```Question 48581
<pre><font size = 5 color = "blue"><b>use the discriminant to determine the number of solutions
of the quadratic equation, and whether the solutions are
real or complex. Note: It is not necessary to find the
roots; just determine the number and types of solutions.

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Just learn these rules:

Ax² + Bx + C =0 (or any letter besides x)

has the discriminant  B² - 4AC. Calculate it.

1. If the discriminant is 0, there is one real solution.

2. If the discriminant is negative, there are two conjugate
complex solutions.

3. If the discriminant is positive, there are two real
solutions.
A. If the discriminant is a perfect square, the two real
solutions are rational.
B. If the discriminant is not a perfect square, the two
real solutions are irrational.

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2x² + x - 1 = 0

Give the x a coefficient of 1:

2x² + 1x - 1 = 0

Compare that to

Ax² + Bx + C = 0

A = 2, B = 1, and C = -1

Calculate discriminant     B² - 4AC
(1)² - 4(2)(-1)
1 + 8
9

9 is positive, so there by rule 3 above there are two real
solutions.  9 is also a perfect square, so by rule 3A the
two real solutions are rational.

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4/3x² - 2x + 3/4 = 0

Compare that to

Ax² + Bx + C = 0

A = 4/3, B = -2, and C = 3/4

Calculate discriminant     B² - 4AC
(-2)² - 4(4/3)(3/4)
-4

-4 is negative, so there by rule 2 above there are
two conjugate complex solutions.

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2x² + 5x + 5 = 0

Compare that to

Ax² + Bx + C = 0

A = 2, B = 5, and C = 5

Calculate discriminant     B² - 4AC
(5)² - 4(2)(5)
25 - 40
-13

-13 is negative, so there by rule 2 above there are
two conjugate complex solutions.

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3z² + z - 1 = 0

Give the z a coefficient of 1

3z² + 1z - 1 = 0

Compare that to

Ax² + Bx + C = 0

A = 3, B = 1, and C = -1

Calculate discriminant     B² - 4AC
(1)² - 4(3)(-1)
1 + 12
13

13 is positive, so there by rule 3 above there are two real
solutions.  13 is not a perfect square, so by rule 3B the
two real solutions are irrational.

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m² + m + 1 = 0

Give the m² and the m coeficients of 1 each

1m² + 1m + 1 = 0

Compare that to

Ax² + Bx + C = 0

A = 1, B = 1, and C = 1

Calculate discriminant     B² - 4AC
(1)² - 4(1)(1)
1 - 4
-3

-3 is negative, so there by rule 2 above there are
two conjugate complex solutions.

Edwin</pre>

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