Question
4x^2+12x+10<=0 solve each inequality by graphing
Answers
GPT-4o mini
To solve the inequality \(4x^2 + 12x + 10 \leq 0\) by graphing, we first need to analyze the quadratic function \(f(x) = 4x^2 + 12x + 10\).
1. **Find the roots of the equation \(4x^2 + 12x + 10 = 0\):**
We can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 4\), \(b = 12\), and \(c = 10\).
First, we calculate the discriminant \(b^2 - 4ac\):
\[
b^2 - 4ac = 12^2 - 4 \cdot 4 \cdot 10 = 144 - 160 = -16
\]
Since the discriminant is negative, the quadratic function has no real roots, which means the parabola does not intersect the x-axis.
2. **Determine the direction of the parabola:**
The coefficient of \(x^2\) (which is 4) is positive, indicating that the parabola opens upwards.
3. **Evaluate the vertex to check the minimum point:**
The vertex \(x\) coordinate can be found using the formula \(x = -\frac{b}{2a}\):
\[
x = -\frac{12}{2 \cdot 4} = -\frac{12}{8} = -\frac{3}{2}
\]
Now we can find the \(y\) coordinate of the vertex by substituting \(x = -\frac{3}{2}\) back into the function:
\[
f\left(-\frac{3}{2}\right) = 4\left(-\frac{3}{2}\right)^2 + 12\left(-\frac{3}{2}\right) + 10
\]
\[
= 4 \cdot \frac{9}{4} - 18 + 10 = 9 - 18 + 10 = 1
\]
The vertex is at \(\left(-\frac{3}{2}, 1\right)\).
4. **Analyze the inequality \(4x^2 + 12x + 10 \leq 0\):**
Since the parabola opens upward and its minimum point (the vertex) is at \(y = 1\), which is above the x-axis, the function is always positive for all values of \(x\).
Therefore, there are no \(x\) values for which \(4x^2 + 12x + 10 \leq 0\).
In summary, the solution to the inequality is:
\[
\text{No solutions.}
\]
1. **Find the roots of the equation \(4x^2 + 12x + 10 = 0\):**
We can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 4\), \(b = 12\), and \(c = 10\).
First, we calculate the discriminant \(b^2 - 4ac\):
\[
b^2 - 4ac = 12^2 - 4 \cdot 4 \cdot 10 = 144 - 160 = -16
\]
Since the discriminant is negative, the quadratic function has no real roots, which means the parabola does not intersect the x-axis.
2. **Determine the direction of the parabola:**
The coefficient of \(x^2\) (which is 4) is positive, indicating that the parabola opens upwards.
3. **Evaluate the vertex to check the minimum point:**
The vertex \(x\) coordinate can be found using the formula \(x = -\frac{b}{2a}\):
\[
x = -\frac{12}{2 \cdot 4} = -\frac{12}{8} = -\frac{3}{2}
\]
Now we can find the \(y\) coordinate of the vertex by substituting \(x = -\frac{3}{2}\) back into the function:
\[
f\left(-\frac{3}{2}\right) = 4\left(-\frac{3}{2}\right)^2 + 12\left(-\frac{3}{2}\right) + 10
\]
\[
= 4 \cdot \frac{9}{4} - 18 + 10 = 9 - 18 + 10 = 1
\]
The vertex is at \(\left(-\frac{3}{2}, 1\right)\).
4. **Analyze the inequality \(4x^2 + 12x + 10 \leq 0\):**
Since the parabola opens upward and its minimum point (the vertex) is at \(y = 1\), which is above the x-axis, the function is always positive for all values of \(x\).
Therefore, there are no \(x\) values for which \(4x^2 + 12x + 10 \leq 0\).
In summary, the solution to the inequality is:
\[
\text{No solutions.}
\]