3.01: Prerequisites to Numerical Methods for Solving Nonlinear Equations
- Page ID
- 126399
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Lesson 1: Review of Quadratic Equations
Learning Objectives
After successful completion of this lesson, you should be able to:
1) find the solutions of quadratic equations,
2) derive the formula for the solution of quadratic equations,
3) solve simple physical problems involving quadratic equations.
What are quadratic equations, and how do we solve them?
A quadratic equation has the form
\[ax^{2} + bx + c = 0,\ \text{where } a \neq 0 \nonumber\]
The solution to the above quadratic equation is given by
\[x = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a} \nonumber\]
So the equation has two roots, and depending on the value of the discriminant, \(b^{2} - 4ac\), the equation may have real, complex, or repeated roots.
\[\text{If } b^{2} - 4ac < 0,\ \text{the roots are complex.} \nonumber\]
\[\text{If } b^{2} - 4ac > 0,\ \text{the roots are real.} \nonumber\]
\[\text{If } b^{2} - 4ac = 0,\ \text{the roots are real and repeated.} \nonumber\]
Derive the solution to \(ax^{2} + bx + c = 0\).
Solution
\[ax^{2} + bx + c = 0\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E1.1}) \nonumber\]
Dividing both sides by \(a\), \(\left( a \neq 0 \right)\), we get
\[x^{2} + \frac{b}{a}x + \frac{c}{a} = 0 \nonumber\]
Note if \(a = 0\), the solution to
\[ax^{2} + bx + c = 0 \nonumber\]
is
\[x = - \frac{c}{b} \nonumber\]
Rewrite Equation \((\PageIndex{1.E1.1})\) for \(a\neq0\)
\[x^{2} + \frac{b}{a}x + \frac{c}{a} = 0 \nonumber\]
as
\[\left( x + \frac{b}{2a} \right)^{2} - \frac{b^{2}}{4a^{2}} + \frac{c}{a} = 0 \nonumber\]
\[\begin{split} \left( x + \frac{b}{2a} \right)^{2}\ &= \frac{b^{2}}{4a^{2}} - \frac{c}{a}\\ &= \frac{b^{2} - 4ac}{4a^{2}} \end{split} \nonumber\]
\[\begin{split} x + \frac{b}{2a} &= \pm \sqrt{\frac{b^{2} - 4ac}{4a^{2}}}\\ &= \pm \frac{\sqrt{b^{2} - 4ac}}{2a}\end{split} \nonumber\]
\[\begin{split} x &= - \frac{b}{2a} \pm \frac{\sqrt{b^{2} - 4ac}}{2a}\\ &= \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a} \end{split}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E1.2}) \nonumber\]
A ball is thrown down at 50 mph from the top of a building. The building is 420 feet tall. Derive the equation that would let you find the time the ball takes to reach the ground.
Solution
The distance \(s\) covered by the ball is given by
\[s = ut + \frac{1}{2}gt^{2}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{1.E2.1}) \nonumber\]
where
\[u = \text{initial velocity (ft/s)} \nonumber\]
\[g = \text{acceleration due to gravity (ft}/\text{s}^2) \nonumber\]
\[t = \text{time (s)} \nonumber\]
Given
\[\begin{split} u &= 50\ \frac{\text{miles}}{\text{hour}} \times \frac{1\text{ hour}}{3600\text{ s}} \times \frac{5280\text{ ft}}{1\text{ mile}}\\ &= 73.33\ \frac{\text{ft}}{\text{s}}\end{split} \nonumber\]
\[g = \text{32.2} \ \frac{\text{ft}}{\text{s}^{2}} \nonumber\]
\[s = 420\ \text{ft} \nonumber\]
we have from Equation \((\PageIndex{1.E2.1})\)
\[420 = 73.33t + \frac{1}{2}\left( 32.2 \right)\ t^{2} \nonumber\]
\[16.1t^{2} + 73.33t - 420 = 0 \nonumber\]
The above equation is a quadratic equation, the solution of which would give the time it would take the ball to reach the ground. The solution of the quadratic equation is
\[\begin{split} t\ &= \frac{- 73.33 \pm \sqrt{73.33^{2} - 4 \times 16.1 \times ( - 420)}}{2(16.1)} \\ &= 3.315, - 7.870\end{split} \nonumber\]
Since \(t > 0\), the valid value of time \(t\) is \(3.315\ \text{s.}\)
Multiple Choice Test
(1). The value of \(x\) that satisfies \(f\left( x \right) = 0\) is called most suitably the
(A) root of an equation \(f\left( x \right) = 0\)
(B) root of a function \(f\left( x \right)\)
(C) zero of an equation \(f\left( x \right) = 0\)
(D) none of the above
(2). A quadratic equation has _______ root(s).
(A) one
(B) two
(C) three
(D) four
(3). For a certain cubic equation, at least one of the roots is known to be a complex root. How many complex roots does the cubic equation have?
(A) one
(B) two
(C) three
(D) cannot be determined
(4). An equation such as \(\tan x = x\) has _______ root(s).
(A) zero
(B) one
(C) two
(D) infinite
(5). A polynomial of order \(n\) has __________ zeros.
(A) \(n - 1\)
(B) \(n\)
(C) \(n + 1\)
(D) \(n + 2\)
(6). The velocity of a body is given by \(v(t) = 5e^{- t} + 4\), where \(t\) is in seconds and \(v\) is in m/s. The velocity of the body is \(6\) m/s at \(t =\) ____________ seconds.
(A) \(0.1823\)
(B) \(0.3979\)
(C) \(0.9163\)
(D) \(1.609\)
For complete solution, go to
http://nm.mathforcollege.com/mcquizzes/03nle/quiz_03nle_background_solution.pdf