Question

root 3 x square - 8 x + 4 root 3 find the zeros of the quadratic polynomial and verify the relationship between zeros and the coefficient​

Answer 1

AnsWer :

x = 2/√3 and, x = 2√3.

Given :

• General equation ax²+bx+c = 0, a not equal to 0.
• equation - √3x² -8x + 4√3 =0.

Solution :

Let find it's roots, by factorization method apply,

we have equation,

$\tt \implies \sqrt{3} {x}^{2} - 8x + 4 \sqrt{3} = 0.$

$\tt \implies \sqrt{3} {x}^{2} - 6x - 2x + 4 \sqrt{3} = 0.$

$\tt \implies \sqrt{3} x(x - 2 \sqrt{3} ) - 2(x - 2 \sqrt{3} ) = 0.$

$\tt \implies( \sqrt{3} x - 2)(x - 2 \sqrt{3} ) = 0.$

Either,

$\tt \implies \sqrt{3} x - 2 = 0. \\ \tt \implies x = \frac{2}{ \sqrt{3} }$

And,

$\tt \implies x - 2 \sqrt{3} = 0. \\ \tt \implies x = 2 \sqrt{3} .$

Now,

Let Both roots be,

$\tt \alpha = \frac{2}{ \sqrt{3} } \: \: and \: \: \beta = 2 \sqrt{3}$

Compare with given equation,and Verify

Sum of roots,

$\tt \implies\alpha + \beta = \frac{ - b}{a}$

$\tt \implies \frac{2}{ \sqrt{3} } + 2 \sqrt{3} = \frac{8}{ \sqrt{3} }$

$\tt \implies\frac{8}{ \sqrt{3} } = \frac{8}{ \sqrt{3} }$

Product of roots,

$\tt \implies \alpha \beta = \frac{c}{a}$

$\implies \tt \frac{2}{\cancel{ \sqrt{3}} } \times 2 \cancel{\sqrt{3}} = \frac{4 {\sqrt{3} }}{{ \sqrt{3}} }$

$\tt \implies4 = 4.$

Therefore,the roots of this given equation be 2√3 and 2/√3.

Answer 2

Answer:

× = 2 /3 and × = 2 /3.

Step-by-step explanation:

general equation o