Find the zeros of quadratic polynomials and verify the relationship be

1.	Find the zeros of quadratic polynomials and verify the relationship between the zeros and their coefficients:p(x) = x2 + 2√2x – 6
Answer» 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 = \(\frac{– coefficient\, of\, x}{coefficient\, of\, x^2}\) √2 + (-3√2) = – \(\frac{(2\sqrt2)}{1}\) -2√2 = -2√2 Product of roots = \(\frac{constant}{coefficient\, of\, x^2}\) √2 x (-3√2) = \(\frac{(-6)}{ 2\sqrt2}\) -3 x 2 = -6/1 -6 = -6 Therefore, the relationship between zeros and their coefficients is verified.

Find the zeros of quadratic polynomials and verify the relationship between the zeros and their coefficients:p(x) = x2 + 2√2x – 6

Answer»

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 = \(\frac{– coefficient\, of\, x}{coefficient\, of\, x^2}\)

√2 + (-3√2) = – \(\frac{(2\sqrt2)}{1}\)

-2√2 = -2√2

Product of roots = \(\frac{constant}{coefficient\, of\, x^2}\)

√2 x (-3√2) = \(\frac{(-6)}{ 2\sqrt2}\)

-3 x 2 = -6/1

-6 = -6

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