Question

Q22. Find the zeros of the polynomial f(x) = x² + 2√2 x-6 = 0 and verify
the relationship between zeros and its coefficients.​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

Given,

p(x) = x2 + 2√2x – 6

We put p(x) = 0

⇒ x2 + 2√2x – 6 = 0

⇒ x2 + 3√2x – √2x – 6 = 0

⇒ x(x + 3√2) – √2 (x + 3√2) = 0

⇒ (x – √2)(x + 3√2) = 0

This gives us 2 zeros, for x = √2 and x = -3√2

Hence, the zeros of the quadratic equation are √2 and -3√2.

Now, for verification

Sum of zeros = –coefficientofxcoefficientofx2–coefficientofxcoefficientofx2

√2 + (-3√2)

= – (22√)1(22)1 -2√2

= -2√2

Product of roots = constantcoefficientofx2constantcoefficientof

x2 √2 x (-3√2)

= (−6)22√(−6)22 -3 x 2

= -6/1 -6

= -6

Therefore, the relationship between zeros and their coefficients is verified.

hope it will help you