# How do I find all intersection points of f(x)=ax^4 and g(x)=bx^2 where a,b > 0?

Dec 8, 2017

Solve $a {x}^{4} = b {x}^{2}$

#### Explanation:

$a {x}^{4} = b {x}^{2}$ if and only if

$a {x}^{4} - b {x}^{2} = 0$ so we need

${x}^{2} \left(a {x}^{2} - b\right) = 0$

Solutions $x = 0$ or $\pm \sqrt{\frac{b}{a}}$

Dec 8, 2017

$x = 0$ and $x = \pm \sqrt{\frac{b}{a}}$

#### Explanation:

To find intersection points, we just set the two functions equal to each other and solve the equation:

$a {x}^{4} = b {x}^{2}$

Firstly, we can quite quickly see that $0$ will always be a solution:
$a \cdot {0}^{4} = b \cdot {0}^{2}$

To find the other solutions, we first divide through by $a$:
$\frac{\cancel{a} {x}^{4}}{\cancel{a}} = \frac{b {x}^{2}}{a}$

Next we divide through by ${x}^{2}$:
${x}^{4} / {x}^{2} = \frac{b \cancel{{x}^{2}}}{a \cancel{{x}^{2}}}$

We can then take the square roots of both sides to get our solution:
$\sqrt{{x}^{2}} = \pm \sqrt{\frac{b}{a}}$

$x = \pm \sqrt{\frac{b}{a}}$

Since we know both $b$ and $a$ are positive, we can tell that the number inside the square root is always positive, so both solutions are valid.

I've also made this graph in which you can play around with $a$ and $b$ to see that the solutions always work for $a , b > 0$:
https://www.desmos.com/calculator/ziku0su5wn