Question

xsq -2x -8 find the zeroes

Answer 1

In p(x) = x² + 4 x - 3

Then,

Sum of zeroes = -b/a ⇒ -4/1 ⇒ -4

Product of zeroes = c/a ⇒ -3/1 ⇒ -3
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New sum of zeroes = (1+α)/β + (1+β)/α

= α(1+α) + β(1+β) / αβ

= (α + α² + β + β²) / αβ

= ((α+β) + (α+β)² - 2αβ) / αβ

= (-4 + (-4)² - 2(-3) ) / (-3)

= (-4 + 16 + 6) / (-3)

= -18/3 ⇒ -6

New product of zeroes = ((1+α) / β) × ((1+β) / α)

= (1+α)(1+β) / αβ

= ((α+β) + αβ + 1) / αβ

= (-4 + (-3) + 1) / (-3)

= -6/-3

= 2
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Then new polynomial = x² - (sum of zeroes)x + (product of zeroes)

= x² -(-6)x + 2

= x² + 6 x + 2
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Answer 2

$x {}^{2} - 2x - 8 = 0$
$x {}^{2} - 4x + 2x - 8 =0$
$x(x - 4) + 2(x - 4) = 0$
$(x - 4)(x + 2) = 0 \\( x - 4) =0 and(x + 2) = 0$
$x = 4andx = - 2$