SME
Grade 9 Math · Algebra 1
Solving quadratic equations: factoring, square roots, completing the square and the quadratic formula
Use this for “Solve 2x² − 3x − 5 = 0 with the quadratic formula.”
Chart
Vertex (0.75, −6.125), axis of symmetry x = 0.75
Zeros: x = −1, x = 2.5
Discriminant: 49
Root with −√D: −1
Root with +√D: 5/2
This is an example. Type your own values to start a new problem.
a
x²
+
(
b
)
x
+
(
c
)
=
0
D
Discriminant
worked out (not typed)
x₁
Root with −√D
worked out (not typed)
x₂
Root with +√D
worked out (not typed)
Formulas
D
=
b
2
−
4
×
a
×
c
49
=
(−3)
2
−
4
×
2
×
(−5)
x₁
=
−
b
− √
D
2 ×
a
−1
=
−(−3) −
√
49
2 × 2
x₂
=
−
b
+ √
D
2 ×
a
5
2
=
−(−3) +
√
49
2 × 2
Assumptions
•
First write the equation as ax² + bx + c = 0.
•
D = b² − 4ac: when D > 0 there are two roots, when D = 0 just one, and when D < 0 none that are real.
•
Leave √D exact when D is not a perfect square; round only at the end.
Step-by-step
WE KNOW
a = 2, b = −3, c = −5
FIND
discriminant (D), root with −√D (x₁), root with +√D (x₂)
Step 1 · Find discriminant (D)
Use D = b² − 4 × a × c
The discriminant counts the roots: positive two, zero one, negative none.
D
=
(−3)
2
−
4
×
2
×
(−5)
D
=
9
−
4
×
2
×
(−5)
D
=
9
−
8
×
(−5)
D
=
9
−
(−40)
D
=
49
Step 2 · Find root with −√D (x₁)
Use x₁ = (−b − √D) ÷ (2 × a)
The
quadratic
formula,
x
=
−
b
± √
D
2
a
,
with
the
minus
sign.
x₁
=
−(−3) −
√
49
2 × 2
x₁
=
3 −
√
49
4
x₁
=
3 − 7
4
x₁
=
−
4
4
x₁
=
−1
Step 3 · Find root with +√D (x₂)
Use x₂ = (−b + √D) ÷ (2 × a)
The quadratic formula with the plus sign.
x₂
=
−(−3) +
√
49
2 × 2
x₂
=
3 +
√
49
4
x₂
=
3 + 7
4
x₂
=
10
4
x₂
=
5
2
Check
Put the numbers back in. Both sides should match.
49
=
(−3)
2
−
4
×
2
×
(−5)
✓
−1
=
−(−3) −
√
49
2 × 2
✓
5
2
=
−(−3) +
√
49
2 × 2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.