3. Review · Equations and systems

Polynomial equations

1 Solve the polynomial equations

a) $3x^{3} + 2x - 1 = 2\,(x^{3} - 2x^{2}) + 6$

Move everything to the left-hand side:

3x^{3}+2x-1-2x^{3}+4x^{2}-6 = 0 \;\Longrightarrow\; x^{3}+4x^{2}+2x-7 = 0.

We try $a=1$ with Ruffini's rule:

	$1$	$4$	$2$	$-7$
$1$		$1$	$5$	$7$

	$1$	$5$	$7$	$0$

Therefore, $\;(x-1)(x^{2}+5x+7) = 0$. The factor $\,x^{2}+5x+7\,$ has $\Delta = 25-28 = -3 < 0$, so it is irreducible.

\boxed{\;x = 1.\;}

b) $x^{3} - 2x^{2} = 3x - 6$

Everything to the left-hand side and we factor by grouping:

x^{3}-2x^{2}-3x+6 = x^{2}(x-2) - 3(x-2) = (x-2)(x^{2}-3) = 0.

$x-2 = 0 \Rightarrow x = 2$. $x^{2}-3 = 0 \Rightarrow x = \pm\sqrt{3}$.

\boxed{\;x \in \{-\sqrt{3},\,\sqrt{3},\,2\}.\;}

c) $(x+1)^{2} - (x-2)^{2} = (x+3)^{2} + x^{2} - 20$

We expand both sides:

\text{LHS} = (x^{2}+2x+1)-(x^{2}-4x+4) = 6x-3.

\text{RHS} = (x^{2}+6x+9)+x^{2}-20 = 2x^{2}+6x-11.

We set them equal and simplify:

6x-3 = 2x^{2}+6x-11 \;\Longrightarrow\; 2x^{2} = 8 \;\Longrightarrow\; x^{2}=4.

\boxed{\;x = \pm 2.\;}

d) $x^{4} + x^{2} = 2x^{3}$

Everything to the left-hand side and we take out the common factor $x^{2}$:

x^{4}-2x^{3}+x^{2} = x^{2}(x^{2}-2x+1) = x^{2}(x-1)^{2} = 0.

\boxed{\;x = 0\ \text{(double)},\ \ x = 1\ \text{(double)}.\;}

e) $x^{3} - 5x^{2} + 7x - 3 = 0$

We try $a=1$ with Ruffini's rule:

	$1$	$-5$	$7$	$-3$
$1$		$1$	$-4$	$3$

	$1$	$-4$	$3$	$0$

So $\;(x-1)(x^{2}-4x+3) = 0$. We factor the second factor: $x^{2}-4x+3 = (x-1)(x-3)$.

x^{3}-5x^{2}+7x-3 = (x-1)^{2}(x-3) = 0.

\boxed{\;x = 1\ \text{(double)},\ \ x = 3.\;}

f) $x^{3} = 2\,(2x + 4 - x^{2})$

Everything to the left-hand side:

x^{3}+2x^{2}-4x-8 = 0.

By grouping:

x^{2}(x+2)-4(x+2) = (x+2)(x^{2}-4) = (x+2)^{2}(x-2) = 0.

\boxed{\;x = -2\ \text{(double)},\ \ x = 2.\;}

Biquadratic equations

2 Solve the biquadratic equations

a) $x^{4} + 4x^{2} = 0$

Common factor $x^{2}$:

x^{2}(x^{2}+4) = 0.

$x^{2}+4$ has no real roots (a sum of positives), so only $x^{2}=0$ remains.

\boxed{\;x = 0\ \text{(double)}.\;}

b) $x^{4} - 8x^{2} = 9$

Substitution $x^{2} = t$: $\;t^{2}-8t-9 = 0$.

t = \dfrac{8 \pm \sqrt{64+36}}{2} = \dfrac{8\pm 10}{2} \;\Rightarrow\; t = 9\ \text{or}\ t = -1.

If $t=-1$: $x^{2}=-1$ has no real roots. If $t=9$: $x = \pm 3$.

\boxed{\;x = \pm 3.\;}

c) $x^{6} + x^{2} + 216 = x^{2}\,(1 + 35x)$

We expand the right-hand side and move everything to the left:

x^{6}+x^{2}+216 = x^{2}+35x^{3} \;\Longrightarrow\; x^{6}-35x^{3}+216 = 0.

Substitution $x^{3} = t$: $\;t^{2}-35t+216 = 0$.

t = \dfrac{35 \pm \sqrt{1225-864}}{2} = \dfrac{35 \pm 19}{2} \;\Rightarrow\; t = 27\ \text{or}\ t = 8.

$x^{3}=27 \Rightarrow x = 3$. $x^{3}=8 \Rightarrow x = 2$.

\boxed{\;x \in \{2,\,3\}.\;}

d) $36 = x^{2}\,(13 - x^{2})$

We expand and move everything to one side:

36 = 13x^{2}-x^{4} \;\Longrightarrow\; x^{4}-13x^{2}+36 = 0.

Substitution $x^{2} = t$: $\;t^{2}-13t+36 = 0$.

t = \dfrac{13 \pm \sqrt{169-144}}{2} = \dfrac{13 \pm 5}{2} \;\Rightarrow\; t = 9\ \text{or}\ t = 4.

$x^{2}=9 \Rightarrow x = \pm 3$. $x^{2}=4 \Rightarrow x = \pm 2$.

