how can i find x value from given y (6th degree polynomial)

y = [7 12 18 27 35 39];
p1 = 3.7739e-21
p2 = -6.5969e-17
p3 = 2.2715e-13
p4 = 1.1034e-09
p5 = -4.8986e-06
p6 = 0.010416
p7 = -0.70896
y = p1*x^6 + p2*x^5+p3*x^4+p4*x^3+p5*x^2+p6*x+p7;

Answer 1

[Matt J]

x=roots([p1,p2,p3,p4,p5,p6,p7-y])

[firrou bouteflika]

[Matt J]

Get rid of line 9. You don't need it.

[firrou bouteflika]

i'm not sure about those answers

[Matt J]

Please do not post screenshots of pure text. If you just copy/paste the text, it will consume much less disk space in the forum.

I don't know why you would doubt the output of roots, but you can plug it back into the original polynomial to verify that it is satisfied.

x=roots([p1,p2,p3,p4,p5,p6,p7-y])

polyval([p1,p2,p3,p4,p5,p6,p7],x)

[Steven Lord]

You will need to call roots once per element of y. You cannot find the root for a vector of y vectors all at once.

[Matt J]

p1 = 3.7739e-21;
p2 = -6.5969e-17;
p3 = 2.2715e-13;
p4 = 1.1034e-09;
p5 = -4.8986e-06;
p6 = 0.010416;
p7 = -0.70896;

y = [7 12 18 27 35 39];

x=nan(numel(y),6);
for i=numel(y):-1:1

    x(i,:)=roots([p1,p2,p3,p4,p5,p6,p7-y(i)]);
    
end

x

x = 
   1.0e+03 *

   9.3101 + 1.8774i   9.3101 - 1.8774i  -4.5565 + 0.0000i   1.1038 + 1.7008i   1.1038 - 1.7008i   1.2089 + 0.0000i
   9.2800 + 1.8478i   9.2800 - 1.8478i  -4.5894 + 0.0000i   2.1232 + 0.0000i   0.6932 + 1.8383i   0.6932 - 1.8383i
   9.2421 + 1.8102i   9.2421 - 1.8102i  -4.6272 + 0.0000i   2.7430 + 0.0000i   0.4401 + 2.0518i   0.4401 - 2.0518i
   9.1810 + 1.7489i   9.1810 - 1.7489i  -4.6807 + 0.0000i   3.3617 + 0.0000i   0.2187 + 2.3009i   0.2187 - 2.3009i
   9.1215 + 1.6884i   9.1215 - 1.6884i  -4.7256 + 0.0000i   3.7931 + 0.0000i   0.0849 + 2.4753i   0.0849 - 2.4753i
   9.0896 + 1.6555i   9.0896 - 1.6555i  -4.7472 + 0.0000i   3.9881 + 0.0000i   0.0301 + 2.5515i   0.0301 - 2.5515i

[firrou bouteflika]

i'm sorry can i just ask you another time what is this unswer, i don't understand it

x = 
   1.0e+03 *

   9.3101 + 1.8774i   9.3101 - 1.8774i  -4.5565 + 0.0000i   1.1038 + 1.7008i   1.1038 - 1.7008i   1.2089 + 0.0000i
   9.2800 + 1.8478i   9.2800 - 1.8478i  -4.5894 + 0.0000i   2.1232 + 0.0000i   0.6932 + 1.8383i   0.6932 - 1.8383i
   9.2421 + 1.8102i   9.2421 - 1.8102i  -4.6272 + 0.0000i   2.7430 + 0.0000i   0.4401 + 2.0518i   0.4401 - 2.0518i
   9.1810 + 1.7489i   9.1810 - 1.7489i  -4.6807 + 0.0000i   3.3617 + 0.0000i   0.2187 + 2.3009i   0.2187 - 2.3009i
   9.1215 + 1.6884i   9.1215 - 1.6884i  -4.7256 + 0.0000i   3.7931 + 0.0000i   0.0849 + 2.4753i   0.0849 - 2.4753i
   9.0896 + 1.6555i   9.0896 - 1.6555i  -4.7472 + 0.0000i   3.9881 + 0.0000i   0.0301 + 2.5515i   0.0301 - 2.5515i

because my answer should be just a one number like 2.49e+03

i'm sorry that i have to use another screenshot to explain it whit new y values

these are the y values [16.27 27.9 41.86 62.79 81.39 90.69]

[Walter Roberson]

format long g
p = [3.7769e-21, -6.6055e-17, 2.2806e-13, 1.0989e-09, -4.8887e-06, 0.010408, -0.71229];
R = roots(p)

R = 
           9350.21416809902 +      1915.74457506144i
           9350.21416809902 -      1915.74457506144i
           -4499.3555970064 +                     0i
           1608.69399831728 +      1978.77107818079i
           1608.69399831728 -      1978.77107818079i
           70.7499925489447 +                     0i

realroots = R(imag(R)==0)

realroots = 2×1

1.0e+00 * -4499.3555970064 70.7499925489447

syms x
poly = poly2sym(p)

poly = 

fplot([poly,0], [-4750 250])

It is a degree 6 polynomial in real coefficients. There are always going to be an even number of real roots: in this case there are two real roots. You can never get just one real root for a polynomial of even order that has real coefficients.

[firrou bouteflika]

i see thank you and i just realized i can use polyval

h = polyval(p,[16.27 27.9 41.86 62.79 81.39 90.69])