How do I find the x-intercepts of -x^6+7x^4+3x^2, without using Desmos/computer software?
I$${f}{\left({\mathtt{x}}\right)} = {\mathtt{\,-\,}}{{\mathtt{x}}}^{{\mathtt{6}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{4}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}$$
+23254 First, factor out -x2, giving you -x2(x4 - 7x2 + 3)
So, one intercept occurs when -x2 = 0, so x = 0.
Now, it would be nice if (x4 - 7x2 + 3) factors, but it doesn't, so use the quadratic formula with the change that the quadratic formula gives you values for x2, not x:
x2 = [7 ± √( (-7)2 - 4·1·3)] / [2·1] ---> x2 = [7 ± √( 37)] / [2]
You will need to take the ± square roots of that answer.
Now, these x-values will be: + √( [7 + √( 37)] / 2 )
- √( [7 + √( 37)] / 2 )
+ √( [7 - √( 37)] / 2 )
- √( [7 - √( 37)] / 2 )
+23254 First, factor out -x2, giving you -x2(x4 - 7x2 + 3)
So, one intercept occurs when -x2 = 0, so x = 0.
Now, it would be nice if (x4 - 7x2 + 3) factors, but it doesn't, so use the quadratic formula with the change that the quadratic formula gives you values for x2, not x:
x2 = [7 ± √( (-7)2 - 4·1·3)] / [2·1] ---> x2 = [7 ± √( 37)] / [2]
You will need to take the ± square roots of that answer.
Now, these x-values will be: + √( [7 + √( 37)] / 2 )
- √( [7 + √( 37)] / 2 )
+ √( [7 - √( 37)] / 2 )
- √( [7 - √( 37)] / 2 )
