Question

Find the zeros of the polynomial 4 root 3 X square + 4 root 3 x minus 3 root 3 also verify the relationship between the zeros and coefficient

Answer 1

$given \: equation \: is \: 4 \sqrt{3} x {}^{2} 4 \sqrt{3} x - 3 \sqrt{3} = 0 \\ \\ on \: dividing \: equation \: by \sqrt{3} \\ \\ 4x {}^{2} + 4x - 3 = 0 \\ \\ 4x {}^{2} + 6x - 2x - 3 = 0 \\ \\ 2(2x + 3) - 1(2x + 3) = 0 \\ \\ (2x - 1)(2x + 3) = 0 \\ \\ therefore \: roots \: are \: \: x = \frac{1}{2} and \: x = \frac{ - 3}{2}$

$verfication ; \: putting \: x = \frac{1}{2} \: in l.h.s. \: of \: \: given \: equation\: \\ \\ \: \: 4 \sqrt{3} \times( \frac{1}{2} ) {}^{2} + 4 \sqrt{3} \times \frac{1}{2} - 3 \sqrt{3} \\ = 0 = r.h.s. \\ \\ \\ now \: putting \: x = \frac{ - 3}{2} \: in \: l.h.s. of \: the \: given \: equation\\ \\ 4 \sqrt{3} ( \frac{ - 3}{2} ) {}^{2} + 4 \sqrt{3} \frac{( - 3)}{2} - 3 \sqrt{3} \\ \\ = 0 = r.h.s.\\ \\ \\ hope \: it \: helps...$

Answer 2

The Sum of zeros is $-1$ and Product of zeroes is $-\frac{3}{4}$

Step-by-step explanation:

Given,

Find the zeros of the polynomial $4\sqrt{3}x^{2} +4\sqrt{3}x-3\sqrt{3}$

Let,

∴ $4\sqrt{3}x^{2} +4\sqrt{3}x-3\sqrt{3}=0$

⇒$4\sqrt{3}x^{2} +6\sqrt{3}x-2\sqrt{3}x-3\sqrt{3}=0$

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

⇒$\sqrt{3}(2x-1)(2x+3)=0$

∴ $2x-1=0$

⇒$x=\frac{1}{2}$

And,

$(2x+3)=0$

⇒$2x=-3$

⇒$x=\frac{-3}{2}$

Let,

$Alpha=-\frac{3}{2}$ and $Beta=\frac{1}{2}$

Relationship between zeroes and Coefficient,

$Sum of zeroes = Alpha + Beta$

⇒$Sum of zeroes = -\frac{3}{2} + \frac{1}{2}$

∴ $Sum of zeroes = -1$

And,

$Product of zeroes = Alpha \times Beta$

⇒$Product of zeroes = -\frac{3}{2}  \times \frac{1}{2}$

∴ $Product of zeroes = -\frac{3}{4}$

So, The Sum of zeros is $-1$ and Product of zeroes is $-\frac{3}{4}$