What are the x-coordinates of the solutions to this system of equations? x2 y2 = 16 y = x 4
Solving Equations Algebraically |
Contents: This page corresponds to § 2.four (p. 200) of the text.
Suggested Bug from Text:
p. 212 #7, viii, 11, xv, 17, 18, 23, 26, 35, 38, 41, 43, 46, 47, 51, 54, 57, lx, 63, 66, 71, 72, 75, 76, 81, 87, 88, 95, 97
Quadratic Equations
Equations Involving Radicals
Polynomial Equations of College Caste
Equations Involving Partial Expressions or Accented Values
Quadratic Equations
A quadratic equation is ane of the form ax2 + bx + c = 0, where a, b, and c are numbers, and a is not equal to 0.
Factoring
This approach to solving equations is based on the fact that if the product of 2 quantities is cypher, then at least i of the quantities must exist zero. In other words, if a*b = 0, then either a = 0, or b = 0, or both. For more on factoring polynomials, run into the review section P.3 (p.26) of the text.
Example i.
2x2 - 5x - 12 = 0.
(2x + iii)(x - 4) = 0.
2x + 3 = 0 or x - 4 = 0.
x = -three/two, or x = 4.
Foursquare Root Principle
If x2 = k, so x = ± sqrt(k).
Example ii.
x2 - 9 = 0.
102 = 9.
10 = 3, or x = -3.
Example 3.
Example 4.
xii + 7 = 0.
ten2 = -7.
ten = ±
.
Note that
=
, then the solutions are
x = ±
Completing the Square
The thought behind completing the square is to rewrite the equation in a form that allows us to apply the foursquare root principle.
Example 5.
x2 +6x - i = 0.
x2 +6x = i.
x2 +6x + 9 = one + ix.
The 9 added to both sides came from squaring one-half the coefficient of 10, (six/2)2 = 9. The reason for choosing this value is that now the left hand side of the equation is the square of a binomial (two term polynomial). That is why this procedure is called completing the square. [ The interested reader can see that this is true by considering (x + a)2 = 102 + 2ax + atwo. To go "a" ane need only carve up the 10-coefficient by two. Thus, to consummate the square for x2 + 2ax, one has to add a2.]
(ten + 3)ii = ten.
Now we may employ the foursquare root principle and so solve for x.
10 = -three ± sqrt(10).
Case 6.
2x2 + 6x - 5 = 0.
2x2 + 6x = 5.
The method of completing the square demonstrated in the previous instance only works if the leading coefficient (coefficient of xii) is one. In this example the leading coefficient is ii, but we tin can modify that by dividing both sides of the equation past 2.
102 + 3x = 5/ii.
Now that the leading coefficient is one, nosotros accept the coefficient of x, which is now iii, divide it past 2 and square, (iii/two)ii = nine/4. This is the constant that nosotros add together to both sides to complete the square.
x2 + 3x + 9/4 = 5/2 + ix/iv.
The left hand side is the foursquare of (ten + 3/2). [ Verify this!]
(x + 3/2)2 = xix/4.
Now nosotros use the square root principle and solve for x.
ten + 3/2 = ± sqrt(nineteen/iv) = ± sqrt(19)/2.
x = -three/2 ± sqrt(19)/2 = (-3 ± sqrt(19))/2
So far nosotros have discussed three techniques for solving quadratic equations. Which is best? That depends on the trouble and your personal preference. An equation that is in the correct course to apply the foursquare root principle may be rearranged and solved by factoring equally nosotros run across in the next example.
Example 7.
10two = 16.
xtwo - sixteen = 0.
(x + 4)(10 - 4) = 0.
x = -4, or x = four.
In some cases the equation can exist solved past factoring, only the factorization is not obvious.
The method of completing the square will e'er work, even if the solutions are complex numbers, in which case we will accept the foursquare root of a negative number. Furthermore, the steps necessary to consummate the foursquare are ever the same, so they can be applied to the general quadratic equation
ax2 + bx + c = 0.
The event of completing the square on this general equation is a formula for the solutions of the equation called the Quadratic Formula.
Quadratic Formula
The solutions for the equation ax2 + bx + c = 0 are
We are saying that completing the foursquare always works, and we have completed the square in the full general example, where we have a,b, and c instead of numbers. So, to find the solutions for any quadratic equation, nosotros write it in the standard form to notice the values of a, b, and c, and so substitute these values into the Quadratic Formula.
Ane consequence is that you never accept to complete the square to discover the solutions for a quadratic equation. Even so, the process of completing the foursquare is of import for other reasons, and so you still demand to know how to do it!
Examples using the Quadratic Formula:
Example viii.
2x2 + 6x - 5 = 0.
In this case, a = 2, b = 6, c = -five. Substituting these values in the Quadratic Formula yields
Observe that nosotros solved this equation earlier by completing the square.
Notation: In that location are two real solutions. In terms of graphs, at that place are ii intercepts for the graph of the office f(x) = 2x2 + 6x - five.
Example 9.
