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#6**+10 **

You are right up until the very last step shaniab29544. On the next to last step you have x3 = 9. Because the 3 comes after the x we assume this means x to the power 3 (at least that is what heureka and I have assumed), so x-cubed = 9. This means that x is the cube root of 9 or ^{3}√9.

If you prefer to interpret this as just x times 3 then we would have x times 3 = 9 so that x would be 9 divided by 3, or x = 3.

.

Alan May 16, 2015

#1**+10 **

**What is the solution to the equation 1−5(x3+x2)=-5x2−4(x3+2) ?**

$$\small{\text{$

\begin{array}{rcl}

1-5(x^3+x^2) &=& -5x^2-4(x^3+2) \\\\

1-5(x^3+x^2) + 5x^2 +4(x^3+2) &=& 0\\\\

1 - 5x^3 - 5x^2 + 5x^2 + 4x^3 + 8 &=& 0 \\\\

-5x^3 - 5x^2 + 5x^2 + 4x^3 + 9 &=& 0 \\\\

-5x^3 \underbrace{- 5x^2 + 5x^2}_{=0} + 4x^3 + 9 &=& 0 \\\\

-5x^3 + 4x^3 + 9 &=& 0 \\\\

-x^3 + 9 &=& 0 \\\\

9 &=& x^3 \\\\

x^3 &=& 9 \qquad | \qquad \sqrt[3]{}\\\\

x &=& \sqrt[3]{9}\\\\

x &=& 2.0800838231

\end{array}

$}}\\\\

\mathbf{ x = 2.0800838231 }$$

heureka May 13, 2015

#4**+5 **

What makes you think it's wrong shaniab29544?

heureka has given the correct real number root (there are two other roots, but they are complex).

.

Alan May 15, 2015

#5**+10 **

**it like this i think **

1−5(x3+x2)=-5x2−4(x3+2)1−5x3−5x2=-5x2−4x3−81−5x2=-5x2+x3−89−5x2=-5x2+x39=x3 27=x

shaniab29544 May 15, 2015

#6**+10 **

Best Answer

You are right up until the very last step shaniab29544. On the next to last step you have x3 = 9. Because the 3 comes after the x we assume this means x to the power 3 (at least that is what heureka and I have assumed), so x-cubed = 9. This means that x is the cube root of 9 or ^{3}√9.

If you prefer to interpret this as just x times 3 then we would have x times 3 = 9 so that x would be 9 divided by 3, or x = 3.

.

Alan May 16, 2015