+961 What is the solution to the equation 1−5(x3+x2)=-5x2−4(x3+2)?
+33661 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.
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+26393 What is the solution to the equation 1−5(x3+x2)=-5x2−4(x3+2) ?
$$\small{\text{$
\begin{array}{rcl}
1-5(x^3+x^2) &=& -5x^2-4(x^3+2) \\\\
1-5(x^3+x^2) + 5x^2 +4(x^3+2) &=& 0\\\\
1 - 5x^3 - 5x^2 + 5x^2 + 4x^3 + 8 &=& 0 \\\\
-5x^3 - 5x^2 + 5x^2 + 4x^3 + 9 &=& 0 \\\\
-5x^3 \underbrace{- 5x^2 + 5x^2}_{=0} + 4x^3 + 9 &=& 0 \\\\
-5x^3 + 4x^3 + 9 &=& 0 \\\\
-x^3 + 9 &=& 0 \\\\
9 &=& x^3 \\\\
x^3 &=& 9 \qquad | \qquad \sqrt[3]{}\\\\
x &=& \sqrt[3]{9}\\\\
x &=& 2.0800838231
\end{array}
$}}\\\\
\mathbf{ x = 2.0800838231 }$$
+33661 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).
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+961 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
+33661 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.
.
