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use the discriminant $x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}$ $b^2 -…

Question

use the discriminant

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$b^2 - 4ac$ is called the discriminant.

  • $b^2 - 4ac$ is zero $\

ightarrow$ 1 real solution

  • $b^2 - 4ac$ is positive $\

ightarrow$ 2 real solutions

  • $b^2 - 4ac$ is negative $\

ightarrow$ 2 imaginary solutions

equations:

  1. $x^2 - 10x + 25 = 0$
  2. $x^2 - 9 = 0$
  3. $x^2 + 6x + 2 = 0$
  4. $x^2 + 3x + 5 = 0$

handwritten calculations (partial):

  • for $x^2 - 10x + 25 = 0$: $\frac{10 \pm \sqrt{10^2 - 4(1)(25)}}{2(1)} = \frac{10 \pm 0}{2} = 5$
  • for $x^2 - 9 = 0$: factored as $(x - 3)(x + 3)$
  • for $x^2 + 6x + 2 = 0$: $\frac{-6 \pm \sqrt{6^2 - 4(1)(2)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 8}}{2} = \frac{-6 \pm \sqrt{28}}{2} = \frac{-6 \pm 2\sqrt{7}}{2} = -3 \pm \sqrt{7}$
  • for $x^2 + 3x + 5 = 0$: $\frac{-3 \pm \sqrt{3^2 - 4(1)(5)}}{2(1)} = \frac{-3 \pm \sqrt{9 - 20}}{2} = \frac{-3 \pm \sqrt{-11}}{2}$ (imaginary solutions)

Explanation:

Response

To analyze the quadratic equations using the discriminant \( D = b^2 - 4ac \), we follow these steps for each equation:

1. Equation: \( \boldsymbol{x^2 - 10x + 25 = 0} \)

Step 1: Identify \( a, b, c \)

For \( ax^2 + bx + c = 0 \), here \( a = 1 \), \( b = -10 \), \( c = 25 \).

Step 2: Calculate discriminant \( D \)

\( D = b^2 - 4ac = (-10)^2 - 4(1)(25) \)
\( D = 100 - 100 = 0 \)

Step 3: Determine solutions

If \( D = 0 \), there is 1 real solution (a repeated root).
Using the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \):
\( x = \frac{-(-10) \pm \sqrt{0}}{2(1)} = \frac{10}{2} = 5 \)

2. Equation: \( \boldsymbol{x^2 - 9 = 0} \)

Step 1: Identify \( a, b, c \)

Rewrite as \( x^2 + 0x - 9 = 0 \), so \( a = 1 \), \( b = 0 \), \( c = -9 \).

Step 2: Calculate discriminant \( D \)

\( D = 0^2 - 4(1)(-9) = 0 + 36 = 36 \)

Step 3: Determine solutions

If \( D > 0 \), there are 2 real solutions.
Using the quadratic formula:
\( x = \frac{-0 \pm \sqrt{36}}{2(1)} = \frac{\pm 6}{2} \)
Thus, \( x = 3 \) or \( x = -3 \) (or factor as \( (x - 3)(x + 3) = 0 \)).

3. Equation: \( \boldsymbol{x^2 + 6x + 2 = 0} \)

Step 1: Identify \( a, b, c \)

Here \( a = 1 \), \( b = 6 \), \( c = 2 \).

Step 2: Calculate discriminant \( D \)

\( D = 6^2 - 4(1)(2) = 36 - 8 = 28 \)

Step 3: Determine solutions

If \( D > 0 \), there are 2 real (irrational) solutions.
Using the quadratic formula:
\( x = \frac{-6 \pm \sqrt{28}}{2(1)} = \frac{-6 \pm 2\sqrt{7}}{2} = -3 \pm \sqrt{7} \)

4. Equation: \( \boldsymbol{x^2 + 3x + 5 = 0} \)

Step 1: Identify \( a, b, c \)

Here \( a = 1 \), \( b = 3 \), \( c = 5 \).

