Algebra 1 · Unit 9
Count real solutions with the discriminant, and verify with algebra, a graph, or a table. Interpret answers in a story.
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D = b² − 4ac
D > 0 two · D = 0 one · D < 0 none
Discriminant
D = b² − 4ac. D > 0 two real solutions, D = 0 one real solution, D < 0 no real solution.
Verify algebraically
Substitute each proposed x back into the original equation.
Verify on a graph
The solutions are the x-intercepts. A graphing tool or a sketch shows how many times the parabola meets the x-axis.
A table is a technology check
Look for y = 0, or a sign change. A table can miss an intercept that sits between listed x-values.
Count solutions. Integer discriminants.
D = 25 − 24 = 1 > 0
Two real solutions.
Check: (x − 2)(x − 3) = 0, x = 2 and x = 3.
Answer: two
D = 0 − 16 = −16 < 0
No real solution.
Answer: none
One real solution, x = 5 (double root).
D = 0.
Answer: one
9 − 6 − 3 = 0. Yes.
The other solution is x = −1.
Answer: yes
Two x-intercepts, at ±2.
Answer: two
One real solution, x = 0.
Answer: one
Rewrite first, or check a table.
2x² − 8x = 0
2x(x − 4) = 0
Two real solutions, 0 and 4.
D = 64 − 0 = 64 > 0.
Answer: two
D = 9 − 16 = −7 < 0
None.
Answer: none
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| y | 2 | 0 | 0 | 2 |
y = 0 at x = 2 and x = 3.
Two real solutions. The table caught both because they are integers.
Answer: two; x = 2, 3
1 + 1 − 2 = 0. Yes.
The parabola crosses at x = 1 and x = −2.
Answer: yes
A square cannot be negative. None.
Answer: none
Only w = 3. Negative width is rejected.
One solution in context, two real solutions of the equation.
Answer: one in the story
A table that misses a root, or justify with two methods.
D = 4 + 4 = 8 > 0, two real solutions.
x = 1 ± √2
1 + √2 is between 2 and 3 (sign change −1 to 2).
1 − √2 is between 0 and −1, not in the listed x-values.
The table skipped the negative intercept.
Answer: two; table skipped one
D = 36 − 4k = 0
k = 9
Answer: k = 9
x − 1 = √2
(1 + √2)² − 2(1 + √2) − 1
1 + 2√2 + 2 − 2 − 2√2 − 1 = 0
Answer: yes
No x-intercepts.
D = 16 − 40 = −24 < 0. None.
Answer: none
Only the nonnegative root is a landing time after launch.
Check h(0) = 8 > 0, so it starts above ground. One positive zero.
Answer: one in the story
D = 4 − 4 = 0, one real solution.
(x + 1)² = 0, x = −1.
The vertex sits on the x-axis.
Answer: one
D = 1 + 4 = 5 > 0, two real solutions.
The roots 1 ± √52 are not integers, so an integer table misses them.
A graph still crosses the x-axis twice.
Answer: two real solutions, not integers
D = k² − 64 < 0
k² < 64
−8 < k < 8
Answer: −8 < k < 8
Rounding √2 cuts a little off. Then x² − 2 may show 0.001 instead of 0.
Check with exact algebra, or use more digits. A tiny leftover is a rounding error, not a wrong root.
Answer: rounding; check exactly
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