The graphs of $y=x^4$ and $y=5x^2-6+4x^2$ intersect at four points with $x$-coordinates $\pm \sqrt{m}$ and $\pm \sqrt{n}$, where $m > n$. What is $m-n$?

 Sep 1, 2023

To find the $x$-coordinates of the points of intersection between the graphs of $y=x^4$ and $y=5x^2-6+4x^2$, we need to set the two equations equal to each other and solve for $x$. 

Setting $x^4 = 5x^2-6+4x^2$, we can combine like terms to obtain $x^4 - 9x^2 + 6 = 0$. This is a quadratic equation in terms of $x^2$, so we can use the quadratic formula to solve for $x^2$:

$$x^2 = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(6)}}{2(1)}$$

Simplifying further, we have:

$$x^2 = \frac{9 \pm \sqrt{81 - 24}}{2}$$
$$x^2 = \frac{9 \pm \sqrt{57}}{2}$$

Since we are only interested in real solutions, we can discard the negative square root. Thus, we have:

$$x^2 = \frac{9 + \sqrt{57}}{2}$$

To find the values of $m$ and $n$, we need to determine which value is greater. Let's denote $\sqrt{m}$ as the larger solution and $\sqrt{n}$ as the smaller solution. Therefore, we have:

$$\sqrt{m} = \sqrt{\frac{9 + \sqrt{57}}{2}}$$
$$\sqrt{n} = \sqrt{\frac{9 - \sqrt{57}}{2}}$$

To simplify these expressions, we can rationalize the denominators by multiplying both the numerator and denominator by the conjugate of the denominator. This gives us:

$$\sqrt{m} = \sqrt{\frac{9 + \sqrt{57}}{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$$
$$\sqrt{n} = \sqrt{\frac{9 - \sqrt{57}}{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$$

Simplifying further, we have:

$$\sqrt{m} = \frac{\sqrt{(9 + \sqrt{57}) \cdot 2}}{\sqrt{2}}$$
$$\sqrt{n} = \frac{\sqrt{(9 - \sqrt{57}) \cdot 2}}{\sqrt{2}}$$

Now, let's simplify the expressions under the square roots:

$$\sqrt{(9 + \sqrt{57}) \cdot 2} = \sqrt{18 + 2\sqrt{57}}$$
$$\sqrt{(9 - \sqrt{57}) \cdot 2} = \sqrt{18 - 2\sqrt{57}}$$

Therefore, we have:

$$\sqrt{m} = \frac{\sqrt{18 + 2\sqrt{57}}}{\sqrt{2}}$$
$$\sqrt{n} = \frac{\sqrt{18 - 2\sqrt{57}}}{\sqrt{2}}$$

To determine the value of $m-n$, we need to subtract $\sqrt{n}$ from $\sqrt{m}$:

$$m-n = \left(\frac{\sqrt{18 + 2\sqrt{57}}}{\sqrt{2}}\right)^2 - \left(\frac{\sqrt{18 - 2\sqrt{57}}}{\sqrt{2}}\right)^2$$

Expanding and simplifying, we have:

$$m-n = \frac{(18 + 2\sqrt{57})}{2} - \frac{(18 - 2\sqrt{57})}{2}$$
$$m-n = \frac{2\sqrt{57}}{2}$$
$$m-n = \sqrt{57}$$

Therefore, the value of $m-n$ is $\sqrt{57}$.


 Sep 2, 2023
edited by Guest  Sep 2, 2023

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