Question

Find the zeros of the polynomial f(x) =
$6 {x}^{2} - 3$
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Solution needed with explanation​

Answer 1

Answer:

$x = ± \sqrt{ \frac{1}{2} }$

Step-by-step explanation:

We are given that $6 {x}^{2} - 3$

To find : Zeroes of the polynomial

$\implies$ 6x² - 3 = 0

Taking out 3 as common

$\implies$ 3(2x² - 1 ) = 0

Dividing both sides by 3

$\implies$ 2x² - 1 = 0

$\implies$ 2x² = 1

${x}^{2} = \frac{1}{2} \\ \\ <em><u>Taking \: square \: root</u></em> \\ \\ x = ± \sqrt{ \frac{1}{2} }$

By taking square root , the gained value will be both negative & positive.

So, here we get

•°• X = + √1/2 & - √1/2

are the values or zeroes of the polynomial which satisfies it.

Answer 2

Answer:

Step-by-step explanation:

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