\boxed{\;x \in \{-3,\,-2,\,2,\,3\}.\;}

Rational equations

3 Solve the rational equations

Remember the 5 steps: factor, LCM, build common denominators, clear the denominators and solve. At the end, check that no solution makes the original denominators zero.

a) $\dfrac{1}{x} + \dfrac{1}{x+3} = \dfrac{3}{10}$

We add the left-hand side and cross-multiply:

\dfrac{(x+3)+x}{x(x+3)} = \dfrac{3}{10} \;\Longrightarrow\; 10\,(2x+3) = 3x(x+3).

20x+30 = 3x^{2}+9x \;\Longrightarrow\; 3x^{2}-11x-30 = 0.

x = \dfrac{11 \pm \sqrt{121+360}}{6} = \dfrac{11 \pm \sqrt{481}}{6}.

Neither value makes $x$ or $x+3$ zero, so both are valid:

\boxed{\;x = \dfrac{11+\sqrt{481}}{6} \approx 5.49,\quad x = \dfrac{11-\sqrt{481}}{6} \approx -1.82.\;}

b) $\dfrac{4}{x} + \dfrac{2x+2}{3x-6} = 4$

We factor $3x-6 = 3(x-2)$. LCM: $3x(x-2)$. We clear the denominators:

$$12(x-2) + x(2x+2) = 12x(x-2).$$

12x-24 + 2x^{2}+2x = 12x^{2}-24x \;\Longrightarrow\; 10x^{2}-38x+24 = 0.

We divide by $2$: $\;5x^{2}-19x+12 = 0$. $\Delta = 361-240 = 121$:

x = \dfrac{19 \pm 11}{10} \;\Rightarrow\; x = 3\ \text{or}\ x = \tfrac{4}{5}.

Neither makes a denominator zero. Verified:

\boxed{\;x = 3\ \text{or}\ x = \tfrac{4}{5}.\;}

c) $\dfrac{8-x}{2} - \dfrac{2x-11}{x-3} = \dfrac{x+6}{2}$

LCM: $2(x-3)$. We clear the denominators:

(8-x)(x-3) - 2\,(2x-11) = (x+6)(x-3).

We expand each product:

(-x^{2}+11x-24) - (4x-22) = (x^{2}+3x-18).

-x^{2}+7x-2 = x^{2}+3x-18 \;\Longrightarrow\; 2x^{2}-4x-16 = 0 \;\Longrightarrow\; x^{2}-2x-8 = 0.

x = \dfrac{2 \pm \sqrt{4+32}}{2} = \dfrac{2 \pm 6}{2} \;\Rightarrow\; x = 4\ \text{or}\ x = -2.

\boxed{\;x = 4\ \text{or}\ x = -2.\;}

d) $\dfrac{2x}{x-3} - \dfrac{6}{x} = \dfrac{18}{x^{2}-3x}$

We factor $x^{2}-3x = x(x-3)$. LCM: $x(x-3)$. We clear the denominators:

2x \cdot x - 6(x-3) = 18 \;\Longrightarrow\; 2x^{2}-6x+18 = 18.

2x^{2}-6x = 0 \;\Longrightarrow\; 2x(x-3) = 0 \;\Rightarrow\; x = 0\ \text{or}\ x = 3.

Careful! Both $x=0$ and $x=3$ make the original denominators zero ($x$ and $x-3$). Both are extraneous solutions.

\boxed{\;\text{No solution.}\;}

This is a very typical example of the extraneous-solutions trap — you must always check.

e) $\dfrac{x-1}{x-2} - \dfrac{2x-2}{x^{2}+3x} = \dfrac{5x-5}{x^{2}+x-6}$

We factor: $\;x^{2}+3x = x(x+3)$, $\;x^{2}+x-6 = (x+3)(x-2)$, $2x-2=2(x-1)$, $5x-5=5(x-1)$. We rewrite the equation:

\dfrac{x-1}{x-2} - \dfrac{2(x-1)}{x(x+3)} = \dfrac{5(x-1)}{(x+3)(x-2)}.

LCM: $x(x-2)(x+3)$. We clear the denominators:

(x-1)\cdot x(x+3) - 2(x-1)(x-2) = 5(x-1)\cdot x.

We take out the common factor $(x-1)$:

(x-1)\bigl[\,x(x+3) - 2(x-2) - 5x\,\bigr] = 0.

(x-1)\bigl[\,x^{2}+3x-2x+4-5x\,\bigr] = (x-1)(x^{2}-4x+4) = (x-1)(x-2)^{2} = 0.

Therefore $x = 1$ or $x = 2$. But $x = 2$ makes the denominator $x-2$ zero (extraneous). $x = 1$ does not make any denominator zero.

\boxed{\;x = 1.\;}

f) $\dfrac{x+4}{x-3} - \dfrac{1-2x}{x^{2}-x-6} = 0$

We factor $x^{2}-x-6 = (x-3)(x+2)$. LCM: $(x-3)(x+2)$:

$$(x+4)(x+2) - (1-2x) = 0.$$

x^{2}+6x+8 - 1 + 2x = x^{2}+8x+7 = 0 \;\Rightarrow\; (x+1)(x+7) = 0.

$x = -1$ and $x = -7$. Neither makes a denominator zero:

\boxed{\;x = -1\ \text{or}\ x = -7.\;}

g) $\dfrac{3}{x} + \dfrac{4}{2x^{2}-4x} = \dfrac{1}{2x-4}$

We factor: $\;2x^{2}-4x = 2x(x-2)$, $\;2x-4 = 2(x-2)$. LCM: $2x(x-2)$. We clear the denominators:

3 \cdot 2(x-2) + 4 = x.

6x-12+4 = x \;\Longrightarrow\; 6x-8 = x \;\Longrightarrow\; 5x = 8.

$x = \tfrac{8}{5}$. It does not make any denominator zero:

\boxed{\;x = \tfrac{8}{5}.\;}