Equations with fractions, with a root and by substitution

Algebra and calculus, grade 10 (4 units) Lesson 3 of 20Learning

Equations with fractions, with a root and by substitution

An equation with fractions

In an equation with fractions, first fix the domain: a denominator cannot be 0. Then multiply both sides by the common denominator, solve, and check that every solution is in the domain.

\frac{x^2}{x-3}=\frac{9}{x-3}

The domain: $\displaystyle x\neq3$ .
We multiply by $\displaystyle x-3$ : $\displaystyle {x^2}={9}$ , so $\displaystyle {x}={3}$ or $\displaystyle {x}={-3}$ .
$\displaystyle {x}={3}$ is not in the domain – there both denominators are 0. The only solution: $\displaystyle {x}={-3}$ .

An equation with a root

An equation with the unknown under a root is solved like this:

isolate the root on one side;
square both sides, and solve the equation you get;
put each solution into the original equation and check it.

The check is a must: squaring can add a solution. A root $\displaystyle \sqrt{\ }$ is never negative, while after squaring even $\displaystyle {2}={-2}$ becomes $\displaystyle {4}={4}$ .

The example below solves $\displaystyle {\sqrt{x+5}}={x-1}$ :

We square: $\displaystyle {x+5}={x^2-2x+1}$ , that is $\displaystyle {x^2-3x-4}={0}$ .
We factor: $\displaystyle {(x-4)(x+1)}={0}$ , so $\displaystyle {x}={4}$ or $\displaystyle {x}={-1}$ .
We check $\displaystyle {x}={4}$ : $\displaystyle {\sqrt{9}}={3}$ , and $\displaystyle {4-1}={3}$ too. It is a solution.
We check $\displaystyle {x}={-1}$ : $\displaystyle {\sqrt{4}}={2}$ , but $\displaystyle {-1-1}={-2}$ . It is not a solution.

The only solution: $\displaystyle {x}={4}$ .

An equation with two roots

When an equation has two roots in it, leave one root on each side and square; if a root is still left, isolate it and square again.

The example below solves $\displaystyle {\sqrt{x+5}-\sqrt{x}}={1}$ :

We move $\displaystyle \sqrt{x}$ to the other side: $\displaystyle {\sqrt{x+5}}={1+\sqrt{x}}$ .
We square: $\displaystyle {x+5}={1+2\sqrt{x}+x}$ , that is $\displaystyle {2\sqrt{x}}={4}$ , and $\displaystyle {\sqrt{x}}={2}$ .
We square again: $\displaystyle {x}={4}$ .
We check: $\displaystyle {\sqrt{9}-\sqrt{4}}={3-2}={1}$ . It is a solution.

A biquadratic equation

A biquadratic equation $\displaystyle {ax^4+bx^2+c}={0}$ has only $\displaystyle x^4$ , $\displaystyle x^2$ and a number in it. Put $\displaystyle {t}={x^2}$ , and you get a quadratic equation in $\displaystyle t$ .

x^4-5x^2+4=0

We put $\displaystyle {t}={x^2}$ : $\displaystyle {t^2-5t+4}={0}$ , so $\displaystyle {t}={1}$ or $\displaystyle {t}={4}$ .
Back to $\displaystyle x$ : $\displaystyle {x^2}={1}$ gives $\displaystyle {x}={\pm1}$ , and $\displaystyle {x^2}={4}$ gives $\displaystyle {x}={\pm2}$ .
Four solutions: $\displaystyle -2$ , $\displaystyle -1$ , 1, 2.

A negative solution for $\displaystyle t$ gives no $\displaystyle x$ , because $\displaystyle x^2$ is never negative. The same substitution works in any equation in which one expression repeats, like $\displaystyle (x^2-2x)^2$ and $\displaystyle x^2-2x$ .

The number of solutions on a graph

The solutions of $\displaystyle {f(x)}={g(x)}$ are the $\displaystyle x$ -coordinates of the meeting points of the two graphs, so the number of solutions is the number of meeting points. On the board – $\displaystyle {f(x)}={\sqrt{x}}$ , whose graph starts at $\displaystyle (0,0)$ because a negative number has no root, and the line $\displaystyle {g(x)}={x-2}$ (the dashed one):

\displaystyle f(x)=\sqrt{x}

\displaystyle g(x)=x-2

\displaystyle (4,2)

The graphs meet at one point, $\displaystyle (4,2)$ : $\displaystyle {\sqrt{x}}={x-2}$ has one solution, $\displaystyle {x}={4}$ .
Squaring gives $\displaystyle {x^2-5x+4}={0}$ , which also has the solution $\displaystyle {x}={1}$ . The graph shows why it is rejected: at $\displaystyle {x}={1}$ the root is 1, and the line is at height $\displaystyle -1$ .

Exercise 1

Solve the equations.

Write the domain of each of the equations in parts 1–3.

Exercise 2

Solve the equations, and check every solution in the original equation.

Explain without calculating why the equation in part 4 has no solution.

Exercise 3

Solve the equations with two roots.

Exercise 4

Solve by substitution.

