Mathematics

Equations

Resolution of equations: 2nd degree (complete and incomplete), biquadratic, higher degree polynomials using synthetic division, rational equations, radical equations, logarithmic, exponential (3 types) and absolute value equations.

They are of the form ax2+bx+c=0ax^2 + bx + c = 0
They are solved using the quadratic formula:
x=−b±b2−4ac2ax = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Ex.: x2−3x+2=0x^2 - 3x + 2 = 0
x=3±(−3)2−4⋅1⋅22x = \dfrac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 2}}{2}
⇒x=1  ;x=2\Rightarrow \quad x=1\;; \quad x=2
Incomplete quadratic equations:
  • If the linear term is missing, ax2+c=0ax^2 + c = 0, isolate the variable x. Ex.:
x2−4=0⇒x=±4⇒x=±2x^2 - 4 = 0 \Rightarrow x = \pm\sqrt{4} \Rightarrow x = \pm 2
  • If the constant term is missing, ax2+bx=0ax^2 + bx = 0. Factor out the common variable x and solve for each factor. Ex.:
x2−2x=0⇒x(x−2)=0⇒x=0  ;x=2x^2 - 2x = 0 \Rightarrow x(x-2) = 0 \Rightarrow x=0\;; \quad x=2
Biquadratic: Equations of the form ax4+bx2+c=0ax^4 + bx^2 + c = 0
They can be transformed into a quadratic equation using the substitution x2=tx^2 = t and solving for t with the quadratic formula.
t=−b±b2−4ac2at = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}
For each obtained value of t, we find the corresponding values of x by reversing the substitution.
Ex.: x4−3x2−4=0⇒{x2=t}⇒t2−3t−4=0x^4 - 3x^2 - 4 = 0 \Rightarrow \{x^2 = t\} \Rightarrow t^2 - 3t - 4 = 0
t=3±(−3)2−4⋅1⋅(−4)2t = \dfrac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2}
t=−1⇒x2=−1⇒x=−1∉Rt = -1 \Rightarrow x^2 = -1 \Rightarrow x = \sqrt{-1} \notin \mathbb{R}
t=4⇒x2=4⇒x=±4⇒x=±2t = 4 \Rightarrow x^2 = 4 \Rightarrow x = \pm\sqrt{4} \Rightarrow x = \pm 2
Bicubic: Equations of the form ax6+bx3+c=0ax^6 + bx^3 + c = 0
We use the substitution x3=tx^3 = t and solve for t with the quadratic formula. For each value of t, we find the corresponding value of x.
Ex.: x6−7x3−8=0⇒{x3=t}⇒t2−7t−8=0x^6 - 7x^3 - 8 = 0 \Rightarrow \{x^3 = t\} \Rightarrow t^2 - 7t - 8 = 0
t=7±(−7)2−4⋅1⋅(−8)2t = \dfrac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-8)}}{2}
t=−1⇒x3=−1⇒x=−13⇒x=−1t = -1 \Rightarrow x^3 = -1 \Rightarrow x = \sqrt[3]{-1} \Rightarrow x = -1
t=8⇒x3=8⇒x=83⇒x=2t = 8 \Rightarrow x^3 = 8 \Rightarrow x = \sqrt[3]{8} \Rightarrow x = 2
In general, for equations of the form ax2k+bxk+c=0ax^{2k} + bx^k + c = 0, we use the substitution xk=tx^k = t. Ex.:
x8+2x4+6=0⇒{x4=t}⇒t2+2t+6=0x^8 + 2x^4 + 6 = 0 \Rightarrow \{x^4 = t\} \Rightarrow t^2 + 2t + 6 = 0
Equations expressed as a product of factors, of the form:
P(x)⋅Q(x)⋅R(x)⋅…⋅T(x)=0P(x) \cdot Q(x) \cdot R(x) \cdot \ldots \cdot T(x) = 0
They are solved immediately by setting each factor to 0. (Zero Product Property). Ex.:
(x+1)⋅(x−2)⋅(x2−9)=0(x+1) \cdot (x-2) \cdot (x^2-9) = 0
Set each factor to 0:
x+1=0⇒x=−1x + 1 = 0 \Rightarrow x = -1
x−2=0⇒x=2x - 2 = 0 \Rightarrow x = 2
x2−9=0⇒x=−3  ;x=3x^2 - 9 = 0 \Rightarrow x = -3\;; \quad x = 3
Factor the polynomial and solve it as in the previous case.
x4+x3−5x2+x−6=0x^4 + x^3 - 5x^2 + x - 6 = 0
We factor the polynomial using synthetic division:
11−51−622626131301313−3−30−31010\begin{array}{r|rrrrr} & 1 & 1 & -5 & 1 & -6 \\ 2 & & 2 & 6 & 2 & 6 \\ \hline & 1 & 3 & 1 & 3 & 0 \end{array} \qquad \begin{array}{r|rrrr} & 1 & 3 & 1 & 3 \\ -3 & & -3 & 0 & -3 \\ \hline & 1 & 0 & 1 & 0 \end{array}
(x2+1)(x−2)(x+3)=0(x^2+1)(x-2)(x+3) = 0
x2+1=0⇒x∉Rx^2 + 1 = 0 \Rightarrow x \notin \mathbb{R}
x−2=0⇒x=2x - 2 = 0 \Rightarrow x = 2
x+3=0⇒x=−3x + 3 = 0 \Rightarrow x = -3
Operate on the fractions and find a common denominator. The common denominator can then be eliminated (by multiplying the entire equation by it), converting the equation into a polynomial one.
Ex.:
3−x1−x2−2+x1+x=11−x\dfrac{3-x}{1-x^2} - \dfrac{2+x}{1+x} = \dfrac{1}{1-x}
3−x1−x2−(2+x)(1−x)(1+x)(1−x)=(1+x)(1−x)(1+x)\dfrac{3-x}{1-x^2} - \dfrac{(2+x)(1-x)}{(1+x)(1-x)} = \dfrac{(1+x)}{(1-x)(1+x)}
3−x−(2−x−x2)=1+x3-x - (2-x-x^2) = 1+x
x2−x=0x^2 - x = 0
x(x−1)=0x(x-1) = 0
We obtain the possible solutions:
x=0;x=1x = 0 \quad ; \quad x = 1
The solutions must be checked, and any extraneous solutions are discarded.
In this example, the solution x=1 results in a division by 0, so it is invalid. The valid solution is:
x=0x = 0
The variable appears inside a radical (root).
Isolate the radical term and square both sides of the equation. Ex.:
7−3x−x=7\sqrt{7-3x} - x = 7
7−3x=x+7\sqrt{7-3x} = x + 7
(7−3x)2=(x+7)2\left(\sqrt{7-3x}\right)^2 = (x+7)^2
7−3x=x2+14x+497 - 3x = x^2 + 14x + 49
x2+17x+42=0x^2 + 17x + 42 = 0
The solutions must be checked in the original equation to discard any extraneous solutions. In this example, the answer is:

