Lesson Objectives

Dear learner;

At the end of this lesson you will be able to:

• Solve simple equations involving absolute values.
• Illustrate the use of absolute value properties to solve equations involving linear terms inside the absolute value.
• Solve quadratic equations.
• Identify relationships between roots and coefficients of a quadratic equation.
• Compute equations involving exponents and radicals.

Key Terms

• Absolute value
• Factorization
• Completing the square method
• General quadratic formula
• Quadratic equations
• Exponential equation
• Logarithmic equations

1. An Ethiopian farmer wants to create two fields for different crops. The first field should be located a certain distance from a water source, while the second field should be at a distance such that the total distance from both fields to the water source is constant. If the absolute value of the difference in the distance between the two fields and the water source is 5 meters, how can the farmer calculate the possible distances from each field to the water source?

Solving Non-linear Equations

Solving Non-linear Equations: Finding the values of variables that satisfy equations where the variables are raised to powers other than one or involve non-linear functions, such as (x^2 + y^2 = 1) or (xy = 3).

3.1 Equations Involving Absolute Value

i. Solving equations involving absolute value

Properties of absolute value.

1. For any real number p, |p| = |−p|.
2. For any real number p, |p| ≥ 0.
3. For any non- negative number p, |x| = p means x = p or x = −p.

Example 1:

Solve the equation, |x – 7| = 13

Solution:

For any non- negative number 𝑝, |x| = 𝑝 means x = 𝑝 or x = −𝑝.
So, you begin by making it into two separate equations and then solve them
separately.

| x – 7| = 13 means:
x – 7 = 13 𝑜𝑟 x – 7 = −13
x = 13 + 7 𝑜𝑟 x = −13 + 7

x = 20 𝑜𝑟 x = − 6
Hence, x = 20, x = – 6 are solutions of the equation.

Remark: An absolute value equation has no solution if the absolute value
expression equals a negative number since an absolute value can never be negative.
For example, |x| = −1 has no solution (explain the reason?)

Example 2:

Solve the absolute value equation, $3|2x + 4|-4 = 8$

Solution:

3|2x + 4| – 4 = 8 (add 4 to both sides)
3|2x + 4| = 12 (divide both sides by 3)
|2x + 4| = 4
2x + 4 = 4 𝑜𝑟 2x + 4 = – 4 (subtract 4 to both sides)
2x = 0 𝑜𝑟 2x = − 8 (divide both sides by 2)
Hence, x = 0, or x = − 4 are solutions to the equation.

3.2 Quadratic Equations

Definition: An equation of the form $ax^{2} + bx + c = 0$ where $a, b, c \in \mathbb{R}$ and $a \ne 0$ is a quadratic equation. Here, 𝑎 is called the leading coefficient, 𝑏 is the coefficient of middle term and 𝑐 is the constant term.

3.2.1. Solving quadratic equations

There are three basic methods for solving quadratic equations: factorization (if possible), completing the square and the quadratic formula method.

i. Factorization

To solve a quadratic equation $ax^{2} +bx+c = 0$ by factorization method:

1. Put all terms on one side of the equal sign, leaving zero on the other side.
2. Factorize the equation. you need to find two numbers p and q such that $ p + q = b$ and $p\times q =c$
3. Set each factor equal to zero.
4. Solve each of these equations.
5. Check by inserting your answer in the original equation.

Example 3:

Solve the quadratic equation

a) $x^{2} + 8x = 0 $

b) $x^{2} – 6x – 16 = 0$

c) $x^{2} + 25 = 10x$

Solution

a) To solve the equation $x^{2} + 8x = 0$ using the factorization method, we can first factor out an x from the equation:
$x(x + 8) = 0$
Next, we set each factor equal to zero to find the solutions:
$x = 0 or x + 8 = 0$
Solving for x in the second equation:
$x + 8 = 0$
$x = -8$
Therefore, the solutions for the equation $x^2 + 8x = 0$ using the factorization method are $x = 0 and x = -8$.

