Question

${x}^{2} + 2 \sqrt{2x} - 6$
find the zeroes of a quadratic polynomial and verify the relationship between them​

Answer 1

To find the zeroes of polynomial p(x)

$p(x) = {x}^{2} + 2 \sqrt{2}x - 6 = 0$

$\implies {x}^{2} + 2\sqrt{2}x - 6 =0$

$\implies {x}^{2} + 3\sqrt{2}x - \sqrt{2}x - 6 = 0$

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

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

$x = \sqrt{2} \\ x = -3\sqrt{2}$

Zeroes of $p(x) = {x}^{2} + 2 \sqrt{2}x - 6 = 0$ are $\sqrt{2}$ and $-3\sqrt{2}$

To verify the relationship:

Sum of zeroes = $-3\sqrt{2} + \sqrt{2}= - 2\sqrt{2}$

Product of zeroes = $-3\sqrt{2} × \sqrt{2}= -6$

Sum of zeroes = $\dfrac{-Coefficient \; of\; x}{Coefficient\; of \; {x}^{2}}$

$= \dfrac{-2\sqrt{2}}{1} =-2\sqrt{2}$

Product of zeroes = $\dfrac{Constant \; term}{Coefficient \;of \;{x}^{2}}$ $=\dfrac {-6}{1} = -6$

Hence verified.

Answer 2

Answer:-

Quadratic Equation- x² + 2√2x - 6

= x² + 2√2x - 6

= x² + ( 3√2 - √2 )x - 6

= x² + 3√2x - √2x - 6

= x ( x + 3√2) -√2 ( x + 3√2)

= ( x + 3√2) and ( x - √2)

= ( x = - 3√2) and ( x = √2)

Verification:-

• Sum of Roots = -b/a

=> -3√2 + √2 = -2√2/1

=> -2√2 = -2√2

• Product of Roots = c/a

=> ( -3√2) × ( √2) = -6/1

=> -6 = -6

Hence Verified!