4x2 + 4x + ane = 0
In this example a = iv, b = four, and c = i.
At that place are ii things to notice virtually this instance.
- In that location is but one solution. In terms of graphs, this means there is only ane ten-intercept.
- The solution simplified so that there is no square root involved. This means that the equation could take been solved by factoring. (All quadratic equations tin be solved by factoring! What I mean is information technology could have been solved easily past factoring.)
4x2 + 4x + one = 0.
(2x + 1)2 = 0.
x = -1/two.
Example ten.
tenii + x + 1 = 0
a = 1, b = one, c = ane
Annotation: There are no existent solutions. In terms of graphs, there are no intercepts for the graph of the function f(x) = xii + 10 + i. Thus, the solutions are complex because the graph of y = 102 + x + 1 has no x-intercepts.
The expression under the radical in the Quadratic Formula, bii - 4ac, is called the discriminant of the equation. The terminal 3 examples illustrate the three possibilities for quadratic equations.
one. Discriminant > 0. Two real solutions.
2. Discriminant = 0. 1 real solution.
3. Discriminant < 0. 2 complex solutions.
Notes on checking solutions
None of the techniques introduced then far in this section can introduce inapplicable solutions. (See example 3 from the Linear Equations and Modeling department.) Withal, it is still a adept thought to bank check your solutions, because it is very easy to make careless errors while solving equations.
The algebraic method, which consists of substituting the number dorsum into the equation and checking that the resulting statement is true, works well when the solution is "simple", just it is not very practical when the solution involves a radical.
For instance, in our adjacent to last example, 4x2 + 4x + one = 0, nosotros plant i solution x = -ane/2.
The algebraic check looks like
4(-1/2)2 +4(-1/2) + 1 = 0.
four(one/4) - 2 + 1 = 0.
1 - 2 + 1 = 0.
0 = 0. The solution checks.
In the example before that, 2x2 + 6x - v = 0, we found two real solutions, ten = (-three ± sqrt(19))/2. It is certainly possible to check this algebraically, but it is not very easy. In this instance either a graphical check, or using a calculator for the algebraic check are faster.
Showtime, find decimal approximations for the two proposed solutions.
(-3 + sqrt(nineteen))/2 = 0.679449.
(-3 - sqrt(19))/2 = -3.679449.
Now utilise a graphing utility to graph y = 2x2 + 6x - 5, and trace the graph to observe approximately where the x-intercepts are. If they are close to the values above, then yous can be pretty sure you have the correct solutions. You can also insert the approximate solution into the equation to run into if both sides of the equation give approximately the same values. However, you still demand to be conscientious in your merits that your solution is correct, since it is non the exact solution.
Note that if you had started with the equation 2xii + 6x - five = 0 and gone directly to the graphing utility to solve it, then y'all would not get the verbal solutions, because they are irrational. Nonetheless, having found (algebraically) two numbers that you remember are solutions, if the graphing utility shows that intercepts are very close to the numbers you found, then you are probably right!
Do ane:
Solve the following quadratic equations.
(a) 3xtwo -5x - 2 = 0. Reply
(b) (10 + ane)2 = three. Answer
(c) xii = 3x + two. Answer
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Equations Involving Radicals
Equations with radicals can often be simplified by raising to the appropriate power, squaring if the radical is a square root, cubing for a cube root, etc. This operation can introduce extraneous roots, so all solutions must be checked.
If there is merely one radical in the equation, then before raising to a power, you should accommodate to take the radical term by itself on 1 side of the equation.
Instance 11.
At present that we have isolated the radical term on the right side, we square both sides and solve the resulting equation for x.
Check:
x = 0
When nosotros substitute x = 0 into the original equation we go the statement 0 = 2, which is not true!
And so, ten = 0 is not a solution.
10 = 3
When nosotros substitute x = iii into the original equation, we get the statement 3 = three. This is true, so x = 3 is a solution.
Solution: x = three.
Notation: The solution is the x-coordinate of the intersection point of the graphs of y = x and y = sqrt(x+one)+1.
Look at what would have happened if we had squared both sides of the equation earlier isolating the radical term.
This is worse than what we started with!
If there is more than one radical term in the equation, then in general, nosotros cannot eliminate all radicals by raising to a power one fourth dimension. Nonetheless, we tin subtract the number of radical terms by raising to a ability.
If the equation involves more than i radical term, and so nosotros still want to isolate 1 radical on one side and raise to a power. And then we echo that process.
Example 12.
Now foursquare both sides of the equation.
This equation has only 1 radical term, and so we take made progress! Now isolate the radical term so square both sides again.
Check:
Substituting x = 5/four into the original equation yields
sqrt(ix/4) + sqrt(1/4) = ii.
3/2 + 1/2 = 2.
This statement is true, so x = 5/4 is a solution.
Note on checking solutions:
The algebraic cheque was easy to do in this case. Nonetheless, the graphical bank check has the advantage of showing that at that place are no solutions that we have not found, at least within the scope of the viewing rectangle. The solution is the ten-coordinate of the intersection bespeak of the graphs of y = 2 and y = sqrt(x+i)+sqrt(10-1).
