Let y = 1− t2 + 2t

The graph of y(t):

solve_1-t2plus2t__4.png

There appear to be 2 roots, one near t = -1 and the other near t = 3. However, if we look more carefully in the region near t = 3 (by zooming in), we see that there is more than one root there.

solve_1-t2plus2t__6.png

By simple substitution, we can see that one of the roots is exactly t = 3:

1− (3)2 + 23 = 0

Now for the root near t = 3.4.

We will use Newton's Method to approximate the root. We need to differentiate y = 1− t2 + 2t. Since we have t as an exponent in our expression, we need to use logarithms when differentiating. [See how to differentiate logarithms in Derivative of the Logarithmic Function].

Let's differentiate 2t by itself first.

Let h = 2t.

Take natural log of both sides:

$\ln h=t\ln 2$

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So

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So for Newton's Method in this example, we would have:

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Initial guess: t1 = 3.4

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We can write this showing the substitution as:

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Now, doing another step of Newton's Method:

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And doing another step:

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And another:

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We can conclude that correct to 7 decimal places, t = 3.4074505.

Using Graphs Instead

Using Scientific Notebook, we can zoom into the root and we can see (from where the graph cuts the y-axis) that t is indeed close to 3.40745.

solve_1-t2plus2t__25.png

Now for the negative case. Let t1 = -1 be our initial guess.

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We could continue until we obtained the required accuracy.

Comparing this to the zoomed in graph, we can see that the solution is t = -1.198250197, correct to 9 decimal places.

solve_1-t2plus2t__30.png

Conclusion

So the solutions for 1− t2 + 2t = 0 are

t = -1.19825,

t = 3, or

t = 3.40745,

correct to 5 decimal places.