Question

Find the quadratic polynomial if the sum and product of the zeroes are √2 and -3/2.Also find the zeroes.

Answer 1

Answer:

Here it's given that....

sum of zeroes = √2

product of zeroes = -3/2

we know that........

p(x) = x² - ( sum of zeroes ) + ( product of zeroes )

x² - √2 + (-3/2)

= x² - √2 - 3/2

= 2x² - 2√2 -3

This your required polynomial......

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Answer 2

Answer: k(2$x^{2}$ -2$\sqrt{2\\}$ x-3)

Step-by-step explanation:

Product of the zeroes = (c/a)

=> (-3/2) = (c/a)

This will give that c = (-3) and a = 2

Now,

Sum of the zeroes = (-b/a)

=> $\sqrt{2}$ = (-b/2)                              (a = 2, found earlier)

=> 2$\sqrt{2\\}$ = (-b)

=>  -(2$\sqrt{2}$ ) = b

This gives that a = 2,b = (-2$\sqrt{2}$) and c = (-3)

So the equation will be 2$x^{2}$ -2$\sqrt{2\\}$ x-3

But, since we derived a,b,c from fractions. They are in the form of ratios. So, we need to multiply any natural no. 'k' to complete the answer

Final answer k(2$x^{2}$ -2$\sqrt{2\\}$ x-3)

Hope it helps you!!

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