4.1 Factoring Quadratics
Problem 4.2. The equation has two integer roots.
(a) How can we tell from the coefficients that both roots are divisible by ?
(b) Find the roots of .
Note (Scratch work). Write a polynomial as and let be its factored form. Then
so and . Since for , every , and , so some .
In , the cannot be odd.
If and , then
Factor pairs of with both factors even:
These are the only possible ones because both factors need to be even, and only and have both factors divisible by .
With and : . Let’s work mod : , so , and , so
so is a multiple of . Since : and .
Let . If then and ; and if with a product of distinct primes, then . So and . Then and .
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(a) First we notice that the roots must be even since, if they are odd, the equation will not be zero, since is even.
Next we know that
So then and , where are such that . So we have:
Now note that we have to show divisibility by , so let us work mod (i.e. ). Since both and are divisible by , and . This means that:
Now since is the product of distinct primes, by Euclid’s Lemma we have .
Finally, since and , then .
(b) I know that both integer roots are divisible by . I also know that they must both be even. So ; in order to have both factors be divisible by :
and clearly is the only combination that satisfies . Thus the roots are and .
Problem 4.3. (a)
(b)
(c)
(d)
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(a) Roots: , .
(b) Roots: , .
(c) , so the roots are .
(d) , roots: , .
Problem 4.4. (a) Factor .
(b) Find if .
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(a) The constant splits as or . By the quadratic formula,
Problem 4.2 (Another solution). has integer roots. How can we tell they are both divisible by ?
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Given : since and are even, is even, so must be even for . So the whole thing is divisible by .
Notice that , and and , so ; since , then . So , , and , so .
Now let and , so and . So and , so now we solve the system of equations:
Then and , so
Problem 4.4 (Part (b) again). Find such that .
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Let , so :
cannot be , so has to be :
So , .
Problem 4.5. How many integers satisfy the equation ?
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If , either , or , or and is even.
Case 1: : . ✓
Case 2: : .
Case 3: , :
, so or .
- : , so and ; here . ✓
- : , so and , not an integer. (Directly: and the exponent is , so the power is .) So is out.
So , , , .
Exercises
Exercise 4.1.2. (a)
(b)
(c)
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(a) . Let . Recall .
By the formula,
Or: needs and . Splitting the middle term,
which gives the factor .
Let’s restart to make it clean:
, and
(b) , and (using ), so and .
(c) . Wait, ur stupid, duh:
Exercise 4.1.3. Find if .
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Let . Then becomes , i.e. , so . Set aside for now: .
Restart: with , so :
Cancelling :
So for all such .
Exercise 4.1.4.
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From : . From : , so .
Exercise 4.1.5.
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Moving everything over: , i.e.
Instead, keep the original sides:
gives , and gives , . Total: .
4.2 Relating Roots and Coefficients
: the roots are .
Problem 4.6. Let have roots and .
(a) Show that for all .
(b) Find in terms of and .
(c) Find in terms of and .
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(a) , and
cannot be , so cancel it:
so and . Since ,
Alternatively: .
(b) .
(c) .
Problem 4.7. Find the sum and the product of the roots of each equation: and .
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Well, and , so we have:
- and .
- and .
Problem 4.8. Let be the roots of .
(a) Relate , , , to and .
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(a) and , so and ; also and (for nonzero roots). Then
So , the same as (that is, at ). Likewise
Adding,
Next, in terms of . Since
and . So we want in terms of and . Trying products:
Squaring the sum: , so , and
Problem 4.9. Find if and are the roots of .
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So , and
With :
Real parts: , so or (check: ✓, ✓).
Imaginary parts: .
So and .
Problem 4.10. Given that the sum of the zeros, the product of the zeros and the sum of the coefficients of are equal, show that all three of these quantities must be equal to .
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, with
Then
so .
Example 4.1. Roots , of , one of which is .
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. Here
With :
Also tried: , so .
4.3 Completing the Square
Problem 4.11. Suppose is a constant. In terms of , what must be added to to produce a quadratic that is the square of a binomial?
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First, two squares:
Now ; compare . In general
so to get , take . (From : let , giving ; let with , giving , so .) So
Check:
Problem 4.13 (MEOW!!!!!!!!!). Write in the form :
(a)
(b)
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From 4.11,
(a) , so
Scratch along the way: ; matching the constant instead, ; and asking for , i.e. . Also , .
(b) , and . First pass:
Expanding to check: , so . Forgot the :
since ; check: . (In decimals: .)
Problem 4.14. (a)
(b)
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(a) Add :
So , , and . With the formula:
(b)
Again: ; adding , , so , , , . With the formula:
Example 4.2 (Completing the square in ). Show solution
Completing the square inside the bracket:
so . Then
.
Problem 4.16. (a)
(b)
(c)
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(a)
By completing the square:
(b)
By completing the square:
(c)
Here and , so the discriminant is :
By completing the square:
Problem 4.17. and .
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Check:
Problem 4.18. Suppose and . We find .
(a) Express as the square of a binomial.
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(a) . Let , so . Trying values:
and then (, , , : with , ).
Attempts: set , i.e. ; write ; or . (Also considered Vieta: , .)
Since , and :
(Matching : ; ; .)
Let , so :
Section 4.3
Exercise 4.3.1. Complete the square of the following quadratics.
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(a)
(b)
(c)
(d)
since .
Exercise 4.3.3. Find such that .
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Squaring gives . Alternatively, , so
so (since ) .
Example 4.3. Solve .
