Completing the square on the general quadratic , with letters in place of numbers, gives a formula for its roots.
Theorem 9.4.1 (Quadratic Formula). For (with ), the solutions of are
Exercise 9.4.2 (Prove the quadratic formula). Show that if , the solutions of are .
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Start from ; its X has on top and on the bottom. Factor out :
Inside the bracket the X has on top, and on the sides, and on the bottom, so complete the square with :
The number under the square root is called the discriminant.
For instance, .
Example 9.4.3. Solve .
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so
Example 9.4.4. Solve .
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Expanding: .
so
Example 9.4.5. Solve .
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Quadratic formula directly:
Completing the square: X: top , sides and , bottom . Adding inside the bracket adds , so subtract outside:
Clearing the fraction first: multiply both sides by :
using .
Example 9.4.6. Solve .
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Directly:
where . So
Clearing fractions first: multiply both sides by :
so : and .
Example 9.4.7. Solve .
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Example 9.4.8. Solve .
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Factoring out first: , so and
Directly:
so
Problem 9.4.9. (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:
The discriminant
In the examples above, a discriminant of gave a single root and a negative discriminant gave two nonreal roots. The problems below use the discriminant to tell what kind of roots a quadratic has without solving it.
Example 9.4.10 (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.
Problem 9.4.11. 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 .
Finding a quadratic from one root
Example 9.4.12 (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 9.4.13 (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,
Exercises
Exercise 9.4.14. . Find over the for which .
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, since
Then
By the quadratic formula on (with ):
so or .
Exercise 9.4.15. Find if and is the reciprocal of .
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so or .
Exercise 9.4.16. Show solution
:
using and . ✓
Exercise 9.4.17. 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 9.4.18. Given that one root of , with , is , find .
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The other root is , so and .
Exercise 9.4.19. . 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 ).
Exercise 9.4.20. 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 …