Math, asked by valancy84762, 6 hours ago

If one root of the equation x2-px+q=0 be double of the other, show that 2p2=9q

Answers

Answered by DeeznutzUwU
0

Answer:

Given: x^{2} -px + q = 0; One of the roots is twice the other

Past knowledge: We know that for a quadratic equation of the form ax^{2} +bx+c=0

Sum of its roots = \frac{-b}{a}

Product of its roots = \frac{c}{a}

To Prove: 2p^{2} = 9q

Proof:

Let \alpha and \beta be the roots of the above equation

The question states that one root is twice the other.

\alpha = 2\beta

Sum of the roots of the equation = \alpha +\beta =  \frac{-(-p)}{1} = p            (Past Knowledge)

\alpha +\beta  = p

We know that \alpha =2\beta                                                                 (Given)

2\beta  + \beta  = p

3\beta = p                                                                                      

\beta  = \frac{p}{3}                                                                                       (1)

Product of roots of the equation = \alpha \beta  = \frac{q}{1} = q

\alpha \beta =q

We know that \alpha =2\beta                                                                 (Given)

(2\beta )\beta  = q

2\beta ^{2} = q

We know that \beta  = \frac{p}{3}                                                                  (1)

2(\frac{p}{3})^{2} = q

2(\frac{p^{2} }{9} )= q

2p^{2} = 9q

Hence Proved

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