Answer is the image linked below.
First, we'll simplify the expression:
\[4x - 3(x^2) \geq 36x - 16\]
Expanding the left side, we get:
\[4x - 3x^2 \geq 36x - 16\]
Now, rearranging terms, we have:
\[-3x^2 + 4x \geq 36x - 16\]
Combining like terms on the right side, we get:
\[-3x^2 + 4x - 36x + 16 \geq 0\]
\[-3x^2 - 32x + 16 \geq 0\]
We now have a quadratic inequality in the form \(-3x^2 - 32x + 16 \geq 0\).
To find the solutions, we first find the roots of the quadratic equation \( -3x^2 - 32x + 16 = 0\). We can use the quadratic formula:
\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]
Where \(a = -3\), \(b = -32\), and \(c = 16\).
Now, let's plug in these values:
\[x = \frac{{-(-32) \pm \sqrt{{(-32)^2 - 4(-3)(16)}}}}{{2(-3)}}\]
\[x = \frac{{32 \pm \sqrt{{1024 + 192}}}}{{-6}}\]
\[x = \frac{{32 \pm \sqrt{{1216}}}}{{-6}}\]
\[x = \frac{{32 \pm 34.85}}{{-6}}\]
This gives us two potential solutions for \(x\):
1. \(x = \frac{{32 + 34.85}}{{-6}} \approx -1.14\)
2. \(x = \frac{{32 - 34.85}}{{-6}} \approx 10.14\)
These solutions divide the number line into three intervals:
1. \(x < -1.14\)
2. \(-1.14 < x < 10.14\)
3. \(x > 10.14\)
Now, we need to determine where the quadratic expression \( -3x^2 - 32x + 16\) is non-negative, i.e., where it is greater than or equal to zero.
We'll test a value from each interval:
1. For \(x < -1.14\), let's test \(x = -2\):
\[-3(-2)^2 - 32(-2) + 16 = -12 - (-64) + 16 = -12 + 64 + 16 = 68 > 0\]
So, this interval satisfies the inequality.
2. For \(-1.14 < x < 10.14\), let's test \(x = 0\):
\[-3(0)^2 - 32(0) + 16 = 0 - 0 + 16 = 16 > 0\]
This interval also satisfies the inequality.
3. For \(x > 10.14\), let's test \(x = 11\):
\[-3(11)^2 - 32(11) + 16 = -363 - 352 + 16 = -699 < 0\]
So, this interval does not satisfy the inequality.
Therefore, the solution set of the inequality \(4x - 3(x^2) \geq 36x - 16\) is \(x \leq -1.14\) and \(x \geq 10.14\).