b) To solve the equation $x^{2} – 6x – 16 = 0$ using the factorization method, we need to find two numbers whose product is -16 and add up to -6. These numbers are -8 and 2.
Therefore, we can rewrite the equation as:
$x^{2} – 8x + 2x – 16 = 0$
x(x – 8) + 2(x – 8) = 0
(x + 2)(x – 8) = 0
Next, we set each factor equal to zero to find the solutions:
x + 2 = 0 or x – 8 = 0
Solving for x in each equation:
x + 2 = 0
x = -2
x – 8 = 0
x = 8
Therefore, the solutions for the equation $x^{2} – 6x – 16 = 0$ using the factorization method are x = -2 and x = 8.

c) To solve the equation $x^{2} + 25 = 10x$ using the factorization method, we first rearrange the equation to set it equal to zero:
$x^{2} – 10x + 25 = 0$
Next, we need to find two numbers whose product is 25 and add up to -10. These numbers are -5 and -5.
Therefore, we can rewrite the equation as:
$x^{2} – 5x – 5x + 25 = 0$
x(x – 5) – 5(x – 5) = 0
(x – 5)(x – 5) = 0
(x – 5)^2 = 0
Next, we set the factor equal to zero to find the solution:
x – 5 = 0
Solving for x:
x = 5
Therefore, the solution for the equation $x^{2} + 25 = 10x$ using the factorization method is x = 5.

ii. Completing the square method

Let $ax^{2} +bx+c = 0$ be a quadratic equation with $a,b,c \ne 0$

Re-write this equation as:

$x^{2} +\frac{b}{a}x + \frac{c}{a} = 0$……divide both sides by a

$x^{2} + \frac{b}{a} x + \frac{b^{2}}{4a^{2}} +\frac{c}{a} = \frac{b^{2}}{4a^{2}}$ ……. take half of the middle term and add square of it to both sides

$x^{2} + \frac{b}{a} x + \frac{b^{2}}{4a^{2}} = \frac{b^{2}}{4a^{2}} – \frac{c}{a}$……. subtraction

The left side $x^{2} + \frac{b}{a} x + \frac{b^{2}}{4a^{2}}$ is a perfect square. Hence it is the aim of completing the square method

$(x + \frac{b}{2a}) (x +\frac{b}{2a}) = \frac{b^{2}}{4a^{2}} – \frac{c}{a}$….. factorization

$(x + \frac{b}{2a}) (x +\frac{b}{2a}) = \frac{b^{2} – 4ac}{4a^{2}}$

$\implies$ $(x +\frac{b}{2a})^{2} = \frac{b^{2} – 4ac}{4a^{2}}$

then the final solution is $ x +\frac{b}{2a} = \pm \sqrt{\frac{b^{2} – 4ac}{4a^{2}}}$

Example 4:

Use completing the square method to solve the following.

a) $x^{2} + 6x + 7 = 0$

b) $x^{2} + 4x + 1 = 0 $

c) $x^{2} + 4x +1 = 0$

Solution

To solve the equation $x^{2} + 6x + 7 = 0$ using the completing the square method, we follow these steps:

1. Move the constant term to the other side of the equation:
$x^{2} + 6x = -7$

2. To complete the square, take half of the coefficient of x (6/2 = 3) and square it (3² = 9). Add it on both sides
$x^{2} + 6x + 9 = -7 + 9$
$(x + 3)^{2} = 2$ ……………. factorization

3. Take the square root of both sides to solve for x:
x + 3 = ±√2
x = -3 ± √2

Therefore, the solutions for the equation $x^{2} + 6x + 7 = 0$ using the completing the square method are x = -3 + √2 and x = -3 – √2.

b) To solve the equation $x^{2} + 4x + 1 = 0$ using the completing the square method, we follow these steps:

1. Move the constant term to the other side of the equation:
$x^{2} + 4x = -1$
$x^{2} + 4x + 4 = -1 + 4$
$(x + 2)^{2} = 3$
x + 2 = ±√3
x = -2 ± √3

Therefore, the solutions for the equation $x^{2} + 4x + 1 = 0$ using the completing the square method are x = -2 + √3 and x = -2 – √3.

