If 2+sqrt 3 is a Polynomial Root

If 2+sqrt 3 is a Polynomial Root.

Solving Equations Algebraically


Contents:
This page corresponds to
§
2.4 (p. 200)
of the text.

Suggested Issues from Text:

p. 212 #7, viii, xi, 15, 17, 18, 23, 26, 35, 38, 41, 43, 46, 47, 51, 54, 57, 60, 63, 66, 71, 72, 75, 76, 81, 87, 88, 95, 97

Quadratic Equations

Equations Involving Radicals

Polynomial Equations of Higher Caste

Equations Involving Fractional Expressions or Absolute Values


Quadratic Equations

A quadratic equation is one of the form axtwo
+ 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 two quantities is zero, then at to the lowest degree one of the quantities must be nix. In other words, if a*b = 0, so either a = 0, or b = 0, or both. For more on factoring polynomials, see the review department P.three (p.26) of the text.

Instance ane.

2xtwo
– 5x – 12 = 0.

(2x + 3)(x – four) = 0.

2x + 3 = 0 or 10 – 4 = 0.

10 = -3/two, or 10 = 4.

Square Root Principle

If x2
= k, then x = ± sqrt(chiliad).

Example two.

tentwo
– nine = 0.

xtwo
= ix.

ten = 3, or 10 = -3.


Example 3.


Instance 4.

x2
+ 7 = 0.

xii
= -7.

x = ±
.

Note that

=

=
, and then the solutions are

x = ±
, ii complex numbers.

Completing the Square

The idea behind completing the square is to rewrite the equation in a form that allows u.s.a. to employ the foursquare root principle.

Example 5.

x2
+6x – i = 0.

x2
+6x = ane.

x2
+6x + 9 = 1 + 9.

The nine added to both sides came from squaring half the coefficient of 10, (6/two)two
= 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
= x2
+ 2ax + aii. To get “a” one need only divide the x-coefficient by 2. Thus, to complete the square for ten2
+ 2ax, one has to add aii.]

(ten + three)two
= 10.

At present we may utilize the foursquare root principle and so solve for x.

10 = -3 ± sqrt(10).


Example 6.

2x2
+ 6x – 5 = 0.

2x2
+ 6x = five.

The method of completing the square demonstrated in the previous case simply works if the leading coefficient (coefficient of x2) is 1. In this example the leading coefficient is ii, merely nosotros tin change that by dividing both sides of the equation by ii.

xii
+ 3x = 5/2.

Now that the leading coefficient is 1, nosotros take the coefficient of x, which is now three, divide it past 2 and foursquare, (3/2)two
= 9/4. This is the constant that we add together to both sides to consummate the foursquare.

x2
+ 3x + 9/iv = 5/2 + nine/iv.

The left hand side is the foursquare of (x + iii/2). [ Verify this!]

(x + iii/2)2
= 19/four.

Now nosotros use the square root principle and solve for x.

x + iii/2 = ± sqrt(xix/4) = ± sqrt(19)/2.

x = -three/2 ± sqrt(19)/two = (-iii ± sqrt(19))/ii

So far nosotros accept discussed three techniques for solving quadratic equations. Which is best? That depends on the problem and your personal preference. An equation that is in the right class to use the foursquare root principle may be rearranged and solved by factoring every bit we run into in the side by side example.

Instance 7.

102
= 16.

tentwo
– 16 = 0.

(10 + 4)(x – 4) = 0.

10 = -iv, or x = 4.

In some cases the equation can be solved by factoring, merely the factorization is not obvious.

The method of completing the square will always piece of work, even if the solutions are complex numbers, in which instance nosotros will take the square root of a negative number. Furthermore, the steps necessary to consummate the square are always the same, so they can exist practical to the full general quadratic equation

axii
+ bx + c = 0.

The effect of completing the square on this full general equation is a formula for the solutions of the equation chosen the Quadratic Formula.

Quadratic Formula

The solutions for the equation ax2
+ bx + c = 0 are

We are maxim that completing the square e’er works, and we have completed the square in the full general case, where nosotros take a,b, and c instead of numbers. And then, to notice the solutions for any quadratic equation, nosotros write information technology in the standard course to detect the values of a, b, and c, then substitute these values into the Quadratic Formula.

One consequence is that y’all never accept to complete the square to observe the solutions for a quadratic equation. However, the process of completing the square is important for other reasons, so y’all still demand to know how to do it!


Examples using the Quadratic Formula:

Example 8.

2x2
+ 6x – 5 = 0.

In this case, a = two, b = 6, c = -5. Substituting these values in the Quadratic Formula yields

Notice that we solved this equation earlier by completing the foursquare.

Note: In that location are two real solutions. In terms of graphs, at that place are 2 intercepts for the graph of the role f(x) = 2x2
+ 6x – 5.


Case 9.

4x2
+ 4x + one = 0

In this example a = four, b = 4, and c = 1.

There are two things to notice about this case.

  • There is just one solution. In terms of graphs, this means there is only one x-intercept.

