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Chapter 10: Problem 10

Show that the equation \(6 x^{4}-7 x+1=0\) does not have more than two distinct real roots.

Short Answer

Expert verified
The equation \(6x^4 - 7x + 1 = 0\) doesn't have more than two distinct real roots, as per Descartes' Rule of Signs.

Step by step solution

01

Understand Descartes' Rule of Signs

Descartes' Rule of Signs says the possible number of positive real roots of a polynomial is either equal to the number of sign changes between consecutive coefficients, or is less than that by an even number. Applied to negative roots, we can change the signs of the function and count the number of sign changes.
02

Count the Number of Positive Roots

Applying Descartes' rule to the function \(f(x) = 6x^4 - 7x + 1\), we count the number of sign changes in the coefficients. We start with a positive 6, change to negative 7, and then to positive 1, so there are 2 sign changes. Therefore, there could be 2 or 0 positive real roots.
03

Count the Number of Negative Roots

Next, we analyze the function \(f(-x)\), the function \(f(x)\) with each \(x\) replaced by \(-x\). This results in \(f(-x) = 6x^4 + 7x + 1\), where the coefficients have no sign changes. Therefore, there are no negative real roots.
04

Conclusion

Putting the results from steps 2 and 3 together, the polynomial can have either 2 or 0 positive real roots and no negative real roots. Thus, it is shown that the equation \(6x^4 - 7x + 1 = 0\) does not have more than two distinct real roots.

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Most popular questions from this chapter

Use a linear approximation of \(f(x)=\sqrt[3]{x}\) at \(x=8\) to approximate \(f(10)\).

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