To solve the polynomial [tex]\( 2r^4 + 23r^2 + 45 = 0 \)[/tex], we can use a method called U-Substitution. Here’s a step-by-step guide to how we approach this problem:
1. Substitute: Let [tex]\( u = r^2 \)[/tex]. This transforms the original polynomial into a quadratic equation in terms of [tex]\( u \)[/tex].
So, substituting [tex]\( u \)[/tex] into the polynomial gives us:
[tex]\[
2u^2 + 23u + 45 = 0
\][/tex]
2. Solve the quadratic equation:
The quadratic equation [tex]\( 2u^2 + 23u + 45 = 0 \)[/tex] can be solved using the quadratic formula:
[tex]\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
where [tex]\( a = 2 \)[/tex], [tex]\( b = 23 \)[/tex], and [tex]\( c = 45 \)[/tex].
3. Discriminant calculation:
Compute the discriminant:
[tex]\[
\Delta = b^2 - 4ac = 23^2 - 4 \cdot 2 \cdot 45 = 529 - 360 = 169
\][/tex]
4. Roots of the quadratic:
Since the discriminant ([tex]\( \Delta \)[/tex]) is positive ([tex]\( \Delta = 169 \)[/tex]), there are two real roots for the equation:
[tex]\[
u_1 = \frac{-23 + \sqrt{169}}{4} = \frac{-23 + 13}{4} = \frac{-10}{4} = -2.5
\][/tex]
[tex]\[
u_2 = \frac{-23 - \sqrt{169}}{4} = \frac{-23 - 13}{4} = \frac{-36}{4} = -9
\][/tex]
5. Revert the substitution:
Recall [tex]\( u = r^2 \)[/tex]. We need to find [tex]\( r \)[/tex] from each [tex]\( u \)[/tex]:
For [tex]\( u_1 = -2.5 \)[/tex]:
[tex]\[
r^2 = -2.5
\][/tex]
Since the square of a real number cannot be negative, there are no real roots for [tex]\( u_1 = -2.5 \)[/tex].
For [tex]\( u_2 = -9 \)[/tex]:
[tex]\[
r^2 = -9
\][/tex]
Similarly, the square of a real number cannot be negative, so there are no real roots for [tex]\( u_2 = -9 \)[/tex].
6. Conclusion:
Since both values derived from the quadratic equation for [tex]\( u \)[/tex] are negative and the square of a real number cannot be negative, there are no real solutions for [tex]\( r \)[/tex] in the original equation [tex]\( 2r^4 + 23r^2 + 45 = 0 \)[/tex].
Therefore, the polynomial [tex]\( 2r^4 + 23r^2 + 45 = 0 \)[/tex] has no real roots.
