = > \frac{90 + 30 \sqrt{5} }{10 + 6 \sqrt{5} } = p$