Download Section 2A – Solving Equations

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Section 2A – Solving Equations
Solving Linear Equations: These are equations that can be written as ax + b = 0.
Move all the variables to one side of the equation and all the constants to the other
side. Divide by the coefficient of x.
Example 1. Solve the equation 4x + 7 = −3x + 13 for x.
Solving Quadratic Equations
A quadratic equation is an equation that can be written as ax2 + bx + c = 0, referred
to as the standard form.
Three methods to solve quadratics:
1. Factoring
2. Completing the Square
3. Using the Quadratic Formula
1. Solving by Factoring
Theorem. If ab = 0, then a = 0 or b = 0 or both.
Strategy for Factoring:
1. Write the equation in standard form: ax2 + bx + c = 0.
2. Factor
3. Set each factor equal to zero and solve
Example 2. Solve the following equations:
(a) x2 + 5x − 3 = 9 − x2
(b) 5x2 − 3x − 30 = x2 − 3x + 6
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2. Completing the Square
2
Note. The solutions
√ to the equation x = c are x =
also write x = ± c.
√
DON’T FORGET THE − c!
√
√
c and x = − c, which we can
Example 3. Solve the following equations:
(a) 5x2 − 3x − 30 = x2 − 3x + 6
(same as 2(b) above)
2
(b) (x − 3) = 13
Method for Completing the Square:
Note: only use this when it does NOT factor!
1. Write the equation in the form: x2 + dx = e. (move the constant to the right
side, divide all terms by the coefficient of x2 )
2
2. Add d2 to both sides.
3. Now the left hand side should factor as x +
d 2
2
4. Solve the equation like Example 3(b)
Example 4. Solve the following equations by completing the square:
(a) x2 + 5x + 3 = 0
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(b) 5x2 + 7x − 2 = 0
3. Using the Quadratic Formula
Theorem. The solution to any quadratic equation of the form ax2 + bx + c = 0 is given
by
√
−b ± b2 − 4ac
x=
2a
• Your equation must be in standard form ax2 + bx + c = 0.
• You have to MEMORIZE this formula!
Example 5. Use the quadratic formula to solve 2x2 + 8x + 5 = 0.
Example 6. Solve the equation 5x2 = x − 1
Equations in Quadratic Form: An equation is in quadratic form when it can
be arranged to look like
a(
)2 + b(
)+c=0
where you have the same expression in both the blanks above. So it’s a quadratic
equation where the x has been replaced by a more complicated expression.
To solve an expression like this do the following:
1. Replace the more complicated expression with u.
2. Solve the quadratic equation au2 + bu + c = 0.
3. Set the expression you replaced with u equal to the solutions from 2 and solve.
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Example 7. Solve the following.
(a) x4 − 4x2 − 13 = 0.
2
1
(b) y 3 − y 3 − 56 = 0
Rational Equations: A rational equation is an equation with rational expressions. To solve, eliminate the fractions by multiplying both sides by the least common
denominator.
Note. Be careful! Solving these equations will sometimes produce an extraneous
solution, which does not solve the original equation and is not really a solution.
Example 8. Solve the following equations.
(a)
−6
x−5
(b) 2 +
−
2
x
5
x−4
=1
=
x+1
x−4
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Radical Equations: To solve an equation that contains one or more radicals do
the following:
1. Isolate one radical on one side of the equation.
2. Raise both sides of the equation to the appropriate power to remove the radical
(for example, square roots should be squared)
3. If necessary, repeat the process until all radicals have been removed, then solve
the equation.
4. This method also produces extraneous solutions. You MUST CHECK your
solutions to make sure they are actually solutions.
Example 9. Solve the following equations.
√
(a) 2 x − 3 + 4 = 11
(b)
√
x+5+1=x
√
√
(c) 2 x − x − 3 = 3
Absolute Value Equations: Recall, that
|x| =




x
if x ≥ 0



−x
if x < 0
Therefore, for the absolute value of an expression to equal a number, then the expression
must equal the number or must equal the negative of the number. Then to solve the
equations we rewrite every absolute value as two separate equations.
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Note. Solving absolute value equations can produce extraneous solutions. You
MUST CHECK your solutions to make sure they are actually solutions.
Example 10. Solve the equations.
(a) 2|x| = 7
(b) |x − 5| = 10
(c) |x2 − 8| = 2x
A geometrical interpretation of absolute value equations: Recall that |a − b|
refers to the distance between a and b on the number line. Therefore, the equation
|x − 5| = 10
is finding numbers that are a distance of 10 from 5.
Equations in Several Variables: When an equation has several variables, we
often solve for one variable in terms of the others, i.e., solve for one variable and get an
expression that contains the other variables.
Example 11. (a) The height of a ball in feet thrown with an initivial velocity v in
feet per second with initial height h in feet is given by s = −16t2 − vt + h. Solve
this equation for v.
(b) The formula for the surface area of a right circular cone is given by S = πr(` + r).
Solve the equation for r.