I need help with Calculas question?

Question

Can some one show me how to solve this.

The equation .2t 3rdpower+.2t 2nd power-8t+1=.1t 2nd power+3t+2


has three solutions. find them(accurate to three decimal places.)

Answer 1

I assume you mean

.2t³ + .2t² - 8t + 1 = .1t² + 3t + 2
.2t³ + .1t² - 11t - 1 = 0

Multiply by ten to get rid of the decimals.

2t³ + t² - 110t - 10 = 0

This doesn't have rational solutions for x. Graph it and zoom in for the solutions.

The solutions are approximately

x = -7.6, -0.1, 7.2

If you zoom in on your graphing calculator you should be able to solve to the required accuracy.

Answer 2

Assuming that you mean:
0.2t^3 + 0.2t^2 -8t + 1 = 0.1t^2 + 3t+2.

The easiest way to do this is to generate numerical solutions with a computer-algebra system or a graphing calculator.

For the left-hand side, factor out a 0.2t from the first three terms and subtract 1 from both sides:
In addition, factor out a 0.1 from the right hand side:
(0.2t)*(t^2 + t - 40)= (0.1)*(t^2 + 30t +10)
2t*(t^2 + t - 40)=t^2 + 30t + 10
Expand the LHS again and set one side to zero
2t^3 + 2t^2 -80t = t^2 + 30t + 10
2t^3 + t^2 - 110t - 10=0

Plot this function on your CAS or calculator and locate the roots. There will be three of them since the equation is a cubic.

r1= -7.626100...
r2= -0.090847...
r3= 7.216948...

Answer 3

Are those decimal points meant to be there? It's a good idea to emphasise them by putting 0 in front:
0.2t^3 +0.2t^2 - 8t + 1 = 0.1t^2 + 3t + 2
In that case, bring all terms to the left side:
0.2t^3 + 0.1t^2 -11t -1 = 0

Using a spreadsheet shows where the solutions are:
f(-7.7) = -1.6776
f(-7.6) = 0.5808, so there is a root somewhere around
-7.63

Likewise we find there are roots between -0.1 and 0,
and between 7.2 and 7.3.

But what methods have you been taught to find the solution more accurately? Since you say it's a calculus question, I'm guessing you are expected to use the Newton-Raphson method, so for example take
x0 = -7.63 as a first approximation
and use the formula
x1 = x0 - (f(x0))/f'(x0)
where f'(x) = 0.6x^2 + 0.2x - 11

This gives
x1 = -7.63 - (-0.0873)/22.40414
.... = -7.6261

Use the same process on the other two values and you have the three solutions you're asked for.

If you haven't done Newton-Raphson, all I can suggest is "halving the interval":
After finding it's positive at -7.6 and negative at -7.7, take the value halfway between them, -7.65
It turns out to be negative here, so take the value at
-7.625,
It turns out to be positive, so take the value halfway between 7.625 and 7.65, i.e. 7.6375
In fact you wouldn't: You would notice that it's much closer to zero for 7.625 than for 7.65, and so you'd try a value closer to 7.625, maybe 7.628.
The value would be negative, so you'd try a value between it and 7.625, maybe 7.626, etc.

Any other questions about it? Email me!