Question

find the zeroes of the polynomial x^2-root2-12​

Answer 1

Answer:

3√2 and (-2√2)

Step-by-step explanation:

Given that-

x² - √2x - 12

We need to find two numbers such that, the product of numbers be (-12) and the sum of numbers be (-√2).

Such two numbers can be (-3√2) and (2√2).

x² - √2x - 12

On splitting it's middle term:

⇒ x² - 3√2x + 2√2x - 12

⇒ x ( x - 3√2 ) + 2√2 ( x - 3√2 )

⇒ (x - 3√2)(x + 2√2)

By zero product rule-

⇒ x - 3√2 = 0 or x + 2√2 = 0

⇒ x = 3√2 or x = -2√2

Hence, the zeroes are 3√2 and (-2√2).

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Verification:

x² - √2x - 12

Here, a = 1, b = √2 and c = - 12

Sum of zeroes = -b/a

⇒ 3√2 - 2√2 = √2/1

⇒ √2 = √2

Product of zeroes = c/a

⇒ 3√2 * (-2√2) = -12/1

⇒ -12 = -12

Hence, it is verified.

Answer 2

Step-by-step explanation:

$\mathsf{let \: the \: polynomial \: be \: p(x).} \\ \\ \mathsf{given \: polynomial : } \\ \mathsf{p(x) = x {}^{2} - \sqrt{2}x - 12 } \\ \\ \mathsf{ \implies \: x {}^{2} - \sqrt{2}x - 12 = 0 } \\ \\ \mathsf{ \implies \: x {}^{2} - 3\sqrt{2}x + 2 \sqrt{2}x - 12 = 0 } \\ \\ \mathsf{ \implies \: x (x - 3 \sqrt{2}) + 2 \sqrt{2}( x - 3 \sqrt{2}) = 0 } \\ \\ \mathsf{ \implies \: (x + 2 \sqrt{2})(x - 3 \sqrt{2} ) = 0 } \\ \\ \implies \: \boxed{ \mathsf{x = 3 \sqrt{2} \: \: or \: \: - 2 \sqrt{2} }}$

The zeros of the given p(x) are 3√2 and -2√2