Exercise two:
Solve the equation sqrt(x+2) + ii = 2x. Answer
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Polynomial Equations of Higher Caste
We accept seen that whatsoever degree two polynomial equation (quadratic equation) in i variable can exist solved with the Quadratic Formula. Polynomial equations of degree greater than two are more complicated. When we encounter such a problem, and so either the polynomial is of a special form which allows us to factor it, or we must approximate the solutions with a graphing utility.
Zero Constant
1 common special instance is where there is no constant term. In this case nosotros may cistron out one or more than powers of 10 to begin the problem.
Example 13.
2x3 + 3x2 -5x = 0.
x (2x2 + 3x -5) = 0.
At present nosotros have a product of x and a quadratic polynomial equal to 0, then we have two simpler equations.
x = 0, or 2xtwo + 3x -5 = 0.
The first equation is little to solve. ten = 0 is the only solution. The second equation may be solved by factoring. Notation: If nosotros were unable to factor the quadratic in the second equation, and then we could have resorted to using the Quadratic Formula. [Verify that you get the same results as beneath.]
10 = 0, or (2x + 5)(x - one) = 0.
So there are three solutions: ten = 0, x = -5/two, x = 1.
Annotation: The solution is found from the intercepts of the graphs of f(x) = 2xthree + 3x2 -5x.
Factor by Group
Case 14.
xthree -2x2 -9x +18 = 0.
The coefficient of tenii is -2 times that of x3, and the same relationship exists between the coefficients of the tertiary and fourth terms. Group terms i and two, and besides terms three and four.
ten2 (x - 2) - 9 (x - ii) = 0.
These groups share the mutual cistron (x - ii), so nosotros can cistron the left manus side of the equation.
(10 - ii)(x2 - 9) = 0.
Whenever we find a production equal to zero, we obtain 2 simpler equations.
x - two = 0, or tentwo - ix = 0.
x = 2, or (10 + 3)(10 - iii) = 0.
And so, there are iii solutions, ten = ii, x = -3, x = 3.
Note: These solutions are found from the intercepts of the graph of f(10) = tenthree -2x2 -9x +eighteen.
Quadratic in Form
Example xv.
x4 - x2 - 12 = 0.
This polynomial is non quadratic, it has degree four. Nevertheless, it tin be idea of as quadratic in xtwo.
(x2) two -(x2) - 12 = 0.
It might help yous to actually substitute z for ten2.
zii - z - 12 = 0 This is a quadratic equation in z.
(z - 4)(z + 3) = 0.
z = iv or z = -3.
Nosotros are non done, because we need to observe values of x that brand the original equation true. Now supplant z by xii and solve the resulting equations.
xii = 4.
x = 2, 10 = -2.
x2 = -three.
x = i
, or 10 = -i
Then there are four solutions, two real and 2 circuitous.
Note: These solutions are establish from the intercepts of the graph of f(x) = x4 - x2 - 12.
A graph of f(x) = 10four - x2 - 12 and a zoom showing its local extrema.
Exercise iii:
Solve the equation xfour - 5x2 + 4 = 0. Answer
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Equations Involving Fractional Expressions or Absolute Values
Example 16.
The least mutual denominator is x(x + 2), so we multiply both sides by this product.
This equation is quadratic. The Quadratic Formula yields the solutions
Checking is necessary considering we multiplied both sides past a variable expression. Using a graphing utility we see that both of these solutions check. The solution is the 10-coordinate of the intersection point of the graphs of y = i and y = 2/ten-one/(10+2).
Example 17.
5 | x - 1 | = x + 11.
The fundamental to solving an equation with accented values is to remember that the quantity inside the absolute value bars could exist positive or negative. We will accept two separate equations representing the different possibilities, and all solutions must be checked.
Case 1. Suppose x - 1 >= 0. So | x - ane | = x - ane, and so nosotros have the equation
5(x - 1) = 10 + 11.
5x - 5 = x + 11.
4x = 16.
x = 4, and this solution checks because 5*3 = 4 + 11.
Example 2. Suppose 10 - ane < 0. So x - 1 is negative, so | x - 1 | = -(x - one). This signal often confuses students, because information technology looks as if we are maxim that the absolute value of an expression is negative, simply we are not. The expression (x - 1) is already negative, and then -(x - one) is positive.
Now our equation becomes
-five(x - 1) = x + 11.
-5x + 5 = x + 11.
-6x = 6.
x = -ane, and this solution checks considering v*two = -1 + 11.
If you use the Java Grapher to check graphically, notation that abs() is absolute value, and then you would graph
5*abs(10 - one) - x - eleven and wait at x-intercepts, or y'all can find the solution as the ten-coordinates of the intersection points of the graphs of y = x+11 and y = 5*abs(x-1).
Exercise 4:
(a) Solve the equation
Reply
(b) Solve the equation | x - 2 | = 2 - x/3 Respond
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