c) To solve $x^{2} + 4x +1 = 0$ re -write

$x^{2} + 4x = -1$ take half of 4 equals 2 and square it which is 4

$x^{2} + 4x +4 = -1 +4$

$(x +2)(x+2) = 3$ …. factorization and simplification

$(x+2)^{2} = 3$

$x+ 2 = \pm \sqrt{3}$

$x = -2 + \pm \sqrt{3}$

Hence the solution is $x = \sqrt{3} -2$ and $x = 2 – \sqrt{3}$

iii. The Quadratic Formula Method

Definition:

The formula $x = \frac{-b\pm \sqrt{b^{2} – 4ac}}{2a}$ is a mathematical formula used to find the solutions of a quadratic equation of the form $ax^{2} + bx + c = 0$. , where a, b, and c are coefficients.

From the completing the square method on the above $ax^{2} + bx + c = 0$ is transformed in to:

$(x + \frac{b}{2a}) (x +\frac{b}{2a}) = \frac{b^{2} – 4ac}{4a^{2}}$

$\implies$ $(x +\frac{b}{2a})^{2} = \frac{b^{2} – 4ac}{4a^{2}}$

then the final solution is $ x = \frac{-b \pm\sqrt{b^{2} – 4ac}}{2a}$

Example 5:

Use quadratic formula to solve the following.

a) $x^{2} – 6x + 4 = 0$

b) $3x^{2} – 5x – 1 = 0$

Solution:

To solve the equation $x^{2} – 6x + 4 = 0$ using the quadratic formula method, we first identify the coefficients a, b, and c in the general quadratic equation form $ax^{2} + bx + c = 0$:

a = 1, b = -6, c = 4

Next, we substitute these values into the quadratic formula:

$ x = \frac{-b \pm\sqrt{b^{2} – 4ac}}{2a}$

Plugging in the values:

$ x = \frac{-(-6) \pm\sqrt{(-6)^{2} – 4(1)(4)}}{2(1)}$
$x = \frac{6 \pm \sqrt{36 – 16}}{2}$
$x = \frac{6 \pm \sqrt{20}}{2}$
$x = \frac{6 \pm 2\sqrt{5}}{2}$

Now, we simplify the expression: $x = \frac{6 \pm 2\sqrt{5}}{2}$

$x = \frac{6 + 2\sqrt{5}}{2}$ or $x = \frac{6 – 2\sqrt{5}}{2}$

Finally, we simplify further to find the solutions:

$x= 3 + \sqrt{5}$ or $x= 3 – \sqrt{5}$

Therefore, the solutions for the equation $x^{2} – 6x + 4 = 0$ using the quadratic formula method are x = 3 + √5 and x = 3 – √5.

b) To solve the equation $3x^{2} – 5x – 1 = 0$ using the quadratic formula method, we first identify the coefficients a, b, and c in the general quadratic equation $ax^{2} + bx + c = 0$:

a = 3, b = -5, c = -1

Plugging in the values:

$ x = \frac{-(-5) \pm\sqrt{(-5)^{2} – 4(3)(-1)}}{2(3)}$
$x = \frac{5 \pm \sqrt{25 +12}}{6}$
$x = \frac{5 \pm \sqrt{37}}{6}$

Therefore, the solutions for the equation $3x^{2} – 5x – 1 = 0$ using the quadratic formula method are:

$x = \frac{5 + \sqrt{37}}{6}$ or $x = \frac{5 – \sqrt{37}}{6}$

Discriminant ($b^{2} – 4ac$)

• The discriminant is the value under the radical sign which is $D = b^{2} – 4 a c$. A quadratix equation with real numbers as coefficients can have the following:
• The solutions of the erquation are $ x = \frac{-b \pm\sqrt{b^{2} – 4ac}}{2a} = \frac{-b \pm\sqrt{D}}{2a}$

Example 6:

a) Check whether $3x^{2} – 4x + 2 = 0$ has distinct real roots, one real root or no real roots. If root exists, find it.

b) Check whether $x^{2} – 4x + 4 = 0$ has distinct real roots, one real root or no real roots. If root exists, find it.

c) Check whether $3x^{2} – 8x + 4 = 0$ has distinct real roots, one real root or no real roots. If root exists, find it.

