solving equations

**william** · March 1st 2009, 03:39 PM

Solve the following algebraically
$a)x^3+5x^2=2x+6$
$b)x^4-13x^2=-36$

**Prove It** · March 1st 2009, 03:47 PM

Originally Posted by william

Solve the following algebraically
$a)x^3+5x^2=2x+6$
$b)x^4-13x^2=-36$

a) $x^3 + 5x^2 - 2x - 6 = 0$

Notice that $f(-1) = (-1)^3 + 5(-1)^2 - 2(-1) - 6 = -1 + 5 + 2 - 6 = 0$

So by the factor and remainder theorems, $x + 1$ is a factor.

Using long division, you should find that

$f(x) = (x + 1)(x^2 + 4x - 6) = 0$

By the null factor law

$x + 1 = 0$ or $x^2 + 4x - 6 = 0$

The first equation is easy to solve, the second is solved using the Quadratic Formula.

So $x = -2 - \sqrt{10}, x = -1, x = -2 + \sqrt{10}$ are the solutions.

**Prove It** · March 1st 2009, 03:50 PM

Originally Posted by william

Solve the following algebraically
$a)x^3+5x^2=2x+6$
$b)x^4-13x^2=-36$

b) $x^4 - 13x^2 + 36 = 0$

Notice that if I write $X = x^2$ the equation becomes

$X^2 - 13X + 36 = 0$

$(X - 4)(X - 9) = 0$

$X = 4$ or $X = 9$ .

Since $X = x^2$

$x^2 = 4$ or $x^2 = 9$

So $x = -3, x = -2, x = 2, x = 3$ are the solutions.

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