Question

How do you solve $ {{x}^{4}}+{{x}^{2}}=1 $ ?

Answer 1

Hint: In this question, we are given a four-degree equation of x and we need to solve it which means we need to find the four values of x which satisfy this equation. For this, we will suppose $ {{x}^{2}} $ as y and form a quadratic equation in terms of y. Then we will use the quadratic formula on the equation of y to find the value pf y. Then we will substitute y for $ {{x}^{2}} $ again and take square root on both sides to find the required roots. For an equation of the form $ a{{x}^{2}}+bx+c $ quadratic formula is given as $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ .

Complete step by step answer:
Here we have the equation as $ {{x}^{4}}+{{x}^{2}}=1 $ . We need to find the value of x which satisfy this equation. As the equation is of degree 4 so we will find four values of x. To simplify, let us substitute $ {{x}^{2}} $ as y. Squaring both sides we get $ {{x}^{4}}={{y}^{2}} $ . So our equation becomes $ {{y}^{2}}+y=1\Rightarrow {{y}^{2}}+y-1=0 $ .
Now let us solve this quadratic equations to find the value of y. Let us use quadratic formula for that. We know that, for an equation of the form $ a{{x}^{2}}+bx+c $ quadratic formula gives the value of x as $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ .
Comparing $ a{{x}^{2}}+bx+c $ with $ {{y}^{2}}+y-1 $ we get a = 1, b = 1 and c = -1.
So our quadratic formula for y becomes,
 $ \begin{align}
  & y=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( -1 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow y=\dfrac{-1\pm \sqrt{1+4}}{2} \\
 & \Rightarrow y=\dfrac{-1\pm \sqrt{5}}{2} \\
\end{align} $
So we have two values of y as $ y=\dfrac{-1-\sqrt{5}}{2}\text{ and }y=\dfrac{-1+\sqrt{5}}{2} $ .
As we had supposed $ {{x}^{2}} $ as y, so these are the values of $ {{x}^{2}} $ we get,
 $ {{x}^{2}}=\dfrac{-1-\sqrt{5}}{2}\text{ and }{{x}^{2}}=\dfrac{-1+\sqrt{5}}{2} $ .
Taking $ {{x}^{2}}=\dfrac{-1-\sqrt{5}}{2} $ .
Taking square root on both sides we get $ x=\pm \sqrt{\dfrac{-1-\sqrt{5}}{2}} $ .
Taking negative common inside the square root we get $ x=\pm \sqrt{-\left( \dfrac{1+\sqrt{5}}{2} \right)} $ .
We know that $ \sqrt{xy}=\sqrt{x}\sqrt{y} $ so we have $ x=\pm \sqrt{-1}\sqrt{\left( \dfrac{1+\sqrt{5}}{2} \right)} $ .
 $ \sqrt{-1} $ can be written as i so, $ x=\pm \sqrt{\left( \dfrac{1+\sqrt{5}}{2} \right)}i $ .
Taking $ {{x}^{2}}=\dfrac{-1+\sqrt{5}}{2} $ .
Taking square root on both sides we get $ x=\pm \sqrt{\dfrac{-1+\sqrt{5}}{2}} $ .
So we have obtained four values of x which are,
 $ x=\sqrt{\left( \dfrac{1+\sqrt{5}}{2} \right)}i,x=-\sqrt{\left( \dfrac{1+\sqrt{5}}{2} \right)}i $ .
 $ x=-\sqrt{\dfrac{-1+\sqrt{5}}{2}},x=-\sqrt{\dfrac{-1+\sqrt{5}}{2}} $ .

Note:
Students should always take care of the signs while solving this sum. Note that we can substitute $ {{x}^{2}} $ as y if the whole equation could change into variable y. Do not forget the square root sign for 5 inside the whole square root sign. Due to $ \pm $ signs, we have four values of x. Out of which two are imaginary roots due to the presence of i.