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\(x^{4}+x-3=0\)

Short Answer

Expert verified
The roots of the equation \(x^{4}+x-3=0\) are \(x= \sqrt{1.5}\) and \(x= -\sqrt{1.5}\).

Step by step solution

01

Identify the quartic equation

The quartic equation provided in the exercise is \(x^{4}+x-3=0\). It can be noticed that this equation is missing \(x^3\), \(x^2\), and quadratic term.
02

Introduce substitution

Since the quartic equation is missing several terms, one trick that can be used to simplify it is by introducing a variable substitution using \(u = x^2\).This transforms the equation into a quadratic form. Substituting \(u = x^2\) into the equation gives us \(u^2 + u - 3 = 0\).
03

Solve the quadratic equation

The equation \(u^2 +u-3 = 0\) is quadratic, and can be solved by using the quadratic formula \(u = \frac{-b \pm sqrt{b^2 - 4ac}}{2a}\). With \(a = 1\), \(b = 1\), and \(c = -3\), we are able to find that \(u = -2\) and \(u = 1.5\).
04

Find values for x

Solving the equation \(u = x^2\) for each value of \(u\) is the next step. When \(u = -2\), we find the equation has no solutions since there's no real square root of a negative number. When \(u = 1.5\), we get \(x = \pm \sqrt{1.5}\), which are the real roots of the equation.

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Most popular questions from this chapter

Multiple Choice Which of the following conditions would enable you to conclude that the graph of \(f\) has a point of inflection at \(x=c ?\) (A) There is a local maximum of \(f^{\prime}\) at \(x=c\) . (B) \(f^{\prime \prime}(c)=0 .\) (C) \(f^{\prime \prime}(c)\) does not exist. (D) The sign of \(f^{\prime}\) changes at \(x=c\) . (E) \(f\) is a cubic polynomial and \(c=0\)

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