9.1 Solving Quadratics by Square Roots and Factoring
A quadratic equation can be solved in several ways. This page starts with what a solution means for the graph, then covers the first two methods: taking square roots and factoring.
Quadratics and their roots
is a quadratic. Set to determine where the quadratic crosses the -axis:
Touching the -axis once: 1 real solution, 1-intercept.
Crossing it twice: 2 real solutions, 2 real -intercepts. That’s the max: a quadratic has at most 2 solutions.
Missing it: no real solution (2 complex solutions), no-intercept.
The square root method
Can I find the -intercepts by taking the square root of both sides?
Set to :.
Solve for .
Note.Remember!!!Always add before the square root when solving for .
Example 9.1.1. Solve .
Show solution
so and .Two real solutions.
Example 9.1.2. Solve .
Show solution
so and .Two real solutions.
Example 9.1.3. Solve .
Show solution
so and .Two real solutions.
Example 9.1.4. Solve .
Show solution
so
Two real solutions.
Example 9.1.5. Solve .
Show solution
, so .One real solution.
Example 9.1.6. Solve .
Show solution
No real solutions. Over the complex numbers, , so
Factoring
Expanding a product of two linear factors,
so factoring comes down to finding two numbers and with and .
The “X” in the solutions below is a factoring aid for finding them: the number on top is the sum , the number on the bottom is the product , and the two numbers on the sides multiply to the bottom and add to the top. When , each side number is divided by , and the reduced fraction gives the factor .
Example 9.1.7. Solve .
Show solution
X: and . So
Checking by grouping:
2 real solutions.
Example 9.1.8. Solve .
Show solution
X: top , bottom , sides and . So
2 real solutions.
Example 9.1.9. Solve .
Show solution
Rearranged: . X: top , bottom , sides and ; dividing each by gives . So
Expanding back: , since .
Check :, so
1 real solution.
Example 9.1.10. Solve .
Show solution
So and .
Example 9.1.11. Solve .
Show solution
X: top , bottom , sides and ; dividing each by gives and . So
Example 9.1.12. Solve .
Show solution
Rearranged: . X: top , bottom , sides and ; dividing by gives and . Splitting the middle term:
so or :,.2 real solutions.
Factoring problems
Problem 9.1.13. 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 .
Show solution
(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 .
Show solution
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 9.1.14. (a)
(b)
(c)
(d)
Show solution
(a) Roots: ,.
(b) Roots: ,.
(c) , so the roots are .
(d) , roots: ,.
Problem 9.1.15. (a) Factor .
(b) Find if .
Show solution
(a) The constant splits as or . By the quadratic formula,
(b) We want such that . Let , so : cannot be , so has to be :
So ,.
Problem 9.1.16. How many integers satisfy the equation ?
Show solution
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 9.1.17. Find if .
Show solution
Let . Then becomes , i.e. , so . Set aside for now: .
Restart: with , so :
Cancelling :
So for all such .
Exercise 9.1.18. Let with and . Find .
Show solution
We have . In the denominator, , and in the numerator . So
Since , and , so
Exercise 9.1.19.
Show solution
From :. From :, so .
Exercise 9.1.20.
Show solution
Moving everything over: , i.e.
Instead, keep the original sides:
gives , and gives ,. Total: .
Exercise 9.1.21. What is the domain of ?
Show solution
We need . Then
so
where and . So also :
Exercise 9.1.22. Find all such that .
Show solution
: 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