Put $\displaystyle {t}={x^2-2x}$ .

Exercise 5

$\displaystyle {f(x)}={\sqrt{x}}$
$\displaystyle {g(x)}={\frac{x+3}{4}}$

Mark on the board the meeting points of the graphs.

\displaystyle f(x)=\sqrt{x}

\displaystyle g(x)=\frac{x+3}{4}

Solve the equation $\displaystyle {f(x)}={g(x)}$ , and check that the number of solutions equals the number of meeting points.

Formulas: algebra and calculus, grade 10 (4 units)

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Algebraic technique

The square of a sum
$\displaystyle (a+b)^2$ $\displaystyle =$ $\displaystyle a^2+2ab+b^2$
The square of a difference
$\displaystyle (a-b)^2$ $\displaystyle =$ $\displaystyle a^2-2ab+b^2$
The difference of squares
$\displaystyle (a+b)(a-b)$ $\displaystyle =$ $\displaystyle a^2-b^2$
Factoring a trinomial – $\displaystyle x_1$ , $\displaystyle x_2$ the roots of $\displaystyle {ax^2+bx+c}={0}$
$\displaystyle ax^2+bx+c$ $\displaystyle =$ $\displaystyle a(x-x_1)(x-x_2)$
The roots of $\displaystyle {ax^2+bx+c}={0}$
$\displaystyle x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
The product of roots
$\displaystyle \sqrt{a}\cdot\sqrt{b}$ $\displaystyle =$ $\displaystyle \sqrt{ab}$ $\displaystyle a\geq0,\ b\geq0$
The quotient of roots
$\displaystyle \frac{\sqrt{a}}{\sqrt{b}}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{a}{b}}$ $\displaystyle a\geq0,\ b>0$

The properties of a function

An even function – its graph symmetric about the $\displaystyle y$ -axis
$\displaystyle f(-x)$ $\displaystyle =$ $\displaystyle f(x)$
An odd function – its graph symmetric about the origin
$\displaystyle f(-x)$ $\displaystyle =$ $\displaystyle -f(x)$
The $\displaystyle x$ -coordinate of the vertex of $\displaystyle {y}={ax^2+bx+c}$
$\displaystyle x=\frac{-b}{2a}$
The absolute value
$\displaystyle |a|=\begin{cases}a & a\geq0\\-a & a<0\end{cases}$

Transforming a graph

A shift of $\displaystyle k$ units up ( $\displaystyle k<0$ – down)
$\displaystyle y=f(x)+k$
A shift of $\displaystyle p$ units to the right ( $\displaystyle p<0$ – to the left)
$\displaystyle y=f(x-p)$
A reflection in the $\displaystyle x$ -axis
$\displaystyle y=-f(x)$
A reflection in the $\displaystyle y$ -axis
$\displaystyle y=f(-x)$
A stretch ( $\displaystyle |a|>1$ ) or a shrink ( $\displaystyle |a|<1$ )
$\displaystyle y=a\cdot f(x)$

The derivative and the rules of differentiation

The average rate of change from $\displaystyle x_1$ to $\displaystyle x_2$ – the slope of the chord
$\displaystyle \frac{f(x_2)-f(x_1)}{x_2-x_1}$
The derivative at a point – the slope of the tangent
$\displaystyle f'(x_0)=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}$
The derivative of a power
$\displaystyle (x^n)'$ $\displaystyle =$ $\displaystyle nx^{n-1}$
The derivative of a constant
$\displaystyle (c)'$ $\displaystyle =$ $\displaystyle 0$
The derivative of a constant multiple
$\displaystyle (c\cdot f(x))'$ $\displaystyle =$ $\displaystyle c\cdot f'(x)$
The derivative of a sum and of a difference
$\displaystyle (f(x)\pm g(x))'$ $\displaystyle =$ $\displaystyle f'(x)\pm g'(x)$
The derivative of a product
$\displaystyle (f(x)\cdot g(x))'$ $\displaystyle =$ $\displaystyle f'(x)\cdot g(x)+f(x)\cdot g'(x)$
The derivative of a composite function
$\displaystyle (f(g(x)))'$ $\displaystyle =$ $\displaystyle f'(g(x))\cdot g'(x)$

The tangent to a graph

The slope of the tangent at the point $\displaystyle (x_1,y_1)$ on the graph
$\displaystyle m=f'(x_1)$
The equation of the tangent
$\displaystyle y-y_1=m(x-x_1)$

The derivative of a square root function

The derivative of $\displaystyle \sqrt{x}$
$\displaystyle (\sqrt{x})'$ $\displaystyle =$ $\displaystyle \frac{1}{2\sqrt{x}}$ $\displaystyle x>0$
The derivative of the root of a function
$\displaystyle (\sqrt{f(x)})'$ $\displaystyle =$ $\displaystyle \frac{f'(x)}{2\sqrt{f(x)}}$ $\displaystyle f(x)>0$

Rising, falling and extremum points

The function rises
$\displaystyle f'(x)>0$
The function falls
$\displaystyle f'(x)<0$
A point that may be an extremum
$\displaystyle f'(x)=0$