Question

Find the zeroes of 3√2x² +13x +6√2 and verify the relation between the zeroes and coefficients of the polynomial.​

Answer 1

$\huge\underline\mathrm{Question-}$

Find the zeroes of 3√2x² +13x +6√2 and verify the relation between the zeroes and coefficients of the polynomial.

$\huge\underline\mathrm{Explanation-}$

Given polynomial :

$\rm{3\sqrt{2}x^2+13x+6\sqrt{2}}$

Equate it to zero.

$\rm{3\sqrt{2}x^2+13x+6\sqrt{2}}$ = 0

Sum = 13

Product = 6√2 × 3√2 => 18 × 2 = 36

Factors = (9, 4)

By using splitting middle term method,

: $\implies$ $\rm{3\sqrt{2}x^2+9x+4x+6\sqrt{2}}$ = 0

By taking as common,

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

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

: $\implies$ $\rm{3x+2\sqrt{2}}$ = 0 and $\rm{\sqrt{2}x+3}$ = 0

Solving it, we get,

$\large{\boxed{\mathrm{\red{x=-\dfrac{2\sqrt{2}}{3}\:and\:-\dfrac{3}{\sqrt{2}}}}}}$

Now,

★ sum of zeroes = $\rm{-\dfrac{2\sqrt{2}}{3}}$ + ( $\rm{-\dfrac{3}{\sqrt{2}}}$ )

: $\implies$ $\rm{\dfrac{-13}{3\sqrt{2}}}$

★ Product of zeroes = $\rm{-\dfrac{2\sqrt{2}}{3}}$ × ( $\rm{-\dfrac{3}{\sqrt{2}}}$ )

: $\implies$ 2

$\rule{200}2$

Verification :

Sum of zeroes = $\dfrac{-b}{a}$

: $\implies$ $\dfrac{-13}{3\sqrt{2}}$

Product of zeroes = $\dfrac{c}{a}$

: $\implies$ = $\cancel{\dfrac{6\sqrt{2}}{3\sqrt{2}}}$

: $\implies$ 2

Hence verified!

Answer 2

$\huge{\underline{\underline{\pink{\mathfrak{AnSwEr :}}}}}$

We are given polynomial

★$\large{\boxed{\sf{3 \sqrt{2} x^2 \: + \: 13x \: + \: 6 \sqrt{2} \: = \: 0}}}$

Use Splitting The Middle Term Method

$\implies \sf{3 \sqrt{2} \: {x}^{2} + 13x \: + \: 6 \sqrt{3} = 0} \\ \\ \implies \sf{3 \sqrt{2} {x}^{2} + 9x \: + 4x \: + 6 \sqrt{2} = 0 } \\ \\ \implies \sf{3x( \sqrt{2}x \: + \: 3x) \: + \: 2 \sqrt{2} ( \sqrt{2} x \: + \: 3) = 0} \\ \\ \implies \sf{( \sqrt{2}x \: + \: 3)(3x \: + \: \sqrt{2} x) \: = \: 0 } \\ \\ \implies \sf{3x \: + \: 2 \sqrt{2} \: \: and \: \: \sqrt{2} x \: + \: 3 = 0 } \\ \\ \implies {\boxed{\sf{x \: = \: - \dfrac{ 2\sqrt{2}}{3} \: \: and \: \: \dfrac{ - 3}{ \sqrt{2} } }}}$

$\rule{200}{1}$

★$\large{\boxed{\sf{Sum \: of \: zeros \: = \: \dfrac{-b}{a}}}}$

$\implies {\sf{\dfrac{-2 \sqrt{2}}{3} \: + \: ( - \dfrac{3}{\sqrt{2}}) \: = \: \dfrac{-(13)}{3 \sqrt{2}}}}$

$\implies {\sf{\dfrac{-13}{3 \sqrt{2}} \: = \: \dfrac{-13}{3 \sqrt{2}}}}$

$\Large{\mathbb{L.H.S \: = \: R.H.S}}$

Hence Proved