Solution:

a) To determine whether the quadratic equation $3x^2 – 4x + 2 = 0$ has distinct real roots, one real root, or no real roots, we can calculate the discriminant (D) of the equation using the formula:

$D = b^{2} – 4ac$

In this case, the coefficients are a = 3, b = -4, and c = 2. Substituting these values into the formula:

$D = (-4)^{2} – 4 \times 3 \times 2$
$D = 16 – 24$
$D = -8$

Since the discriminant is negative (D< 0), the quadratic equation $3x^{2} – 4x + 2 = 0$ has no real roots

b) To determine whether the equation $x^2 – 4x + 4 = 0$ has distinct real roots, one real root, or no real roots, we can analyze the discriminant of the quadratic equation.

In this case, the coefficients are a = 1, b = – 4, and c = 4. Plugging these values into the discriminant formula:

$D = (-4)^{2} – 4\times 1 \times 4$
$D = 16 – 16$
$D = 0$

Since the discriminant is equal to 0, the equation $x^{2} – 4x + 4 = 0$ has one real solution.

Therefore, the equation $x^2 – 4x + 4 = 0$ has one real solution, which is $x = 2$.

c) To determine the nature of the roots of the quadratic equation $3𝑥^2 – 8𝑥 + 4 = 0$, we can calculate the discriminant (D).

In this case, a = 3, b = -8, and c = 4. Substituting these values into the formula:

$D = (-8)^{2} – 4 \times 3 \times 4$
$D = 64 – 48$
$D = 16$

Since the discriminant (D) is positive (D > 0), the equation has two distinct real solutions.

$ x = \frac{-b + \sqrt{b^{2} – 4ac}}{2a}$ or $ x = \frac{-b – \sqrt{b^{2} – 4ac}}{2a}$
$ x = \frac{8 + \sqrt{(-8)^{2} – 4(3)(4}}{2(3)}$ or $ x = \frac{8 – \sqrt{(-8)^{2} – 4(3)(4)}}{2(3)}$

$ x = \frac{8 + \sqrt{16}}{6)}$ or $ x = \frac{8 – \sqrt{16}}{6}$

$ x = \frac{8 + 4}{6)}$ or $ x = \frac{8 – 4}{6}$

$\implies$ $x = 2$ or $x = \frac{2}{3}$

Therefore, the quadratic equation $3𝑥^{2} – 8𝑥 + 4 = 0$ has two distinct real solutions: $x = 2$ and $x = 2/3$.

3.2.2 Relationships Between Roots and Coefficients of a Quadratic Equation

Theorem: Viete’s theorem
If the roots of $ax^{2} + bx + c = 0, a\neq 0$, are

$r_{1}= \frac{-b – \sqrt{b^2 – 4ac}}{2a} \quad \text{and} \quad r_{2} = \frac{-b + \sqrt{b^2 – 4ac}}{2a}$,
then $r_{1} + r_{2} = -\frac{b}{a} \quad \text{and} \quad r_{1} r_{2} = \frac{c}{a}$.

Example 7:

a)Let$ 𝑎𝑥^{2} + 6𝑥 + 𝑐 = 0$. If the sum of the roots is $\frac{-6}{10}$ and the product is $\frac{-3}{5}$, then find the values of 𝑎 and 𝑐.

b) If the sum of the roots of the equation $3x^2 + kx + 1 = 0$ is $7$, then what is the value of $k$?

c) If one of the roots of the equation $x^2 – 4x + k = 0$ exceeds the other by 2, then find the roots and determine the value of k.

Solution

For a quadratic equation $ax^{2} + bx + c = 0$, the sum of the roots is given by $\frac{-b}{a}$.

In this case, the sum of the roots is 7, so we have:

b) To find the value of k when the sum of the roots of the equation $3x^{2} + kx + 1 = 0$ is 7, we can use the sum of roots formula for a quadratic equation:

$\frac{-k}{3} = 7$

Solving for $k$:

$-k = 21$
$k = -21$

Therefore, the value of k is $-21$.

c) Let the roots of the equation $x^{2} – 4x + k = 0$ be α and β, where α exceeds β by 2.

