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What is the solution to the equation 15(x3+x2)=-5x24(x3+2)?

shaniab29544  May 13, 2015

Best Answer 

 #6
avatar+26328 
+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
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6+0 Answers

 #1
avatar+18712 
+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
 #2
avatar+91001 
0

Another beautifully presented answer.  Thank you Heureka  

Melody  May 14, 2015
 #3
avatar+961 
0

it wrong

shaniab29544  May 15, 2015
 #4
avatar+26328 
+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
avatar+961 
+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
avatar+26328 
+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

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