Question

Find the zeroes of the polynomial x² -√2x-12

Answer 1

Answer:

Given, f(x) = x² - √2x - 12  = 0

x² - √2x - 12 = 0

x² - 3√2x + 2√2x - 12 = 0

x.( x - 3√2 ) + 2√2.( x - 3√2 ) = 0

( x + 2√2 ).( x - 3√2 ) = 0

so, -2√2 and 3√2 are the zeros of the given polynomial.

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Answer 2

Given,

• The polynomial x² -√2x-12 is given.

To find,

• We have to find the zeroes of the polynomial x² -√2x-12.

Solution,

We can simply find the zeroes of the polynomial x² -√2x-12 by equating the given polynomial to 0.

            x² -√2x-12 = 0

Now, factorizing the above polynomial, we get

We will factorize the polynomial by the method of splitting the middle term.

          = x² -√2x-12

          = x² -(3√2-2√2)x -12

         = x²-3√2x+2√2x-12

Taking x common from the first two terms and 2√2 common from the last two terms, we get

         = x(x-3√2) +2√2(x-3√2)

Now, taking (x-3√2) common, we get

         (x-3√2)(x+2√2) = 0

         (x-3√2) = 0 , (x+2√2) = 0

                  x = 3√2 ,  x = -2√2

Hence, the zeroes of the polynomial x² -√2x-12 are 3√2 and -2√2.