Question

find the zeros of each of the following quadratic polynomics and verify the relationship X2+2√2x-6​

Answer 1

$a(x-\frac{-b+\sqrt{{b}^{2}-4ac}}{2a})$

$(x-\frac{-2\sqrt{2}+\sqrt{{(2\sqrt{2})}^{2}-4\times -6}}{2})$

$(x-\sqrt{2})(x+3\sqrt{2})$

Answer 2

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 √2and -3√2.

  Now, for verification  Sum of zeros = – c o e f f i c i e n t o f x c o e f f i c i e n t o f x 2 – coefficientofx/coefficientofx2  √2 + (-3√2) = – ( 2 √ 2 ) 1 (22)1  -2√2 = -2√2  Product of roots = c o n s t a n t c o e f f i c i e n t o f x 2 constantcoefficientofx2 √2 x (-3√2) = ( − 6 ) 2 √ 2 (−6)22  -3 x 2 = -6/1 -6 = -6  Therefore, the relationship between zeros and their coefficients is verified.