From the sum and product of roots formula, we know that:
$\alpha + \beta = 4$
$\alpha \times \beta = k$

Given that α exceeds β by 2, we have:
$\alpha = \beta + 2$

Substitute α = β + 2 into the sum formula:
(β + 2) + β = 4
2β + 2 = 4
2β = 2
β = 1

Substitute β = 1 into α = β + 2:
α = 1 + 2
α = 3

Therefore, the roots of the equation $x^{2} – 4x + k = 0$ are α = 3 and β = 1.

To find the value of k, substitute the roots into the product formula:
αβ = k
$3 \times 1 = k$
$k = 3$

Therefore, the roots of the equation are 1 and 3, and the value of k is 3.

3.3 Equations Involving Exponents and Radicals

3.3.1. Equations involving exponents

Note: For $a > 0$, $a^{x} = a^{y}$ if and only if $x = y$

Example 8:

solve

a) $3^{x} = 81$

b) $2^{x} = \frac{1}{16}$

c) $2^{x+4} = 32$

Solution:

a) To solve the equation $3^{x} = 81$, we can rewrite $81$ as a power of $3$:

$81 = 3^{4}$

Now, we can rewrite the equation as:

$3^{x} = 3^{4}$

Since the bases are the same, we can equate the exponents:

$x = 4$

Therefore, the solution to the equation $3^{x} = 81$ is $x = 4$.

b) To solve the equation $2^{x} = \frac{1}{16}$, we can rewrite $\frac{1}{16}$ as $2^{-4}$ since $\frac{1}{16}$ is the same as $2^{-4}$.

Therefore, the equation becomes:

$2^{x} = 2^{-4}$

Since the bases are the same, we can equate the exponents:

$x = -4$

Thus, the solution to the equation $2^{x} = 1/16 is x = -4$.’

c) To solve the equation $( 2^{x + 4} = 32 )$: Express $32$ as a power of 2: $32 = 2^{5}$ So, the equation becomes: $2^{x + 4} = 2^{5}$. Since the bases are the same, set the exponents equal to each other: $x + 4 = 5$. Solve for $x : x + 4 = 5$ Subtract $4$ from both sides: $x = 5 – 4. x = 1$ Thus, the solution to the equation $ 2^{x + 4} = 32 $ is: $x = 1$.

Some rules of exponential equations. (𝑎, 𝑚, and 𝑛 are real number and 𝑎 ≠ 0, and 𝑏 ≠ 0)

a. $\frac{1}{a^{n}} = a^{(-n)}$

b) $a^{m}\ast a^{n} = a^{(m+n)}$

c) $(a^m)^n = a^{mn}$

d) $(ab)^n = a^n b^n$

e) $\left( \frac{a}{b} \right)^n = \frac{a^n}{b^n}$

f) $\frac{a^m}{a^n} = a^{m – n} = \frac{1}{a^{n – m}}$

Example 9:

Solve the following.

a) $( 32^x = 2^{x + 4} )$

b) $(x^2-6x)^{\frac{1}{2}} = 4$

c) $3𝑦^{\frac{3}{2}} + 3 = 27$

d) $\sqrt{3x+7} + \sqrt{x + 2} = 1$

Solution:

a) To solve the equation $( 32^x = 2^{x + 4} )$ : item Express $32$ as a power of $2$:

$32 = 2^5. So, ( 32^x )$ can be rewritten as: $32^x = (2^5)^x$

Rewrite the equation with the same base: $(2^5)^x = 2^{5x}$

Thus, the original equation becomes: $2^{5x} = 2^{x + 4}$

Since the bases are the same, set the exponents equal to each other:

$5x = x + 4$

Solve for $( x ): 5x – x = 4$

$4x = 4$ (Divide both sides by 4)

$x = 1$

Thus, the solution to the equation $(32^x = 2^{x + 4} )$ is: $x = 1$

b) To solve the equation $(x^2 – 6x)^{\frac{1}{2}} = 4$: item Square both sides to eliminate the square root:

