Answer :
We start with the polynomial
[tex]$$
f(x) = x^4 + 4x^2 - 45.
$$[/tex]
Since the polynomial contains only even powers of [tex]$x$[/tex], we substitute
[tex]$$
p = x^2,
$$[/tex]
so that
[tex]$$
x^4 = p^2.
$$[/tex]
This converts the equation into a quadratic in [tex]$p$[/tex]:
[tex]$$
p^2 + 4p - 45 = 0.
$$[/tex]
To solve for [tex]$p$[/tex], we can use the quadratic formula:
[tex]$$
p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
$$[/tex]
where [tex]$a=1$[/tex], [tex]$b=4$[/tex], and [tex]$c=-45$[/tex]. First, calculate the discriminant:
[tex]$$
\Delta = b^2 - 4ac = 4^2 - 4(1)(-45) = 16 + 180 = 196.
$$[/tex]
Since [tex]$\sqrt{196}=14$[/tex], the formula gives:
[tex]$$
p = \frac{-4 \pm 14}{2}.
$$[/tex]
This leads to two solutions:
1. For the positive sign:
[tex]$$
p = \frac{-4 + 14}{2} = \frac{10}{2} = 5.
$$[/tex]
2. For the negative sign:
[tex]$$
p = \frac{-4 - 14}{2} = \frac{-18}{2} = -9.
$$[/tex]
Recall that [tex]$p = x^2$[/tex]. Now, consider each case:
1. For [tex]$p = 5$[/tex]:
[tex]$$
x^2 = 5 \quad \Longrightarrow \quad x = \sqrt{5} \quad \text{or} \quad x = -\sqrt{5}.
$$[/tex]
2. For [tex]$p = -9$[/tex]:
[tex]$$
x^2 = -9.
$$[/tex]
Since a square can never be negative for a real number, there are no real solutions in this case. (The solutions would be complex: [tex]$x=3i$[/tex] and [tex]$x=-3i$[/tex], but we are only considering real zeros.)
Thus, the real zeros of the function are:
[tex]$$
x = \sqrt{5} \quad \text{and} \quad x = -\sqrt{5}.
$$[/tex]
[tex]$$
f(x) = x^4 + 4x^2 - 45.
$$[/tex]
Since the polynomial contains only even powers of [tex]$x$[/tex], we substitute
[tex]$$
p = x^2,
$$[/tex]
so that
[tex]$$
x^4 = p^2.
$$[/tex]
This converts the equation into a quadratic in [tex]$p$[/tex]:
[tex]$$
p^2 + 4p - 45 = 0.
$$[/tex]
To solve for [tex]$p$[/tex], we can use the quadratic formula:
[tex]$$
p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
$$[/tex]
where [tex]$a=1$[/tex], [tex]$b=4$[/tex], and [tex]$c=-45$[/tex]. First, calculate the discriminant:
[tex]$$
\Delta = b^2 - 4ac = 4^2 - 4(1)(-45) = 16 + 180 = 196.
$$[/tex]
Since [tex]$\sqrt{196}=14$[/tex], the formula gives:
[tex]$$
p = \frac{-4 \pm 14}{2}.
$$[/tex]
This leads to two solutions:
1. For the positive sign:
[tex]$$
p = \frac{-4 + 14}{2} = \frac{10}{2} = 5.
$$[/tex]
2. For the negative sign:
[tex]$$
p = \frac{-4 - 14}{2} = \frac{-18}{2} = -9.
$$[/tex]
Recall that [tex]$p = x^2$[/tex]. Now, consider each case:
1. For [tex]$p = 5$[/tex]:
[tex]$$
x^2 = 5 \quad \Longrightarrow \quad x = \sqrt{5} \quad \text{or} \quad x = -\sqrt{5}.
$$[/tex]
2. For [tex]$p = -9$[/tex]:
[tex]$$
x^2 = -9.
$$[/tex]
Since a square can never be negative for a real number, there are no real solutions in this case. (The solutions would be complex: [tex]$x=3i$[/tex] and [tex]$x=-3i$[/tex], but we are only considering real zeros.)
Thus, the real zeros of the function are:
[tex]$$
x = \sqrt{5} \quad \text{and} \quad x = -\sqrt{5}.
$$[/tex]