  • The solution simplified so that at that place is no foursquare root involved. This ways that the equation could have been solved by factoring. (All quadratic equations
    tin can
    be solved past factoring! What I mean is information technology could take been solved
    hands
    by factoring.)

4xtwo
+ 4x + one = 0.

(2x + one)2
= 0.

x = -i/ii.


Example ten.

xtwo
+ x + 1 = 0

a = 1, b = i, c = i

Notation:
There are no real solutions. In terms of graphs, there are no intercepts for the graph of the part f(10) = x2
+ x + 1. Thus, the solutions are complex because the graph of y = xii
+ 10 + 1 has no x-intercepts.

The expression under the radical in the Quadratic Formula, b2
– 4ac, is chosen the discriminant of the equation. The terminal 3 examples illustrate the iii possibilities for quadratic equations.

one. Discriminant > 0. Two existent solutions.

2. Discriminant = 0. I real solution.

3. Discriminant < 0. Ii circuitous solutions.

Notes on checking solutions

None of the techniques introduced so far in this section can introduce extraneous solutions. (Run across case three from the Linear Equations and Modeling department.) However, it is still a good thought to check your solutions, because information technology is very easy to brand careless errors while solving equations.

The algebraic method, which consists of substituting the number back into the equation and checking that the resulting statement is true, works well when the solution is “uncomplicated”, merely it is not very applied when the solution involves a radical.

For example, in our adjacent to last case, 4xii
+ 4x + 1 = 0, nosotros constitute one solution ten = -one/ii.

The algebraic bank check looks like

4(-ane/ii)two
+four(-one/2) + ane = 0.

four(1/iv) – 2 + i = 0.

1 – 2 + ane = 0.

0 = 0. The solution checks.

In the instance earlier that, 2xii
+ 6x – 5 = 0, we found two real solutions, 10 = (-3 ± sqrt(19))/2. It is certainly possible to check this algebraically, simply it is not very easy. In this case either a graphical check, or using a calculator for the algebraic check are faster.

First, discover decimal approximations for the 2 proposed solutions.

(-iii + sqrt(xix))/2 = 0.679449.

(-three – sqrt(nineteen))/2 = -3.679449.

Now use a graphing utility to graph y = 2x2
+ 6x – five, 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 judge solution into the equation to see if both sides of the equation give approximately the same values. However, you lot nevertheless need to be conscientious in your claim that your solution is correct, since it is not the exact solution.

Note that if you had started with the equation 2x2
+ 6x – 5 = 0 and gone directly to the graphing utility to solve it, then you would non become the verbal solutions, considering they are irrational. Nevertheless, having found (algebraically) 2 numbers that you think are solutions, if the graphing utility shows that intercepts are very shut to the numbers you found, and so you are probably right!


Exercise 1:

Solve the following quadratic equations.

(a) 3xtwo
-5x – 2 = 0. Respond

(b) (x + 1)2
= 3. Reply

(c) x2
= 3x + 2. Respond


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Equations Involving Radicals

Equations with radicals tin often be simplified past raising to the advisable power, squaring if the radical is a foursquare root, cubing for a cube root, etc. This operation can introduce extraneous roots, so all solutions must be checked.

If there is only one radical in the equation, and so earlier raising to a power, you lot should arrange to accept the radical term past itself on one side of the equation.

Case 11.

Now that we take isolated the radical term on the right side, we foursquare both sides and solve the resulting equation for ten.

Bank check:

x = 0

When nosotros substitute 10 = 0 into the original equation nosotros get the statement 0 = two, which is not true!

So, x = 0
is not a solution.

x = 3

When we substitute x = 3 into the original equation, nosotros become the statement 3 = 3. This is true, so x = 3
is a solution.

Solution: ten = iii.

Note:
The solution is the x-coordinate of the intersection point of the graphs of y = x and y = sqrt(x+1)+1.

Look at what would have happened if we had squared both sides of the equation
before
isolating the radical term.

This is worse than what we started with!

If there is more than than one radical term in the equation, then in general, nosotros cannot eliminate all radicals past raising to a power one fourth dimension. Yet, nosotros can
decrease
the number of radical terms by raising to a power.

If the equation involves more than ane radical term, then nosotros yet want to isolate one radical on i side and raise to a power. Then we repeat that process.

Case 12.

At present foursquare both sides of the equation.

This equation has just 1 radical term, so we have made progress! Now isolate the radical term and so foursquare both sides over again.

Check:

Substituting 10 = 5/4 into the original equation yields

sqrt(9/4) + sqrt(i/four) = 2.

3/two + 1/ii = 2.

This statement is true, so x = five/4 is a solution.

Notation on checking solutions:

The algebraic check was easy to do in this instance. However, the graphical check has the advantage of showing that there are no solutions that nosotros accept not found, at to the lowest degree within the scope of the viewing rectangle. The solution is the 10-coordinate of the intersection point of the graphs of y = two and y = sqrt(10+1)+sqrt(10-i).


