Alg1, Unit 15, Lesson04_absent-student, page 1
More practice solving quadratics
The discriminant, special cases
Recall that some polynomials resist being factored and are labeled as
“prime.” They still have as many roots as are indicated by the degree of
the polynomial; however, the factoring technique is unable to find
them.
It is possible to find the roots of a prime quadratic polynomial using the
quadratic formula even though the factoring technique fails.
The quantity b2 – 4ac in the quadratic formula is known as the
discriminant. There are three possibilities: this quantity can be positive,
negative, or zero.
b2 – 4ac > 0
Consider when the discriminate is positive. This was the case in all
of the problems of the previous lesson. Also, in the last lesson, the
discriminant was a perfect square root.
How is the problem handled when it is not a perfect square root?
Use a calculator to evaluate the square root.
Example 1: Solve x2 – 3x – 8 = 0
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Alg1, Unit 15, Lesson04_absent-student, page 2
b2 – 4ac = 0
Consider when the discriminant is zero. This gives rise to a double
root.
The quadratic formula “tries” to give us two roots as predicted by
the degree (2)… and it does; however, it’s just that they are the
same. It’s called a double root.
Example 2: Find the roots of –x2 + 16x – 64.
b2 – 4ac < 0
Consider when the discriminant is negative. Since this causes a
negative number to be under the radical (square root sign), we
are unable to proceed. Give the answer as, “No real roots.”
In a subsequent course, you will learn that a negative discriminant
gives rise to imaginary numbers.
Example 3: What are the zeros of f2 – 6f + 10?
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Alg1, Unit 15, Lesson04_absent-student, page 3
Sometimes a quadratic equation is given in which not all the terms are
on the left side.
Move all terms to the left side of the equation (leaving 0 on the
right side) before identifying a, b, and c.
Example 4: Solve 8x2 = -6x + 1
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Alg1, Unit 15, Lesson04_absent-student, page 4
Assignment: Use the quadratic formula in solving these problems.
1. Solve 6x + 8 = x2
2. Find the roots of 2x2 + 4x + 7
3. Find the zeros of 6r2 – 3r + 7
4. Solve x2 + 4x = -4
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Alg1, Unit 15, Lesson04_absent-student, page 5
5. Solve -3x2 + 6x + 5 = 0
6. Solve 6p2 – 3p + 7 = 0
7. Solve -x2 + x = -30
8. What are the zeros of 4x2 + 12x + 9 = 0?
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Alg1, Unit 15, Lesson04_absent-student, page 6
9. What are the roots of 3k2 + k – 14?
10. x2 = 10x – 25
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