x=−3x = -3 is a valid solution
x=−14x = -14 is NOT a solution

In some cases, where there are multiple roots, the process of isolating a root and squaring the equation must be done multiple times. Ex.:
x+x=3x+x2\sqrt{x} + x = \sqrt{3x+x^2}
(x+x)2=(3x+x2)2\left(\sqrt{x} + x\right)^2 = \left(\sqrt{3x+x^2}\right)^2
x2+2xx+x2=3x+x2\sqrt{x}^2 + 2x\sqrt{x} + x^2 = 3x + x^2
x+2xx+x2=3x+x2x + 2x\sqrt{x} + x^2 = 3x + x^2
Isolate the remaining radical and square both sides again:
(2xx)2=(2x)2\left(2x\sqrt{x}\right)^2 = (2x)^2
4x2⋅x=4x24x^2 \cdot x = 4x^2
4x3−4x2=04x^3 - 4x^2 = 0
x2(x−1)=0x^2(x-1) = 0
Check the results.
We obtain x=0x=0 and x=1x=1. By checking them in the original equation, we determine that both are valid solutions.
Split into two equations without the absolute value: one where the expression is positive, and another where it is negative. Ex.:
2x+∣4−6x∣=82x + |4-6x| = 8
Split into 2 equations and solve each one:

2x+(4−6x)=82x + (4-6x) = 8
4x=−44x = -4
x=−1x = -1

2x−(4−6x)=82x - (4-6x) = 8
8x=128x = 12
x=32x = \dfrac{3}{2}

Use the properties of logarithms to condense the terms into a single logarithm on each side of the equation. Once there is a single logarithm on both sides, by the one-to-one property, we set the arguments equal to each other and solve the resulting algebraic equation. It is necessary to check the solutions. Ex.:
log⁡(x+3)−log⁡(x+1)=1−log⁡5\log(x+3) - \log(x+1) = 1 - \log 5
log⁡(x+3)−log⁡(x+1)=log⁡10−log⁡5\log(x+3) - \log(x+1) = \log 10 - \log 5
log⁡(x+3x+1)=log⁡(105)\log\left(\dfrac{x+3}{x+1}\right) = \log\left(\dfrac{10}{5}\right)
Apply the one-to-one property:
x+3x+1=2\dfrac{x+3}{x+1} = 2
We obtain the solution x=1x=1 and verify that it is valid.
Type 1: Both sides of the equation can be expressed as powers of the same base. Apply the one-to-one property and set the exponents equal to each other. Solve the resulting equation. Ex.:
21−x2=18⇒21−x2=2−32^{1-x^2} = \dfrac{1}{8} \Rightarrow 2^{1-x^2} = 2^{-3}
One-to-one property:
1−x2=−31-x^2 = -3
We obtain the solutions x=−2x=-2 and x=2x=2
Type 2: The exponential terms share the same base but require a substitution ax=ta^x = t.
Recalling the properties of exponents, we can perform transformations such as:
2x+2=2x⋅22  ;22x=(2x)2  ;12x=2−x=(2x)−12^{x+2} = 2^x \cdot 2^2 \;; \quad 2^{2x} = (2^x)^2 \;; \quad \dfrac{1}{2^x} = 2^{-x} = (2^x)^{-1}
Example:
9x−2⋅3x+2+81=09^x - 2 \cdot 3^{x+2} + 81 = 0
(32)x−2⋅32⋅3x+81=0(3^2)^x - 2 \cdot 3^2 \cdot 3^x + 81 = 0
(3x)2−18⋅3x+81=0(3^x)^2 - 18 \cdot 3^x + 81 = 0
We use the substitution 3x=t3^x=t, obtaining t2−18t+81=0t^2 - 18t + 81 = 0.
The solution is t=9t=9. We reverse the substitution to find the final solution:
t=9⇒3x=9⇒3x=32⇒x=2t=9 \Rightarrow 3^x=9 \Rightarrow 3^x=3^2 \Rightarrow x=2
Type 3: The exponential terms cannot be written with the same base.
Take the logarithm of both sides, and using the properties of logarithms, solve for the variable. Example:
3x/2=53^{x/2} = 5
log⁡3(3x/2)=log⁡35\log_3\left(3^{x/2}\right) = \log_3 5
x2⋅log⁡33=log⁡35\dfrac{x}{2} \cdot \log_3 3 = \log_3 5
x=2⋅log⁡35x = 2 \cdot \log_3 5