To solve the polynomial equation [tex]\(2s^4 + 13s^2 + 15 = 0\)[/tex] using U-Substitution, follow these steps:
1. Substitute [tex]\(u\)[/tex] for [tex]\(s^2\)[/tex]:
Let [tex]\(u = s^2\)[/tex]. Then, the equation [tex]\(2s^4 + 13s^2 + 15 = 0\)[/tex] becomes:
[tex]\[
2u^2 + 13u + 15 = 0
\][/tex]
2. Solve the quadratic equation [tex]\(2u^2 + 13u + 15 = 0\)[/tex] for [tex]\(u\)[/tex]:
A quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex] can be solved using the quadratic formula:
[tex]\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
In our case, [tex]\(a = 2\)[/tex], [tex]\(b = 13\)[/tex], and [tex]\(c = 15\)[/tex]. Plug these values into the quadratic formula:
[tex]\[
u = \frac{-13 \pm \sqrt{13^2 - 4 \cdot 2 \cdot 15}}{2 \cdot 2}
\][/tex]
Calculate the discriminant:
[tex]\[
\Delta = 13^2 - 4 \cdot 2 \cdot 15 = 169 - 120 = 49
\][/tex]
Now, solve for [tex]\(u\)[/tex]:
[tex]\[
u = \frac{-13 \pm \sqrt{49}}{4}
\][/tex]
3. Find the roots [tex]\(u_1\)[/tex] and [tex]\(u_2\)[/tex]:
The square root of 49 is 7, so:
[tex]\[
u = \frac{-13 \pm 7}{4}
\][/tex]
This gives us two solutions:
[tex]\[
u_1 = \frac{-13 + 7}{4} = \frac{-6}{4} = -\frac{3}{2}
\][/tex]
[tex]\[
u_2 = \frac{-13 - 7}{4} = \frac{-20}{4} = -5
\][/tex]
4. Back-substitute for [tex]\(s\)[/tex]:
Recall that [tex]\(u = s^2\)[/tex]. Therefore, we need to solve for [tex]\(s\)[/tex] from [tex]\(s^2 = u\)[/tex]. We have two [tex]\(u\)[/tex] values:
[tex]\[
s^2 = -\frac{3}{2} \quad \text{and} \quad s^2 = -5
\][/tex]
5. Check the validity of [tex]\(s^2 = -\frac{3}{2}\)[/tex] and [tex]\(s^2 = -5\)[/tex]:
Since both [tex]\(-\frac{3}{2}\)[/tex] and [tex]\(-5\)[/tex] are negative numbers, and the square of a real number cannot be negative, there are no real solutions for [tex]\(s\)[/tex].
Therefore, the polynomial equation [tex]\(2s^4 + 13s^2 + 15 = 0\)[/tex] does not have any real solutions.