Exercise 2:

Solve the equation sqrt(x+ii) + ii = 2x. Answer


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Polynomial Equations of College Degree

We have seen that any caste 2 polynomial equation (quadratic equation) in 1 variable can be solved with the Quadratic Formula. Polynomial equations of degree greater than two are more complicated. When we encounter such a trouble, and then either the polynomial is of a special grade which allows us to factor it, or nosotros must approximate the solutions with a graphing utility.

Zero Constant

One common special instance is where there is no constant term. In this case we may factor out one or more than powers of x to begin the problem.

Case 13.

2xthree
+ 3xii
-5x = 0.

x (2xtwo
+ 3x -five) = 0.

Now we have a product of x and a quadratic polynomial equal to 0, so we have two simpler equations.

x = 0, or 2x2
+ 3x -5 = 0.

The first equation is lilliputian to solve. x = 0 is the just solution. The second equation may exist solved by factoring.
Note:
If nosotros were unable to factor the quadratic in the second equation, then we could have resorted to using the Quadratic Formula. [Verify that yous get the same results as below.]

x = 0, or (2x + 5)(ten – one) = 0.

So there are three solutions: x = 0, x = -five/2, 10 = i.

Note:
The solution is constitute from the intercepts of the graphs of   f(ten) = 2xiii
+ 3x2
-5x.

Cistron by Grouping

Example 14.

x3
-2xii
-9x +18 = 0.

The coefficient of xtwo
is -2 times that of x3, and the aforementioned relationship exists betwixt the coefficients of the third and 4th terms. Group terms one and ii, and likewise terms iii and four.

x2
(x – 2) – 9 (x – 2) = 0.

These groups share the common cistron (x – 2), so we tin can factor the left hand side of the equation.

(ten – 2)(x2
– 9) = 0.

Whenever nosotros find a product equal to nil, we obtain two simpler equations.

ten – 2 = 0, or x2
– 9 = 0.

10 = 2, or (ten + 3)(x – 3) = 0.

So, there are 3 solutions, x = ii, ten = -3, x = iii.

Annotation:
These solutions are found from the intercepts of the graph of   f(x) = x3
-2x2
-9x +18.

Quadratic in Form

Example fifteen.

x4
– tenii
– 12 = 0.

This polynomial is not quadratic, it has degree four. All the same, it can be idea of as quadratic in x2.

(102)
2
-(xii) – 12 = 0.

It might assistance you lot to actually substitute z for x2.

z2
– z – 12 = 0 This is a quadratic equation in z.

(z – 4)(z + 3) = 0.

z = 4 or z = -3.

We are not done, because we need to find values of 10 that brand the original equation true. Now supersede z past ten2
and solve the resulting equations.

xtwo
= 4.

x = ii, x = -2.

x2
= -iii.

10 =
i

, or x = –i.

So there are iv solutions, two real and two complex.

Note:
These solutions are institute from the intercepts of the graph of   f(x) = xiv
– x2
– 12.


A graph of f(x) = xiv
– x2
– 12  and a zoom showing its local extrema.


Practice 3:

Solve the equation xiv
– 5x2
+ iv = 0. Answer


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Equations Involving Partial Expressions or Absolute Values

Example xvi.

The least mutual denominator is x(ten + 2), and then we multiply both sides by this product.

This equation is quadratic. The Quadratic Formula yields the solutions

Checking is necessary because nosotros multiplied both sides by a variable expression. Using a graphing utility we see that both of these solutions check. The solution is the x-coordinate of the intersection indicate of the graphs of y = 1 and y = 2/x-1/(x+ii).

Case 17.

5 | 10 – 1 | = x + 11.

The primal to solving an equation with absolute values is to retrieve that the quantity inside the absolute value bars could be positive or negative. Nosotros will have two separate equations representing the different possibilities, and all solutions must exist checked.


Case 1.

Suppose 10 – 1 >= 0. And so | x – ane | = x – 1, so we accept the equation

5(x – ane) = ten + 11.

5x – v = x + 11.

4x = sixteen.

x = 4, and this solution checks because v*3 = iv + 11.


Case 2.

Suppose x – 1 < 0. Then ten – 1 is negative, then | x – i | = -(x – ane). This bespeak oftentimes confuses students, because it looks as if nosotros are maxim that the absolute value of an expression is negative, only nosotros are not. The expression (x – ane) is already negative, so -(10 – i) is positive.

Now our equation becomes

-5(x – 1) = ten + xi.

-5x + five = 10 + xi.

-6x = 6.

ten = -1, and this solution checks because five*ii = -ane + 11.

If yous employ the Java Grapher to check graphically, notation that abs() is accented value, then you would graph

5*abs(ten - 1) - x - 11
and look at ten-intercepts, or you can find the solution equally the x-coordinates of the intersection points of the graphs of y = x+11 and y = v*abs(10-i).


Exercise iv:

(a) Solve the equation

Answer

(b) Solve the equation | x – 2 | = two – ten/3 Answer


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If 2+sqrt 3 is a Polynomial Root

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