$\left( (x^2 – 6x)^{\frac{1}{2}} \right)^2 = 4^2$

$x^2 – 6x = 16$

Solve the resulting quadratic equation: $x^2 – 6x – 16 = 0$

Factor the quadratic equation: $x^2 – 6x – 16 = (x – 8)(x + 2) = 0$

Set each factor equal to zero and solve for $x: x – 8 = 0$ or $ x + 2 = 0$

$x = 8$ or $x = -2$

Check for extraneous solutions: $( x = 8 ): (8^2 – 6(8))^{\frac{1}{2}} = (64 – 48)^{\frac{1}{2}} = 16^{\frac{1}{2}} = 4$

$\For ( x = -2 ): ((-2)^2 – 6 (-2))^{\frac{1}{2}} = (4 + 12)^{\frac{1}{2}} = 16^{\frac{1}{2}} = 4$

the solutions to the equation $( (x^2 – 6x)^{\frac{1}{2}} = 4 ) are: x = 8 \quad \text{and} \quad x = -2$

c) To solve the equation $(3y^{\frac{3}{2}} + 3 = 27)$: Subtract 3 from both sides to isolate the term with (y):

$3y^{\frac{3}{2}} + 3 – 3 = 27 – 3$

$3y^{\frac{3}{2}} = 24$

Divide both sides by 3 to solve for $y^{\frac{3}{2}}): \frac{3y^{\frac{3}{2}}}{3} = \frac{24}{3}$

$y^{\frac{3}{2}} = 8$

To solve for (y), raise both sides to the power of $(\frac{2}{3})$: the reciprocal of $(\frac{3}{2})$

$\( y^{\frac{3}{2}})^{\frac{2}{3}} = 8^{\frac{2}{3}}$

$y = 8^{\frac{2}{3}}$

$ y = 8^{\frac{2}{3}} = (\sqrt[3]{8})^{2} = (\sqrt[3]{2\ast 2\ast 2})^{2} = (2^{2}) = 4$

Thus, the solution to the equation $3y^{\frac{3}{2}} + 3 = 27$ is
$y = 4$

d) To solve the equation $\sqrt{3x+7} + \sqrt{x+2} = 1$

separate one of the square roots. Let’s separate $\sqrt{3x+7}$

$\sqrt{3x+7} = 1 – \sqrt{x+2}$

Square both sides to eliminate the square root on the left

$(\sqrt{3x+7})^2 = (1 – \sqrt{x+2})^2$

$3x+7 = 1 – 2\sqrt{x+2} + (x+2)$

$3x+7 = 1 + x + 2 – 2\sqrt{x+2}$

$3x+7 = x + 3 – 2\sqrt{x+2}$

Move the terms involving (x) to one side

$3x + 7 – x – 3 = -2\sqrt{x+2}$

$2x + 4 = -2\sqrt{x+2}$

Divide both sides by (-2): $-x – 2 = \sqrt{x+2}$

Square both sides again to eliminate the square root

$(-x – 2)^2 = (\sqrt{x+2})^{2}$

$x^{2} + 4x + 4 = x + 2$

Move all terms to one side to set the equation to zero

$x^{2} + 4x + 4 – x – 2 = 0$

$x^{2} + 3x + 2 = 0$

Factor the quadratic equation: $x^{2} + 3x + 2 = (x + 1)(x + 2) = 0$

Set each factor equal to zero and solve for x: $x + 1 = 0$ or $x + 2 = 0$

$x = -1$ or $x = -2$

Check the solutions in the original equation

For $x = -1$

$\sqrt{3(-1) + 7} + \sqrt{-1 + 2} = \sqrt{4} + \sqrt{1} = 2 + 1 = 3 \neq 1$

This is not a valid solution. $x = -2$

$\sqrt{3(-2) + 7} + \sqrt{-2 + 2} = \sqrt{-6 + 7} + \sqrt{0} = \sqrt{1} + 0 = 1$. This is a valid solution.

Thus, the solution to the equation $\sqrt{3x+7} + \sqrt{x+2} = 1$ is: $x = -2$