powerpoint presentation

2. The number represented by b2 – 4ac = 0 is called the
discriminant of the quadratic equation.
How do we determine the nature of the roots of a
quadratic equation without actually solving the equation?
The nature of the roots can be determined by
finding the value of the discriminant.

3. Case 1. If b2 – 4ac = 0, then the roots of ax2 + bx + c = 0 are
real and equal.
▪ Example 1. Determine the nature of the roots of x2 – 6x + 9 = 0
a = 1, b = -6, c = 9
b2 – 4ac = ( -6)2 – 4(1)(9)
= 36 – 36
= 0
Since b2 – 4ac = 0, then x2 – 6x + 9 = 0 has two real roots which are
equal.

4. Verifying that then x2 – 6x + 9 = 0 has two equal real
roots,
By factoring, we have then x2 – 6x + 9 = 0
Factor: (x – 3)( x – 3) = 0
roots: x = 3 ; x = 3

5. Case 2: If b2 – 4ac > 0, then the roots of ax2 + bx + c = 0
are real and unequal.
Example 1. Determine the nature of the roots of x2 – 8x + 15 = 0.
a = 1, b = -8, c = 15
b2 – 4ac = (-8)2 – 4(1)(15)
= 64 - 60
= 4
Since b2 – 4ac > 0, then x2 – 8x + 15 = 0 has two real roots which
are unequal.

6. Verify that x2 – 8x + 15 = 0 has two unequal real roots:
By factoring we have,
(x – 3)(x - 5) = 0
X = 3 and x = 5

7. Case 3: If b2 – 4ac < 0, then the roots of ax2 + bx + c = 0 are
imaginary and unequal.
Example 3. Determine the nature of the roots of x2 – 2x + 2 = 0.
a = 1, b = -2, c = 2
b2 – 4ac = (-2)2 – 4(1)(2)
= 4 – 8
= -4
Since b2 – 4ac < 0, then x2 – 2x + 2 = 0 has two unequal imaginary
roots.

8. By quadratic formula we have,
x2 – 2x + 2 = 0,
a = 1, b = -2 c = 2
𝑥 =
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
𝑥 =
−(−2)± (−2)2−4(1)(2)
2(1)
x =
2 ± 4 −8
2
x =
2 ± −4
2
x =
2 ±2𝑖
2
x =
2 + 2𝑖
2
and x =
2 −2𝑖
2
x = 1 + 1i and x = 1 - 1i
x = 1 + i and x = 1 - i

9. SMSDGRADE9-MT.
OLYMPUS@ODL2020