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3 years ago

Factor the perfect-square trinomial in y = (x2 + 2x + 1) − 1− 1. y = (x + )2 − 1 −1

Mathematics

2 answers:

dolphi86 [110]3 years ago

4 0

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$y = ( {x}^{2} + 2x + 1) - 1$

$y = ({x + 1})^{2} - 1$

$y = ({x + 1})^{2} - ({1})^{2}$

We're done now but I have something to tell...

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Hint :

${a}^{2} - {b}^{2} = (a + b)(a - b)$

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$y = ( \: (x + 1) + 1)( \: (x + 1) - 1) \\$

$y = (x + 2)(x)$

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Anestetic [448]3 years ago

3 0

Answer:

Step-by-step explanation:

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Problem

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Answer:

$-\dfrac{463}{7}$

Explanation:

Given equation: 2x² - 6x - 7 = 0

In quadratic equation: ax² + bx + c

$Sum \ of \ roots : \alpha + \beta = \dfrac{-b}{a}$

$product \ of \ roots : \alpha \beta = \dfrac{c}{a}$

So, here given:

$Sum : \alpha + \beta = \dfrac{-(-6)}{2} = 3$

$Product : \alpha \beta = \dfrac{-7}{2} = - 3.5$

For finding value:

$\dfrac{\alpha^3}{\beta } + \dfrac{\beta^3 }{\alpha }$

join fractions

$\dfrac{\alpha^4+ \beta^4}{\alpha \beta }$

factor out

$\dfrac{(\alpha^2 + \beta ^2)^2 -2\alpha ^2 \beta ^2 }{\alpha \beta }$

when factored more

$\dfrac{((\alpha + \beta)^2 -2\alpha \beta )^2 -2(\alpha \beta)^2 }{\alpha \beta }$

insert values inside

$\dfrac{((3)^2 -2(-3.5) )^2 -2(-3.5)^2 }{-3.5 }$

calculate for value

$-\dfrac{463}{7}$

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2 years ago