The nature of quadratic roots
(b²-4ac) is the discriminant

If b²-4ac is The equation has:

> 0 and a non perfect square - 2 irrational roots

>0 and a perfect square - 2 rational roots

=0 - i real root

<0 - 0 real root, imaginary roots


Example:

Given x² - 6x + 21 = 0

What type of roots does it have?

a= 1
b= -6
c= 21

Plug it in:

X= (-b +- sqrt(b²-4ac))/(2a)

= -(-6) +- sqrt(-6²)-(4)(1)(21)
= 6 +- sqrt(-48)/2
= 6 +- sqrt((-1)(16)(3))/2
= 3 +- 4i sqrt(3)

Therefore this equation is an example of b²-4ac < 0 and has imaginary roots.

Continuation of our last topic …

Given the equation

If the roots of the equation are known, the equation can be found,it is...

x²+ (-sum of roots)x + (product of roots) = 0

x² + -b/a x + c/a = 0

Example 1

GIVEN THE ROOTS :( 1 ½, 7/5) state the quadratic equation

1. Find the sum
r1 +r2 = 1 ½ + 7/5
= 3/2 +7/5
= 15+14/10
= 29/10
-b/a= 29/10
b/a= -29/10

2. Find the product
r1*r2 = (3/2)(7/5)
= 21/10
c/a = 21/10

3. Plug it in
x² + -b/a x + c/a = 0
x² - 29/10 x + 21/10 = 0

Quadratic Equation = 10x² -29x +21 =0

Example 2

GIVEN THE ROOTS :( 1+2i, 1-2i) state the quadratic equation

1. Find the sum
r1 +r2 = (1+2i)+(1-2i)
= 2
-b/a= 2
b/a= -2

2. Find the product
r1*r2 = (1+2i)(1-2i)
= 1 + 2i - 2i + 4i²
= 1 - 4i²
= 1 – 4 (-1)
= 5
c/a = 5

3. Plug it in
x² + -b/a x + c/a = 0

Quadratic Equation = x² -2x + 5 =0

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