Question

please solve this problem?????​

Answer 1

Answer:

easy

Step-by-step explanation:

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Answer 2

Question :-

Find the zeros of the quadratic polynomial $\sf \sqrt{3}x^2 - 8x + 4 \sqrt{3}$?

⠀

Answer :-

$\sf \sqrt{3}x^2 - 8x + 4 \sqrt{3} = 0$

By using middle term splitting method -

➩ $\sf \sqrt{3}x^2 - 8x + 4 \sqrt{3} = 0$

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

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

➩ $\sf x ( \sqrt{3}x - 2 ) - 2 ( \sqrt{3} \times \sqrt{3} x + 2 \sqrt{3}) = 0$

➩ $\sf x ( \sqrt{3}x - 2 ) - 2\sqrt{3}(\sqrt{3}x - 2) = 0$

➩ $\sf (x - 2\sqrt{3})(\sqrt{3}x - 2) = 0$

⠀

Either $\sf (x - 2\sqrt{3})$ is zero or $\sf (\sqrt{3}x - 2)$ is zero -

• $\sf x - 2\sqrt{3} = 0$

➩ $\sf x= 2\sqrt{3}$

⠀

• $\sf \sqrt{3}x - 2 = 0$

➩ $\sf \sqrt{3}x = 2$

➩ $\sf x = \frac{2}{\sqrt{3}}$

⠀

$\boxed{\sf Zeros = 2\sqrt{3} \: and \: \frac{2}{\sqrt{3}}}$