Question

the above image shows that this equation has solutions
the discriminant is
the parabola opens
because the leading coefficient is
both solutions are positive integers

Explanation:

To solve this problem, we analyze the quadratic equation \( ax^2 + bx + c = 0 \) and its related concepts:

1. Number of Solutions

The graph shows the parabola intersecting the \( x \)-axis at two distinct points (\( x_1 \) and \( x_2 \)). For a quadratic equation, the number of real solutions is determined by the discriminant \( D = b^2 - 4ac \). If \( D > 0 \), there are two distinct real solutions; if \( D = 0 \), one repeated solution; if \( D < 0 \), no real solutions. Here, the graph has two distinct \( x \)-intercepts, so the equation has \(\boldsymbol{2}\) \(\boldsymbol{\text{real}}\) solutions.

2. The Discriminant

The discriminant of \( ax^2 + bx + c = 0 \) is defined as \( \boldsymbol{b^2 - 4ac} \) (from the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)).

3. Direction of the Parabola

The parabola opens upward because the leading coefficient \( a \) (coefficient of \( x^2 \)) is positive (a positive \( a \) makes the parabola open upward; a negative \( a \) makes it open downward).

4. Sign of Leading Coefficient

As the parabola opens upward, the leading coefficient \( a \) is positive.

5. Sign of Solutions

The \( x \)-intercepts (\( x_1 \) and \( x_2 \)) are on the positive \( x \)-axis (since the vertex is on the \( y \)-axis, and the roots are symmetric about it, both to the right of the origin). Thus, both solutions are positive integers: \(\boldsymbol{\text{True}}\) (assuming the context implies this, as the roots lie on the positive \( x \)-axis).

Final Answers (Filling in the Blanks):

• "The equation has" → \( \boldsymbol{2} \), \( \boldsymbol{\text{real}} \)
• "The discriminant is" → \( \boldsymbol{b^2 - 4ac} \)
• "The parabola opens" → \( \boldsymbol{\text{upward}} \)
• "leading coefficient is" → \( \boldsymbol{\text{positive}} \)
• "Both solutions are positive integers" → \( \boldsymbol{\text{True}} \) (or as per the dropdown options, e.g., "Yes", "True", etc.)

(Note: The exact dropdown options may vary, but the reasoning above guides the correct selections.)

Answer:

To solve this problem, we analyze the quadratic equation \( ax^2 + bx + c = 0 \) and its related concepts:

1. Number of Solutions

The graph shows the parabola intersecting the \( x \)-axis at two distinct points (\( x_1 \) and \( x_2 \)). For a quadratic equation, the number of real solutions is determined by the discriminant \( D = b^2 - 4ac \). If \( D > 0 \), there are two distinct real solutions; if \( D = 0 \), one repeated solution; if \( D < 0 \), no real solutions. Here, the graph has two distinct \( x \)-intercepts, so the equation has \(\boldsymbol{2}\) \(\boldsymbol{\text{real}}\) solutions.

2. The Discriminant

The discriminant of \( ax^2 + bx + c = 0 \) is defined as \( \boldsymbol{b^2 - 4ac} \) (from the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)).

3. Direction of the Parabola

The parabola opens upward because the leading coefficient \( a \) (coefficient of \( x^2 \)) is positive (a positive \( a \) makes the parabola open upward; a negative \( a \) makes it open downward).

4. Sign of Leading Coefficient

As the parabola opens upward, the leading coefficient \( a \) is positive.

5. Sign of Solutions

The \( x \)-intercepts (\( x_1 \) and \( x_2 \)) are on the positive \( x \)-axis (since the vertex is on the \( y \)-axis, and the roots are symmetric about it, both to the right of the origin). Thus, both solutions are positive integers: \(\boldsymbol{\text{True}}\) (assuming the context implies this, as the roots lie on the positive \( x \)-axis).

Final Answers (Filling in the Blanks):

• "The equation has" → \( \boldsymbol{2} \), \( \boldsymbol{\text{real}} \)
• "The discriminant is" → \( \boldsymbol{b^2 - 4ac} \)
• "The parabola opens" → \( \boldsymbol{\text{upward}} \)
• "leading coefficient is" → \( \boldsymbol{\text{positive}} \)
• "Both solutions are positive integers" → \( \boldsymbol{\text{True}} \) (or as per the dropdown options, e.g., "Yes", "True", etc.)

(Note: The exact dropdown options may vary, but the reasoning above guides the correct selections.)