Question

If the zeroes of the polynomial x square-px+q are in ratio 2:3 prove that 6p square is equal to 25q

Answer 1

Step-by-step explanation:

Given polynomial is:

$x^2 - px+q$

Let the zeros of the polynomial be $\alpha\: and\: \beta$

Hence,

$\alpha\: = 2x \\ \beta = 3x$

Comparing the polynomial $x^2 - px+q$ with $ax^2 +bx+c$ we find: a = 1, b = - p, c = q.

Now sum of zeros:

$\alpha + \beta = - \frac{b}{a}= - \frac{(-p)}{1}=p\\2x+3x= p\\ \implies p = 5x \\ \implies x = \frac{p}{5}.....(1)$

Product of zeros:

$\alpha \times \beta = \frac{c}{a}= - \frac{(q)}{1}=q\\2x \times 3x= q\\ \implies q = 6x^2......(2)$

From equations (1) & (2), we have:

$q = 6 \times (\frac{p}{5})^2 \\q=6 \times \frac{p^2}{5^2}\\ q= \frac{6p^2}{25}\\ \implies 25q= 6p^2\\ or \\ 6p^2 =25q$

Thus, proved.

Answer 2

Answer:

confusion

Step-by-step explanation:

but but never answer it can be proved