Quadratic Equations

**ecogreen** · July 14th 2010, 04:46 AM

Need help in solving the following quadratic equations to the power of three...

$x^3-y^3=35$
$x-y=5$

Pls explain or show any relevant working. Thanks in advance for your help.

**Ackbeet** · July 14th 2010, 04:58 AM

What have you tried thus far?

**eumyang** · July 15th 2010, 01:13 PM

Use substitution. Solve the 2nd equation for x:
$x = y + 5$

Substitute this into the 1st equation for x:
$\begin{aligned} x^3 - y^3 &= 35 \\ (y + 5)^3 - y^3 &= 35 \\ y^3 + 15y^2 + 75y + 125 - y^3 &= 35 \\ 15y^2 + 75y + 125 &= 35 \\ 15y^2 + 75y + 90 &= 0 \end{aligned}$
Solve this quadratic (you'll get two answers for y). Plug both into the equation
$x = y + 5$
to find the corresponding x values.

Thread: Quadratic Equations

LinkBack

Thread Tools

Display

Quadratic Equations

Similar Math Help Forum Discussions

quadratic equations

quadratic equations?

Quadratic Equations

Converting Quadratic equations to Vertex form and solving quadratic inequalities

5 Quadratic Equations

Search Tags