solve an equation using solve() produces an array instead of a unique value

d = 32*h + 200*(32*h - 8)*((4*h - 1)^2 - 1) - 16

I am trying to solve for h using "solve(d,h)" , however, I keep getting an array for of solutions!

[1/2

1/8 - (7*2^(1/2))/80

(7*2^(1/2))/80 + 1/8 ]

How could I obtain a unique value for h? The solution does not seem right. Is there a better function to be used ?

Thanks.

Answer 1

[Star Strider]

The result is correct. There are three roots.

Plot it to see the result:

syms h 
d = 32*h + 200*(32*h - 8)*((4*h - 1)^2 - 1) - 16;

ds = solve(d)

figure
fplot(d, [-0.1 0.6])
hold on
plot(ds, zeros(size(ds)), 'pg')
hold off
grid

[Walter Roberson]

Yup. There is a term containing h multiplied by a term containing h^2, so you have a cubic polynomial. That is going to have 3 roots.

[Star Strider]

Folllowing up on Walter’s comment, a note added in proof:

d = collect(d,h)

produces:

d =
102400*h^3 - 76800*h^2 + 12832*h - 16

[jojo]

The problem I am running to is the following:

I initially define x(1) = 4, y(1) = 2.

And the h has been expected to be a single digit.

How can I define x(1), y(1) to account for both sinular numbers and tree dimentional arrays?

[Star Strider]

I do not understand.

The ‘x’ and ‘y’ values do not appear anywhere in anything you have posted.

[jojo]

I did not post them initially. I was just explaining that 'h' is to be assigned to some x and y parameters that take singular values. Is there a way of writing x and y such that they can take both single digits and arrays size of h? like x(:,3) something like that. I am knew to matlab and I do not fully grasp array slicing and indexing yet. Thanks.

[Star Strider]

I will need more details.

It depends on how you want to define ‘h’ and what you want to do.