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3 years ago

IM AM SO CONFUSED

Mathematics

1 answer:

Semenov [28]3 years ago

6 0

Hello,

Answer (1/2,-1/4)

$8x^2-2x-1=0\\\\8x^2-4x+2x-1=0\\\\4x(2x-1)+2x-1=0\\\\(2x-1)(4x+1)=0\\\\sol=\{\dfrac{1}{2} ,-\dfrac{1}{4} \}\\$

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Divide. −3/8÷−6/7 . plsssss help

Keith_Richards [23]

Answer:

7/16

Step-by-step explanation:

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Can someone help me with this plz

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Solve (x + 3)2 + (x + 3) – 2 = 0. Let u = Rewrite the equation in terms of u. (u2 + 3) + u – 2 = 0 u2 + u – 2 = 0 (u2 + 9) + u –

Verdich [7]

Answer:

The solutions of the original equation are x=-5 and x=-2

Step-by-step explanation:

we have

$(x+3)^2+(x+3)-2=0$

Let

$u=(x+3)$

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$(u)^2+(u)-2=0$

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$u^2+u=2$

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<em>Alternative Method</em>

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$(u)^2+(u)-2=0$

$a=1\\b=1\\c=-2$

substitute in the formula

$u=\frac{-1\pm\sqrt{1^{2}-4(1)(-2)}} {2(1)}$

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$u=\frac{-1\pm3} {2}$

$u=\frac{-1+3} {2}=1$

$u=\frac{-1-3} {2}=-2$

the solutions are

u=-2,u=1

<em>Find the solutions of the original equation</em>

For u=-2

$-2=(x+3)$ ----> $x=-2-3=-5$

For u=1

$1=(x+3)$ ----> $x=1-3=-2$

therefore

The solutions of the original equation are

x=-5 and x=-2

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Answer:

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Maru [420]

Answer:

this is correct

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