Question

find the zeros of the quadratic polynomial p(x) = x^2+2√2x-6 and verify the relationship between the zeros and its coefficient ​

Answer 1

AnswEr:

• x² + 2√2x - 6

Finding the Zeroes of the polynomial by middle term splitting.

$:\implies\sf\: x^2 + 2\sqrt{2}x - 6$

$:\implies\sf\: x^2 + 3\sqrt{2}x - \sqrt{2} x - 6$

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

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

So, The Zeroes are $\sf\: -3\sqrt{2} \: and \: \sqrt{2}$

$\rule{200}2$

$\diamond\: \large\bold{\underline{\sf{\red{Now\; Verification}}}}$

• Sum of Zeroes = $\sf\: (\alpha \: + \; \beta)$

$:\implies\sf\: (\alpha \: + \; \beta) = \dfrac{-b}{a}$

$:\implies\sf\: -3\sqrt{2} + \sqrt{2} = -2\sqrt{2}$

$:\implies\sf\: -2\sqrt{2} = -2\sqrt{2}$

• Product of Zeroes = $\sf\:(\alpha\:\beta)$

$:\implies\sf\: (\alpha\:\beta) = \dfrac{c}{a}$

$:\implies\sf\: (-3 \sqrt{2} ) ( \sqrt{2}) = -6$

$:\implies\sf\: -6 = -6$

$\bold{\underline{\sf{\blue{Hence\: Verified!}}}}$

Answer 2

$\mathtt{ \huge{ \fbox{Solution :) }}}$

Given ,

The polynomial is (x)² + 2√2x - 6

By prime factorisation method

(x)² + 2√2x - 6 = 0

(x)² + 3√2x - √2x - 6 = 0

x(x + 3√2) - √2(x + 3√2) = 0

(x - √2)(x + 3√2) = 0

x = √2 or x = -3√2

Hence , the zeroes of the polynomial are √2 and -3√2

$\mathtt{ \huge{ \fbox{Verification :) }}}$

We know that ,

$\mathtt{ \large \fbox{Sum \: of \: roots = \frac{ - b}{a} }}$

Thus ,

√2 + (-3√2) = -2√3/1

-2√3 = - 2√3

and

$\large \mathtt {\fbox{Product \: of \: roots = \frac{c}{a}}}$

Thus ,

√2 × (-3√2) = -6/1

-6 = -6

Hence verified

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