Question

If $x + \sqrt[4]{5-x^4} = 2$, then find $x\sqrt[4]{5-x^4} = ?$ (x is a real Number)

• A. $\frac{8 + \sqrt{42}}{2}$
• B. $\frac{8 - \sqrt{42}}{2}$
• C. $\frac{8 + \sqrt{168}}{4}$
• D. $\frac{8 - \sqrt{168}}{4}$

Answer 1

Answer: (B)

$\frac{8 - \sqrt{42}}{2}$

Answer 2

Let the given equation be $x + \sqrt[4]{5-x^4} = 2$. Let $y = \sqrt[4]{5-x^4}$. From this definition, we must have $y \ge 0$ and $5-x^4 \ge 0$. The equation becomes: $x + y = 2$

Also, raising $y = \sqrt[4]{5-x^4}$ to the fourth power gives: $y^4 = 5-x^4$ $x^4 + y^4 = 5$

We have a system of two equations with $x$ and $y$:

1. $x + y = 2$
2. $x^4 + y^4 = 5$

We want to find the value of $x\sqrt[4]{5-x^4}$, which is $xy$. We can express $x^4+y^4$ in terms of $x+y$ and $xy$. First, $x^2 + y^2 = (x+y)^2 - 2xy = (2)^2 - 2xy = 4 - 2xy$. Then, $x^4 + y^4 = (x^2+y^2)^2 - 2(xy)^2 = (4-2xy)^2 - 2(xy)^2$.

Substitute this into equation (2): $5 = (4-2xy)^2 - 2(xy)^2$ $5 = (16 - 16xy + 4(xy)^2) - 2(xy)^2$ $5 = 16 - 16xy + 2(xy)^2$

Rearranging this into a quadratic equation for $xy$: $2(xy)^2 - 16xy + 16 - 5 = 0$ $2(xy)^2 - 16xy + 11 = 0$

Let $P = xy$. The quadratic equation is $2P^2 - 16P + 11 = 0$. Using the quadratic formula to solve for $P$: $P = \frac{-(-16) \pm \sqrt{(-16)^2 - 4(2)(11)}}{2(2)}$ $P = \frac{16 \pm \sqrt{256 - 88}}{4}$ $P = \frac{16 \pm \sqrt{168}}{4}$ $P = \frac{16 \pm \sqrt{4 \times 42}}{4}$ $P = \frac{16 \pm 2\sqrt{42}}{4}$ $P = \frac{8 \pm \sqrt{42}}{2}$

So, $xy$ can be $\frac{8 + \sqrt{42}}{2}$ or $\frac{8 - \sqrt{42}}{2}$.

Now we must consider the constraints that $x$ and $y$ must be real numbers, and $y \ge 0$. $x$ and $y$ are the roots of the quadratic equation $t^2 - (x+y)t + xy = 0$, which is $t^2 - 2t + P = 0$. For $x$ and $y$ to be real, the discriminant of this quadratic must be non-negative: $D = (-2)^2 - 4(1)(P) = 4 - 4P \ge 0$ $4 \ge 4P \implies P \le 1$.

Let's check the two possible values for $P$:

1. $P_1 = \frac{8 + \sqrt{42}}{2}$. Since $\sqrt{42} > 0$, $8+\sqrt{42} > 8$, so $P_1 > \frac{8}{2} = 4$. This value is not less than or equal to 1, so it does not yield real solutions for $x$ and $y$.

2. $P_2 = \frac{8 - \sqrt{42}}{2}$. Let's check if $P_2 \le 1$: $\frac{8 - \sqrt{42}}{2} \le 1$ $8 - \sqrt{42} \le 2$ $6 \le \sqrt{42}$ Squaring both sides (which are positive): $36 \le 42$. This is true. So, $P = \frac{8 - \sqrt{42}}{2}$ is the only possible value for $xy$ that yields real solutions for $x$ and $y$.

The roots of $t^2 - 2t + P = 0$ are $t = \frac{2 \pm \sqrt{4-4P}}{2} = 1 \pm \sqrt{1-P}$. We need to ensure $y \ge 0$. The roots are $1 + \sqrt{1-P}$ and $1 - \sqrt{1-P}$. Since $P \le 1$, $1-P \ge 0$, so $\sqrt{1-P}$ is real. Also, $P = xy = \frac{8-\sqrt{42}}{2} \approx \frac{8-6.48}{2} \approx 0.76$. So $1-P \approx 1-0.76 = 0.24$, and $\sqrt{1-P} \approx \sqrt{0.24} < 1$. Thus, both roots $1 + \sqrt{1-P}$ and $1 - \sqrt{1-P}$ are positive. Therefore, the condition $y \ge 0$ is satisfied for either assignment of $x$ and $y$.

The value of $x\sqrt[4]{5-x^4} = xy$ is $P = \frac{8 - \sqrt{42}}{2}$.