Solve Equations of the Quadratic Form
Learn how to solve equations that can be transformed into quadratic equations using substitution methods. This page provides Grade 12 math problems with step-by-step solutions and graphical interpretations to help students master quadratic equation solving techniques.
Question 1
Solve the equation:
\[
0.1x^4 - 1.3x^2 + 3.6 = 0
\]
Solution
Let \( u = x^2 \) which gives \( u^2 = x^4 \), and rewrite the given equation in terms of \( u \):
\[
0.1u^2 - 1.3u + 3.6 = 0
\]
Solve the above quadratic equation to find \( u \).
\[
u = 4 \quad \text{and} \quad u = 9
\]
We now use the substitution \( u = x^2 \) to solve for \( x \).
\[
x = \pm 2, \quad x = \pm 3
\]
The four x-intercepts of the graph of \( y = 0.1x^4 - 1.3x^2 + 3.6 \) are the graphical solutions to the equation as shown below.
Question 2
Solve the equation:
\[
\sqrt{x} = 3 - \dfrac{1}{4}x
\]
Solution
Let \( u = \sqrt{x} \) which gives \( u^2 = x \), and rewrite the equation in terms of \( u \):
\[
u = 3 - \dfrac{u^2}{4}
\]
Multiply all terms by 4, simplify, write the quadratic in standard form and solve:
\[
u^2 + 4u - 12 = 0
\]
\[
u = -6 \quad \text{and} \quad u = 2
\]
Use the substitution \( u = \sqrt{x} \) to solve for \( x \).
\[
u = -6 \Rightarrow \text{no solution}, \quad u = 2 \Rightarrow x = 4
\]
The x-intercept of the graph is the graphical solution to the equation as shown below.
Question 3
Solve the equation:
\[
\left(3 - \dfrac{4}{x}\right)^2 - 6\left(3 - \dfrac{4}{x}\right) = 16
\]
Solution
Let \( y = 3 - \dfrac{4}{x} \) which gives \( y^2 = \left(3 - \dfrac{4}{x}\right)^2 \), and rewrite the equation in terms of \( y \):
\[
y^2 - 6y = 16
\]
Simplify and solve:
\[
y^2 - 6y - 16 = 0
\]
\[
y = -2 \quad \text{and} \quad y = 8
\]
Solve for \( x \).
\[
y = - 2, \text{gives the equation in x: } \quad 3 - \dfrac{4}{x} = -2 \Rightarrow x = \dfrac{4}{5}
\]
\[
y = 8, \text{gives the equation in x: } \quad 3 - \dfrac{4}{x} = 8 \Rightarrow x = -\dfrac{4}{5}
\]
The x-intercepts of the graph are the graphical solutions to the equation as shown below.
Question 4
Solve the equation:
\[
2(x - 1)^{2/3} + 3(x - 1)^{1/3} - 2 = 0
\]
Solution
Let \( y = (x - 1)^{1/3} \) which gives \( y^2 = (x - 1)^{2/3} \), and rewrite the equation in terms of \( y \):
\[
2y^2 + 3y - 2 = 0
\]
\[
y = -2 \quad \text{and} \quad y = \dfrac{1}{2}
\]
Solve for \( x \).
\[
(x - 1)^{1/3} = -2 \Rightarrow x = -7, \quad (x - 1)^{1/3} = \dfrac{1}{2} \Rightarrow x = \dfrac{9}{8}
\]
The x-intercepts of the graph are the graphical solutions to the equation as shown below.
Question 5
Find all real solutions for the equation:
\[
2\left(\dfrac{2}{x - 3}\right)^2 - \dfrac{2}{x - 3} - 3 = 0
\]
Solution
Let \( u = \dfrac{2}{x - 3} \) which gives \( u^2 = \left(\dfrac{2}{x - 3}\right)^2 \), and rewrite the equation in terms of \( u \):
\[
2u^2 - u - 3 = 0
\]
\[
u = -1 \quad \text{and} \quad u = \dfrac{3}{2}
\]
Solve for \( x \).
\[
\dfrac{2}{x - 3} = -1 \Rightarrow x = 1, \quad \dfrac{2}{x - 3} = \dfrac{3}{2} \Rightarrow x = \dfrac{13}{3}
\]
The x-intercepts of the graph are the graphical solutions to the equation as shown below.
More References and links