Step 2: Calculate discriminant \( D \)

\( D = 3^2 - 4(1)(5) = 9 - 20 = -11 \)

Step 3: Determine solutions

If \( D < 0 \), there are 2 complex (imaginary) solutions.
Using the quadratic formula (with \( \sqrt{-11} = i\sqrt{11} \)):
\( x = \frac{-3 \pm i\sqrt{11}}{2} \)

Summary of Results:
Equation\( a \)\( b \)\( c \)Discriminant (\( D \))Number of SolutionsSolutions (if real)
\( x^2 - 9 = 0 \)10-9362 real\( x = 3, -3 \)
\( x^2 + 6x + 2 = 0 \)162282 real (irrational)\( x = -3 \pm \sqrt{7} \)
\( x^2 + 3x + 5 = 0 \)135-112 imaginary\( x = \frac{-3 \pm i\sqrt{11}}{2} \)

The discriminant determines the nature of the roots:

  • \( D > 0 \): 2 distinct real roots.
  • \( D = 0 \): 1 real (repeated) root.
  • \( D < 0 \): 2 complex (imaginary) roots.

Answer:

To analyze the quadratic equations using the discriminant \( D = b^2 - 4ac \), we follow these steps for each equation:

1. Equation: \( \boldsymbol{x^2 - 10x + 25 = 0} \)

Step 1: Identify \( a, b, c \)

For \( ax^2 + bx + c = 0 \), here \( a = 1 \), \( b = -10 \), \( c = 25 \).

Step 2: Calculate discriminant \( D \)

\( D = b^2 - 4ac = (-10)^2 - 4(1)(25) \)
\( D = 100 - 100 = 0 \)

Step 3: Determine solutions

If \( D = 0 \), there is 1 real solution (a repeated root).
Using the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \):
\( x = \frac{-(-10) \pm \sqrt{0}}{2(1)} = \frac{10}{2} = 5 \)

2. Equation: \( \boldsymbol{x^2 - 9 = 0} \)

Step 1: Identify \( a, b, c \)

Rewrite as \( x^2 + 0x - 9 = 0 \), so \( a = 1 \), \( b = 0 \), \( c = -9 \).

Step 2: Calculate discriminant \( D \)

\( D = 0^2 - 4(1)(-9) = 0 + 36 = 36 \)

Step 3: Determine solutions

If \( D > 0 \), there are 2 real solutions.
Using the quadratic formula:
\( x = \frac{-0 \pm \sqrt{36}}{2(1)} = \frac{\pm 6}{2} \)
Thus, \( x = 3 \) or \( x = -3 \) (or factor as \( (x - 3)(x + 3) = 0 \)).

3. Equation: \( \boldsymbol{x^2 + 6x + 2 = 0} \)

Step 1: Identify \( a, b, c \)

Here \( a = 1 \), \( b = 6 \), \( c = 2 \).

Step 2: Calculate discriminant \( D \)

\( D = 6^2 - 4(1)(2) = 36 - 8 = 28 \)

Step 3: Determine solutions

If \( D > 0 \), there are 2 real (irrational) solutions.
Using the quadratic formula:
\( x = \frac{-6 \pm \sqrt{28}}{2(1)} = \frac{-6 \pm 2\sqrt{7}}{2} = -3 \pm \sqrt{7} \)

4. Equation: \( \boldsymbol{x^2 + 3x + 5 = 0} \)

Step 1: Identify \( a, b, c \)

Here \( a = 1 \), \( b = 3 \), \( c = 5 \).

Step 2: Calculate discriminant \( D \)

\( D = 3^2 - 4(1)(5) = 9 - 20 = -11 \)

Step 3: Determine solutions

If \( D < 0 \), there are 2 complex (imaginary) solutions.
Using the quadratic formula (with \( \sqrt{-11} = i\sqrt{11} \)):
\( x = \frac{-3 \pm i\sqrt{11}}{2} \)

Summary of Results:
Equation\( a \)\( b \)\( c \)Discriminant (\( D \))Number of SolutionsSolutions (if real)
\( x^2 - 9 = 0 \)10-9362 real\( x = 3, -3 \)
\( x^2 + 6x + 2 = 0 \)162282 real (irrational)\( x = -3 \pm \sqrt{7} \)
\( x^2 + 3x + 5 = 0 \)135-112 imaginary\( x = \frac{-3 \pm i\sqrt{11}}{2} \)

The discriminant determines the nature of the roots:

  • \( D > 0 \): 2 distinct real roots.
  • \( D = 0 \): 1 real (repeated) root.
  • \( D < 0 \): 2 complex (imaginary) roots.