Question

How do you find the solutions to the quadratic equation ${{x}^{4}}-5{{x}^{2}}+4=0$?

Answer 1

This is the question of algebraic expression as it consists of variables, coefficients, constants, and mathematical operations such as addition, subtraction, multiplication and division. In the given question of an expression, you just need to simplify the expression by taking the common factor using mathematical operations and evaluate further. Later we substitute such that we will get quadratic formula provides the solution for the quadratic equation:
$a{{x}^{2}}+bx+c=0$. In which a, b and c are the coefficient of respectively terms in the quadratic equation, as follows: Roots of the quadratic equation= $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Formula used:
The quadratic formula provides the solution for the quadratic equation:
$a{{x}^{2}}+bx+c=0$
In which a, b and c are the coefficient of respectively terms in the quadratic equation, as follows:
Roots of the quadratic equation= $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

Complete step by step solution:
We have given that,
${{x}^{4}}-5{{x}^{2}}+4=0$
Taking out ${{x}^{2}}$as a common factor, we get
${{x}^{2}}\left( {{x}^{2}}-5 \right)+4=0$
Substituting $y={{x}^{2}}$ in the above equation, we get
$y\left( y-5 \right)+4=0$
Simplifying further we get
${{y}^{2}}-5y+4=0$
Splitting the middle term, we obtained
${{y}^{2}}-4y-y+4=0$
Taking out common factor by making pairs, we get
$y\left( y-4 \right)-1\left( y-4 \right)=0$
$\left( y-1 \right)\left( y-4 \right)=0$
Equating each common factor equal to 0, we get
$y-1=0\ \ and\ \ y-4=0$
Solving for the value of ‘y’, we get
$y=1,4$
Undo the substitution i.e. $y={{x}^{2}}$,
${{x}^{2}}=1\ and\ {{x}^{2}}=4$
Now solving
${{x}^{2}}=1\$
$x=\sqrt{1\ }=\pm 1$
Now solving
${{x}^{2}}=4$
$x=\sqrt{4\ }=\pm 2$
$x=\pm 2$

Therefore,
The possible value of $x$ is $\pm 1\ and\ \pm 2$.

Note: To solve or evaluation these types of expression, we need to know about the:
Solving quadratic equations using the formula
Simplifying radicals
Find prime factors
The general form of quadratic equation is $a{{x}^{2}}+bx+c=0$, where a b and c are the numerical coefficients or constants, and the value of $x$ is unknown one fundamental rule is that the value of a, the first constant can never be zero.