Question

verify the relation between sum and product of zeroes and coefficients of quadratic polynomial 3√2 + 13x +6√2​

Answer 1

Answer:

sry don't know check the question properly

Answer 2

Answer:

Let us first factorize the given equation $3\sqrt{2} x^{2} +13x +6\sqrt{2}$ as shown below:

⇒$3\sqrt{2} x^{2} +13x +6\sqrt{2}$

⇒$3\sqrt{2} x^{2} + 9x + 4x + 6\sqrt{2}$

⇒$3x(\sqrt{2} x + 3) + 2\sqrt{2} (\sqrt{2} x +3)$

⇒$(3x + 2\sqrt{2} ) (\sqrt{2} x + 3)$

Therefore, the zeroes of the given polynomial are:

$(3x + 2\sqrt{2} ) = 0$

$x = \frac{-2\sqrt{2} }{3} = \alpha$

$(\sqrt{2} x +3) = 0$

$x = \frac{-3}{\sqrt{2} } = \beta$

Now, the sum and product of the roots is as follows:

$\alpha + \beta = -\frac{2\sqrt{2} }{3} - \frac{3}{\sqrt{2} } = - ( \frac{2\sqrt{2} }{3} +\frac{3}{\sqrt{2} }) = \frac{-4-9}{3\sqrt{2} } = \frac{-13}{3\sqrt{2} } = -\frac{b}{a}$

$\alpha \beta = (-\frac{2\sqrt{2} }{3} )(-\frac{3}{\sqrt{2} } ) = 2 = \frac{6\sqrt{2} }{3\sqrt{2} } = \frac{c}{d}$

Hence verified.

Hope this helps you!!!