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Factor out of the first two terms:
and ; half the coefficient is . To complete the square on we need
So
that is, or .
Example 4.4 (Nature of the roots). Show solution
Discriminants: the first is negative, so no real roots; the second is , with , so two (irrational) roots; for the last, , a perfect square.
The roots are, in order: nonreal, irrational, rational, rational.
Section 4.4
Example 4.5 (A real quadratic with root ). Show solution
We want , , such that is a root of . By Vieta,
so and (from ). Let : then
Plugging in (the term is ):
In factored form, .
An attempt to eliminate gave , i.e. , so
which is not real.
But , and nonreal roots mean . With we have , and the only way this is true (with real) is if .
With : . Since and ,
so the quadratic is
Check:
and completing the square, , so and .
Example 4.6 (A quadratic with root ). Show solution
Attempt: evaluate the -terms at :
which led to and, multiplying through by the conjugate,
Testing this:
whose -coefficient is , not rational. So the other root should be the conjugate,
Problem 4.21. Find all such that for some .
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has a real solution , so its discriminant is nonnegative:
Check with , : has discriminant . ✓
Answer: all such that .
Exercises
Exercise 4.4.1. Show solution
:
using and . ✓
Exercise 4.4.2. Find all such that has only one distinct root.
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If the double root is , then by Vieta and , so . Then
Or, with the discriminant:
Both work.
Exercise 4.4.5. Given that one root of , with , is , find .
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The other root is , so and .
Exercise 4.4.6. . Find all such that both roots are real.
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When , the factors vanish. So, trying : . Trying : ; from , .
In fact every real works: expanding, , with discriminant
so the roots are always real (for the equation is just ).
Problems 4.25–4.44
Example 4.7. . Find over the for which .
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, since
Then
By the quadratic formula on (with ):
so or .
Problem 4.25. , where are the roots of . Find .
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By Vieta, and .
Then .
Substituting for (and using ):
Now , and from ,
so , and .
Problem 4.27. Find if and is the reciprocal of .
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so or .
Problem 4.31. What is the domain of ?
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We need . Then
so
where and . So also :
Problem 4.32. The roots of satisfy , ; and for , and . Solve for .
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For fun:
so the two equations are and .
Example 4.8. Find all such that .
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. Completing the square, , so
Taking square roots, , so :
(Equivalently, gives : and .)
Problem 4.37. Find if and .
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Substituting :
and , where has no real roots. So and .
Problem 4.38. Show solution
Price (value at a given : ); tickets per game; each one-dollar increase loses 100 tickets, so the tickets sold are
which is at . At the revenue is . We want
Problem 4.39. Find all such that .
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so .
Problem 4.40. Let be a quadratic polynomial such that , , . Find .
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, so . Completing the square, . By Vieta, and . From : , and also
where , so
which only gives back .
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We know , , , and want . for some .
. We also know ; together,
Now since .
(Overthinking it.)
From , , so and .
Check: , and :)
So , and at : .
Problem 4.41. Find all such that .
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: can be anything (nonzero) if .
The base can also be : . Or the base can be with an even exponent: , and the exponent is at and at , both even. So
Problem 4.43. Show solution
The roots satisfy , i.e. , so . So and : indeed and .
By the quadratic formula,
where
so , i.e. or .
Answer: , .
Problem 4.44. If has solutions for such that each solution is the negative of the other, find in terms of and .
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We want . By Vieta,
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We want for the roots in , with ; find in terms of .
Also , i.e. , .
If the roots are , the quadratic is a multiple of , with no term: it should look like , i.e. .
Expanding,
so the -term is , and its coefficient must be . By Vieta,
gives , so , i.e. , and
Also , so .
Further problems
Example 4.9. Find all such that .
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and , so
Completing the square, , so and : either or .
If : , so , , , so or , with (check: and ).
If : , , and .
Another route: in gives
With : , so
i.e. , together with .
So
Example 4.10. Let be the real roots of . Find .
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.
Let (zero at ). Then , and the left side is .
where
so
Example 4.11. Find such that is as small as possible and the entire graph of the equation lies within .
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and (the denominator never vanishes for real ). So
and …
Example 4.12. Find all :
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First manipulations: , so ; ; ; .
Adding all five equations:
so
Completing each square:
and , so
Since all are squared, all , but since they are added, each has to be , so
To verify, we plug into the equations:
So for all .
Example 4.13. Find all such that , with .
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Both terms vanish (they become like and on each side) when and :
with , (and , ).
Clearing denominators: the numerators become and , with
Another try: set the numerators’ sum to zero,
where . Then
Factoring the sum of the numerators instead gives , so , : again or .
Properly, factor out from both numerators:
The bracket vanishes when
so , i.e. , and besides ,
So the solutions are , , and .
Example 4.14. Find all for which the zeros of differ by .
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(So , i.e. .) Then
and . Doubling,
and with :
(Equivalently, and give , so , i.e. .)
Substituting :
For : , so ; for : .
Checks of with : ✓, and ✓.
Labelling the pairs , and , : from , , so
Example 4.15. Find all ordered pairs with such that and .
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Let , so :
So either , which means , or , which means , i.e. .
With :
so or (check: ).
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(Solution continued.) Substitute each value of back into .
If :
so
If :
so
giving or .
Checks: : ✓, and : ✓. But has to hold too: ✓, while . So from these substitutions only and survive. The case makes and the roots of , i.e. , which gives and .
Final answer:
YAY
Example 4.16. Let with and . Find .
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We have . In the denominator, , and in the numerator . So
Since , and , so