Question

Find the zeros and verify its relationship with coefficients:
$x {}^{2} - 11x - 42$
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Answer 1

Answer:

Question:

Find the zeros and verify its relationship with coefficients of x² - 11x - 42

Solution:

$: { \implies{ \sf{ {x}^{2} - 11x - 42}}} \\ \\ : { \implies{ \sf{ {x}^{2} - 14x + 3x - 42 }}} \\ \\ : { \implies{ \sf{ x(x - 14) + 3(x - 14) }}} \\ \\ : { \implies{ \sf{(x + 3)(x - 14) = 0}}} \\ \\ : { \implies{ \sf{x = - 3 \: or \: 14}}}$

So, The Zeroes of polynomial are -3, 14

_______________________

Verifying Relation between zeroes & coefficients:

• $\alpha = - 3$
• $\beta = 14$

${ \sf{Sum \: of \: Zeroes( \alpha + \beta ) = </p><p> \frac{ - b}{a} }} \\ \\ - 3 + 14 = \frac{ - ( - 11)}{1} \\ \\ 11 = \frac{11}{1} \\ \\ 11 = 11$

${ \sf{Product \: of \: Zeroes( \alpha \beta ) = \frac{c}{a} }} \\ \\ - 3(14) = \frac{ - 42}{1} \\ \\ - 42 = - 42$

Hence Proved

Answer 2

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