Question 1050453
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35x^4+2x^2-1=0 
What is the value of x?
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<pre>
This equation is called <U>biquadratic</U> (the degrees of the unknown "x" are 4, 2 and zero).
The method of solving such equations is as follows:

Introduce new variable y = x^2.  Then your equation takes the form

35y^2 + 2y - 1 = 0.

It is just a quadratic equation, and you can apply the quadratic formula to solve it:

{{{y[1,2]}}} = {{{(-2 +- sqrt(2^2 - 4*35*(-1)))/(2*35)}}} = {{{(-2 +- sqrt(144))/70}}} = {{{(-2 +- 12)/70}}}.

{{{y[1]}}} = {{{10/70}}} = {{{1/7}}}.  {{{y[2]}}} = {{{-14/70}}} = {{{-1/5}}}.

Since y = x^2, we are interested in positive "y" only.

Thus we have x^2 = {{{1/7}}} and, hence, x = +/-{{{sqrt(1/7)}}} = +/-{{{1/sqrt(7)}}}.

<U>Answer</U>. The two real roots of the original equation are +/-{{{1/sqrt(7)}}}.
</pre>

See the lesson 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/Solving-polynomial-equations-of-high-degree-by-introducing-a-new-variable.lesson>Solving polynomial equations of high degree by introducing a new variable</A>

in this site.