Question

1. determine the number of solutions for each equation.

equation	exactly one real solution	exactly two real solutions	no real solutions
$x^2 - 2x + 3 = 0$	$\bigcirc$	$\bigcirc$	$\bigcirc$
$x^2 + 5x - 4 = 0$	$\bigcirc$	$\bigcirc$	$\bigcirc$
$x^2 - 8x + 16 = 0$	$\bigcirc$	$\bigcirc$	$\bigcirc$

Explanation:

To determine the number of real solutions for a quadratic equation \( ax^2 + bx + c = 0 \), we use the discriminant \( D = b^2 - 4ac \):

• If \( D > 0 \), there are exactly two real solutions.
• If \( D = 0 \), there is exactly one real solution.
• If \( D < 0 \), there are no real solutions.

Step 1: Analyze \( x^2 + 1 = 0 \)

Here, \( a = 1 \), \( b = 0 \), \( c = 1 \).
Discriminant: \( D = 0^2 - 4(1)(1) = -4 \).
Since \( D < 0 \), no real solutions.

Step 2: Analyze \( x^2 - 2x + 3 = 0 \)

Here, \( a = 1 \), \( b = -2 \), \( c = 3 \).
Discriminant: \( D = (-2)^2 - 4(1)(3) = 4 - 12 = -8 \).
Since \( D < 0 \), no real solutions.

Step 3: Analyze \( x^2 + 5x - 4 = 0 \)

Here, \( a = 1 \), \( b = 5 \), \( c = -4 \).
Discriminant: \( D = 5^2 - 4(1)(-4) = 25 + 16 = 41 \).
Since \( D > 0 \), exactly two real solutions.

Step 4: Analyze \( x^2 - 8x + 16 = 0 \)

Here, \( a = 1 \), \( b = -8 \), \( c = 16 \).
Discriminant: \( D = (-8)^2 - 4(1)(16) = 64 - 64 = 0 \).
Since \( D = 0 \), exactly one real solution.

Answer:

• For \( \boldsymbol{x^2 + 1 = 0} \): Select "No Real Solutions".
• For \( \boldsymbol{x^2 - 2x + 3 = 0} \): Select "No Real Solutions".
• For \( \boldsymbol{x^2 + 5x - 4 = 0} \): Select "Exactly Two Real Solutions".
• For \( \boldsymbol{x^2 - 8x + 16 = 0} \): Select "Exactly One Real Solution".