• anonymous
solve x^4-34x^2=-225
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • jamiebookeater
I got my questions answered at in under 10 minutes. Go to now for free help!
  • apoorvk
let x^2 be t then the eqn. becomes: t^2-34t+225=0 now solve the quadratic and get values of t, and eliminate nay negative values of t, since t is x^2, which cannot be negative.
  • anonymous
\[x ^{4}-34x ^{2}=-225\]First, let's set this equal to 0. \[x ^{4}-34x ^{2}+225=0\] From there, examine the terms and identify it as being quadratic-like because it has three terms that mimic a quadratic (as the power of x on the second term is half the power of x on the first term, and the third term has no x). Then, as others have said, we want to make a substitution: \[u=x ^{2}\] In order to sub it in for the first term, we can do a little re-write first. \[(x ^{2})^2-34x ^{2}+225=0\] Now you can sub it in: \[(u)^2-34(u)+225=0\] Now, factor like usual: \[(u-25)(u-9)=0\] And then use the Zero-Product Property: \[u-25=0\] \[u-9=0\]Solve for u u = 9 and u = 25 At this point, don't forget, you were solving for x to begin with, so now you have to substitute back. \[u=x ^{2}\]Therefore: \[x ^{2}=9\]and \[x ^{2}=25\] To solve for x, take the square root of both sides for each equation. \[\sqrt{x ^{2}}=\pm \sqrt{9}\]\[x=\pm3\] and \[\sqrt{x ^{2}}=\pm \sqrt{25}\]\[x=\pm5\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.