Question

How do you solve \[{{x}^{4}}-3{{x}^{2}}+4=0\] using the quadratic formula?

Answer 1

Hint: Substitute \[{{x}^{2}}=X\] and convert the biquadratic equation into a quadratic equation in ‘X’. Now, consider the coefficient of \[{{X}^{2}}\] as ‘a’, coefficient of X as b and the constant term as c. Apply the discriminant formula given as \[X=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] to get the two values of X. Finally, consider the relation \[x=\pm \sqrt{X}\] to get four values of x.

Complete step by step answer:
Here, we have been provided with the biquadratic equation: - \[{{x}^{4}}-3{{x}^{2}}+4=0\] and we are asked to solve this equation. That means we have to find the values of x using the quadratic formula. But first we need to know about the quadratic formula.
Now, in elementary algebra, the quadratic formula is an expression that provides the solution of a quadratic equation. There are some other methods also by which we can solve a quadratic equation like factoring, graphing, completing the square method etc. But sometimes it is difficult to factor as we are not able to split the middle term, in such cases we use the quadratic formula.
Let us come to the question. We have the biquadratic equation: - \[{{x}^{4}}-3{{x}^{2}}+4=0\]. To convert this into a quadratic equation we need to substitute \[{{x}^{2}}=X\]. So, using this substitution, we get,
\[\Rightarrow {{x}^{4}}-3{{x}^{2}}+4=0\]
Considering the coefficient of \[{{X}^{2}}\] as ‘a’, coefficient of X as ‘b’ and the constant term as ‘c’, we have,
\[\Rightarrow \] a = 1, b = -3, c = 4
Using the quadratic formula given as: - \[X=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\], we get,
\[\begin{align}
  & \Rightarrow X=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-4\times 1\times 4}}{2\times 1} \\
 & \Rightarrow X=\dfrac{3\pm \sqrt{9-16}}{2} \\
 & \Rightarrow X=\dfrac{3\pm \sqrt{-7}}{2} \\
 & \Rightarrow X=\dfrac{3\pm \sqrt{7}i}{2} \\
\end{align}\]
Therefore, the two values of X are \[\dfrac{3+\sqrt{7}i}{2}\] and \[\dfrac{3-\sqrt{7}i}{2}\].
Now, we have to find the values of x for each value of X. So, considering the two values of X one – by – one, we get,
(i) When \[X=\left( \dfrac{3+\sqrt{7}i}{2} \right)\].
Since, \[{{x}^{2}}=X\]
\[\begin{align}
  & \Rightarrow x=\pm \sqrt{X} \\
 & \Rightarrow x=\pm \sqrt{\dfrac{3+\sqrt{7}i}{2}} \\
\end{align}\]
The above expression can be written as: -
\[\begin{align}
  & \Rightarrow x=\pm \sqrt{\dfrac{3+\sqrt{7}i}{2}\times \dfrac{2}{2}} \\
 & \Rightarrow x=\pm \dfrac{\sqrt{6+2\sqrt{7}i}}{2} \\
 & \Rightarrow x=\pm \dfrac{\sqrt{{{\left( \sqrt{7}+i \right)}^{2}}}}{2} \\
 & \Rightarrow x=\pm \left( \dfrac{\left( \sqrt{7}+i \right)}{2} \right) \\
\end{align}\]
(ii) When \[X=\left( \dfrac{3-\sqrt{7}i}{2} \right)\].
\[\begin{align}
  & \Rightarrow X={{x}^{2}} \\
 & \Rightarrow x=\pm \sqrt{X} \\
 & \Rightarrow x=\pm \sqrt{\dfrac{3-\sqrt{7}i}{2}} \\
\end{align}\]
The above expression can be written as: -
\[\begin{align}
  & \Rightarrow x=\pm \sqrt{\dfrac{3-\sqrt{7}i}{2}\times \dfrac{2}{2}} \\
 & \Rightarrow x=\pm \sqrt{\dfrac{6-2\sqrt{7}i}{2}} \\
 & \Rightarrow x=\pm \dfrac{\sqrt{{{\left( \sqrt{7}-i \right)}^{2}}}}{2} \\
\end{align}\]
\[\Rightarrow x=\pm \left( \dfrac{\sqrt{7}-i}{2} \right)\]

Hence, our solutions are \[x=\pm \left( \dfrac{\sqrt{7}+i}{2} \right)\] and \[x=\pm \left( \dfrac{\sqrt{7}-i}{2} \right)\].

Note: One may note that here we have obtained four values of x as our solution because the given equation was a polynomial of degree 4. Here, all the values of x were complex numbers. There were three cases possible for the values of x, they were: - all solutions real, two real and two complex solutions, all solutions complex. Here, we have the third case. Note that here the coefficients of \[{{x}^{3}}\] and x were both 0 and that is why \[{{x}^{2}}=X\] was used otherwise we would have required to use the hit – and – trial method for finding the two values of x. You must remember the quadratic formula, i.e., the discriminant formula, to solve the question.