Question

x^{10}-2x^5+1=0 Step 1. Identify the quadratic form Let t= Answer . We now have: t^2-2t+1=0 Step 2. Factor Factor this and solve for t to get t=Answer Step 3. Solve for x We have solved for t now we need to use this value for t to help us solve for x. Revisit step 1 to remind you of the relationship between t and x. Type your real solutions (no extraneous) from smallest to largest. x= Answer

Answer 1

Answer:

1) t = x⁵

2) t = 1

3) x = 1

Step-by-step explanation:

Given equation:

$x^{10}-2x^5+1=0$

To express the given equation in quadratic form, we can make a substitution to simplify it.

As we need the leading term to be raised to the power of 2 rather than 10, we can let t = x⁵, as:

$t^2 = (x^5)^2 = x^{10}$

Therefore, the equation becomes:

$t^2-2t+1=0$

To factor this, we can use the method of grouping.

Find two numbers that multiply to the product of the leading coefficient and the constant term, and sum to the coefficient of the middle term.

Since the leading coefficient and constant term are both 1, and the middle term is -2, the two numbers are -1 and -1 (since their product is 1 and their sum is -2).

Rewrite the middle term as the sum of these two numbers:

$t^2-t-t+1=0$

Group the terms:

$(t^2-t)+(-t+1)=0$

Factor out the greatest common factor from each group:

$t(t-1)-1(t-1)=0$

Factor out the common factor (t - 1):

$\begin{aligned}(t-1)(t-1)&=0\\\\(t-1)^2&=0\end{aligned}$

Now, solve for t:

$\begin{aligned}\sqrt{(t-1)^2}&=\sqrt{0}\\\\t-1&=0\\\\t&=1\end{aligned}$

To solve for x, substitute t = x⁵ back in:

$\begin{aligned}x^5&=1\\\\\sqrt[5]{x^5}&=\sqrt[5]{1}\\\\x&=1\end{aligned}$

Additional Information

Please see the attached graph for proof that x¹⁰ - 2x⁵ + 1 = 0 has only one real solution at x = 1.