Question

plz solve this answer

Answer 1

$\sqrt{3} {x}^{2} - 8x + 4 \sqrt{3}$

Here,

$a = \sqrt{3} \\ b = - 8 \\ c = 4 \sqrt{3}$

Now,

$\sqrt{3} {x}^{2} - 8x + 4 \sqrt{3} \\ = \sqrt{3} {x}^{2} - (6 + 2)x + 4 \sqrt{3} \\ = \sqrt{3} {x}^{2} - 6x - 2x + 4 \sqrt{3} \\ = \sqrt{3} x(x - 2 \sqrt{3} ) - 2(x - 2 \sqrt{3} ) \\ = ( \sqrt{3} x - 2)(x - 2 \sqrt{3} )$

Therefore, zeroes are =

$\alpha = 2 \sqrt{3} \\ \beta = \frac{2}{ \sqrt{3} }$

Now, relationship between coefficients and zeroes:

$\alpha + \beta = \frac{ - b}{a} \\ \alpha \times \beta = \frac{c}{a}$

1.

LHS=

$\alpha + \beta \\ = 2 \sqrt{3} + \frac{2}{ \sqrt{3} } \\ = \frac{6 + 2}{ \sqrt{3} } \\ = \frac{8}{ \sqrt{3} }$

RHS=

$\frac{ - b}{a} \\ = \frac{ - ( - 8)}{ \sqrt{3} } \\ = \frac{8}{ \sqrt{3} }$

LHS=RHS

2.LHS=

$\alpha \times \beta \\ = 2 \sqrt{3} \times \frac{2}{ \sqrt{3} } \\ = 2 \times 2 \\ = 4$

RHS=

$\frac{c}{a} \\ = \frac{4 \sqrt{3} }{ \sqrt{3} } \\ = 4$

LHS = RHS

Hence, verified.