What substitution should be used to rewrite 16(x^3+1)^2-22(x^3+1)-3=0 as a quadratic equation?
2 answers:
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Answer:
Step-by-step explanation:
THe interpretation of the given question is as follows:
y'' + ky + ry³ = A cos ωt
Let k = 4, r = 3, A = 7 and ω = 8
The objective is to find the first three non zero terms in the Taylor polynomial approximation to the solution with initial values y(0) = 0 ; y' (0) = 1
SO;
y'' + ky " ry³ = A cos ωt
where;
k = 4, r = 3, A = 7 and ω = 8
y(0) = 0 ; y' (0) = 1
y'' + 4y + 3y³ = 7 cos 8t
y'' = - 4y - 3y³ + 7 cos 8t ---- (1)
∴
y'' (0) = -4y(0) - 3y³(0) + 7 cos (0)
y'' (0) = - 4 × 0 - 3 × 0 + 7
y'' (0) = 7
Differentiating equation (1) with respect to t ; we have:
y''' = - 4y' - 9y² × y¹ - 56 sin 8t
y''' (0) = -4y'(0) - 9y²(0)× y¹ (0) - 56 sin (0)
y''' (0) = - 4 × 1 - 9 × 0 × 1 - 56 × 0
y''' (0) = - 4
Thus; we have :
y(0) = 0 ; y'(0) = 1 ; y'' (0) = 7 ; y'''(0) = -4
Therefore; the Taylor polynomial approximation to the first three nonzero terms is :
Answer:
Step-by-step explanation:
As the equation in point-slope form
Here:
- m is the slope and
-
is a point on the line
using and
then
Thus
comparing the equation with the slope-intercept form
so,
Therefore,
If a quadratic equation has solutions, the standard procedure will always work: given an equation like
The solutions (if any) are given by
In your case, you have
So the formula becomes
Anyway, this is quite a special case, because you're missing the linear term (since )
This means that you can solve equations like these more easily: rearrange the equation to make it look like . In your case, it becomes
Divide both sides by 9:
Now, you know that the square of a number equals 1/9. By definition, it means that this number is the square root of 1/9. Nevertheless, both the square root and its opposite are solutions of the equation, because the minus sign will cancel out when squaring.
So, in general, you have
which of course makes sense, if you're using real numbers, only if c/a>0.
In your case, this becomes