Questions: Solve the inequality. Write the solution set in interval notation if possible. Simplify your answer, if necessary. d^4 - 37d^2 + 36 ≤ 0 Part 1 of 4 The inequality is already in the form F(d) ≤ 0. Now, find the real solutions to the related equation F(d) = 0. (d^2 □)(d^2 □) = 0

Solution Steps

To solve the inequality \(d^4 - 37d^2 + 36 \leq 0\), we first need to find the real solutions to the related equation \(d^4 - 37d^2 + 36 = 0\). This can be done by factoring the polynomial. Once we have the roots, we can determine the intervals where the inequality holds true.

1. Substitute \(u = d^2\) to transform the quartic equation into a quadratic equation in terms of \(u\).
2. Solve the quadratic equation \(u^2 - 37u + 36 = 0\) to find the values of \(u\).
3. Substitute back \(u = d^2\) to find the values of \(d\).
4. Determine the intervals where the inequality \(d^4 - 37d^2 + 36 \leq 0\) holds true by testing the intervals between the roots.

We start with the equation \(d^4 - 37d^2 + 36 = 0\). By substituting \(u = d^2\), we can rewrite the equation as \(u^2 - 37u + 36 = 0\). The roots of this equation are found to be \(u = 1\) and \(u = 36\).

Next, we substitute back \(u = d^2\) to find the values of \(d\): \[ d^2 = 1 \implies d = \pm 1 \] \[ d^2 = 36 \implies d = \pm 6 \] Thus, the real solutions to the equation are \(d = -6, -1, 1, 6\).

The roots divide the number line into intervals: \[ (-\infty, -6), \quad (-6, -1), \quad (-1, 1), \quad (1, 6), \quad (6, \infty) \] We will test each interval to determine where the inequality \(d^4 - 37d^2 + 36 \leq 0\) holds.

1. Interval \((- \infty, -6)\): Choose \(d = -7\): \[ (-7)^4 - 37(-7)^2 + 36 = 2401 - 37 \cdot 49 + 36 = 2401 - 1813 + 36 = 624 > 0 \]
2. Interval \((-6, -1)\): Choose \(d = -2\): \[ (-2)^4 - 37(-2)^2 + 36 = 16 - 37 \cdot 4 + 36 = 16 - 148 + 36 = -96 < 0 \]
3. Interval \((-1, 1)\): Choose \(d = 0\): \[ 0^4 - 37 \cdot 0^2 + 36 = 36 > 0 \]
4. Interval \((1, 6)\): Choose \(d = 2\): \[ 2^4 - 37 \cdot 2^2 + 36 = 16 - 148 + 36 = -96 < 0 \]
5. Interval \((6, \infty)\): Choose \(d = 7\): \[ 7^4 - 37 \cdot 7^2 + 36 = 2401 - 1813 + 36 = 624 > 0 \]

The inequality \(d^4 - 37d^2 + 36 \leq 0\) holds in the intervals \((-6, -1)\) and \((1, 6)\), including the endpoints where the expression equals zero. Therefore, the solution set in interval notation is: \[ [-6, -1] \cup [1, 6] \]

Final Answer