Question

p(p-3) =1 for class 10

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Answer 1

Step-by-step explanation:

$\tt\implies {p}^{2} - 3p = 1$

$\tt\implies {p}^{2} - 3p - 1 = 0$

• In general, given ax² +bx+c=0, there exists two solutions where:

$\\ \tt \implies x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}$

• In this case, a=1,b=−3 and c=−1.

$\\ \tt \implies{p}^{}=\frac{3+\sqrt{{(-3)}^{2}+4}}{2},\frac{3-\sqrt{{(-3)}^{2}+4}}{2}$

• Simplify.

$\\ \boxed{\red{\tt \implies p=\frac{3+\sqrt{13}}{2},\frac{3-\sqrt{13}}{2}}}$

Answer 2

Answer:

$p = \frac{3 + \sqrt{13} }{2} and \: \frac{3 - \sqrt{13} }{2}$

hope it help you