1.5 Equations
1.5.1 Learning Objectives
By the end of this section, you should be able to:
- solve linear equations and rearrange formulas;
- solve quadratic equations and use the discriminant;
- solve rational and radical equations;
- solve equations of quadratic type and absolute-value equations; and
- identify and reject extraneous solutions.
1.5.2 Linear Equations and Rearranging Formulas
An equation states that two expressions are equal. A solution is a value that makes the equation true. Equivalent equations have the same solutions.
Adding the same expression to both sides, or multiplying both sides by the same nonzero expression, produces an equivalent equation. These operations are used to isolate a variable.
A linear equation is equivalent to
\[ ax+b=0,\qquad a\ne0. \]
Example 1.22 Linear Equation
\[ 5(2x-1)-3=4x+10 \]
\[ 10x-8=4x+10 \quad\Longrightarrow\quad 6x=18 \quad\Longrightarrow\quad \boxed{x=3}. \]
Exercise 1.28 Solve \(7x-9=3x+15\).
Show answer
Answer: \(6\)
The same principles can be used to solve a formula for one variable while treating the remaining variables as constants.
Example 1.23 Rearranging a Formula
Solve \(A=2lw+2lh+2wh\) for \(w\).
\[ A-2lh=2lw+2wh=2(l+h)w, \]
so
\[ \boxed{w=\frac{A-2lh}{2(l+h)}}. \]
1.5.3 Quadratic Equations
A quadratic equation has the form
\[ ax^2+bx+c=0,\qquad a\ne0. \]
Example 1.24 Quadratic by Factoring
\[ x^2-7x+12=0 \]
\[ (x-3)(x-4)=0, \]
so \(\boxed{x=3\text{ or }x=4}\).
Exercise 1.29 Solve \(x^2-2x-15=0\).
Show answer
Answer: \(-3,5\)
Exercise 1.30 Solve \(x^2=13\).
Show answer
Answer: \(\pm\sqrt{13}\)
Possible solution methods include factoring, taking square roots, completing the square, and using the quadratic formula:
\[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \]
The discriminant
\[ D=b^2-4ac \]
describes the number of real solutions:
- \(D>0\): two distinct real solutions;
- \(D=0\): one real solution;
- \(D<0\): no real solutions.
Example 1.25 Quadratic Formula
Solve \(2x^2+x-4=0\).
\[ x=\frac{-1\pm\sqrt{1-4(2)(-4)}}4 =\boxed{\frac{-1\pm\sqrt{33}}4}. \]
Exercise 1.31 Determine the number of real roots of \(3x^2+2x+5=0\).
Show answer
Answer: none
1.5.4 Rational and Radical Equations
For rational equations, multiply by the LCD but retain the original domain restrictions. For radical equations, isolate the radical before raising both sides to a power. These processes may create extraneous solutions, so proposed answers must be checked in the original equation.
Example 1.26 Radical Equation
Solve \(\sqrt{x+6}=x\).
Because the square root is nonnegative, \(x\ge0\). Squaring gives
\[ x+6=x^2 \quad\Longrightarrow\quad (x-3)(x+2)=0. \]
The candidates are \(3\) and \(-2\). Only \(3\) satisfies the original equation, so \(\boxed{x=3}\).
Exercise 1.32 Check whether \(x=4\) solves \(\sqrt{x+5}=x-1\).
Show answer
Answer: yes
1.5.5 Quadratic-Type and Absolute-Value Equations
An equation is of quadratic type when substitution changes it into a quadratic equation. For example, \(x^4-5x^2+4=0\) becomes \(u^2-5u+4=0\) when \(u=x^2\).
An absolute-value equation
\[ |A|=k \]
with \(k\ge0\) is equivalent to \(A=k\) or \(A=-k\).
Example 1.27 Absolute Value
\[ |3x-2|=10 \]
gives
\[ 3x-2=10\quad\text{or}\quad3x-2=-10. \]
Thus
\[ \boxed{x=4\text{ or }x=-\frac83}. \]
Exercise 1.33 Solve \(|2x+1|=7\).
Show answer
Answer: \(3,-4\)
1.5.6 Conceptual Takeaways
- Solving means producing equivalent equations until the variable is isolated.
- The zero-product property applies only when a product equals zero.
- Quadratic equations can be solved by several methods.
- The discriminant predicts the number of real roots.
- Clearing denominators and raising both sides to a power can create extraneous solutions.
- Final answers should be checked in the original equation.
1.5.7 Skills You Should Be Able to Do
- Solve linear equations and rearrange formulas.
- Solve quadratics by factoring, square roots, completing the square, and the quadratic formula.
- Use the discriminant.
- Solve rational and radical equations with appropriate checks.
- Use substitution for equations of quadratic type.
- Solve absolute-value equations by cases.
1.5.8 Practice Problems with Solutions
Solve \(4(2x-3)+5=3x+18\).
Show Solution
\[ 8x-12+5=3x+18 \Longrightarrow 5x=25 \Longrightarrow \boxed{x=5}. \]
Solve \(F=\frac{kmM}{r^2}\) for \(M\).
Show Solution
\[ Fr^2=kmM \Longrightarrow \boxed{M=\frac{Fr^2}{km}}. \]
Solve \(x^2+x-20=0\).
Show Solution
\[ x^2+x-20=(x+5)(x-4)=0, \] so \(\boxed{x=-5,4}\).
Solve \(3x^2-6x-1=0\) using the quadratic formula.
Show Solution
\[ x=\frac{6\pm\sqrt{(-6)^2-4(3)(-1)}}6 =\frac{6\pm\sqrt{48}}6 =\boxed{1\pm\frac{2\sqrt3}{3}}. \]
Determine the number of real solutions of \(5x^2+2x+3=0\).
Show Solution
\[ D=2^2-4(5)(3)=-56<0. \] There are \(\boxed{\text{no real solutions}}\).
Solve \(\frac2x+\frac1{x-1}=3\).
Show Solution
The restrictions are \(x\ne0,1\). Multiply by \(x(x-1)\): \[ 2(x-1)+x=3x(x-1). \] \[ 3x^2-6x+2=0, \] so \[ \boxed{x=1\pm\frac{\sqrt3}{3}}. \] Both values satisfy the restrictions.
Solve \(\sqrt{2x+3}=x\).
Show Solution
Since the square root is nonnegative, \(x\ge0\). Squaring gives \[ 2x+3=x^2 \Longrightarrow (x-3)(x+1)=0. \] Only \(x=3\) checks, so \(\boxed{x=3}\).
Solve \(x^4-13x^2+36=0\).
Show Solution
Let \(u=x^2\): \[ u^2-13u+36=(u-4)(u-9)=0. \] Thus \(x^2=4\) or \(x^2=9\), so \[ \boxed{x=-3,-2,2,3}. \]
Solve \(|4x+3|=11\).
Show Solution
\[ 4x+3=11\quad\text{or}\quad4x+3=-11, \] giving \[ \boxed{x=2\text{ or }x=-\frac72}. \]