Question

Find all real number solutions
x^4/3 - 5x^2/3 + 6 = 0

Answer 1

subs X^2=Y
find solution in terms of Y ... then solve Y=x^2 with each value of Y to get a value of x

Answer 2

this is called a disguised quadratic equation

Answer 3

I think @Hashir meant substitute:\[y=x^{\frac{2}{3}}\]into your equation:\[x^{\frac{4}{3}}-5x^{\frac{2}{3}}+6=0\]to get:\[y^2-5y+6=0\]then solve this quadratic equation in y. finally find the value of x from:\[x=y^{\frac{3}{2}}\]

Answer 4

yup :D

Answer 5

so if we factor and get \[y = -2 or -3\] you just run both solutions through the original?

Answer 6

can you please show me how you factored this?

Answer 7

Sorry, now that I look at it, it should be just 2 or 3
\[(y-2)(y-3) = y^2 - 5y +6\]

Answer 8

that is the correct factorisation, but the answers you stated are not correct.

Answer 9

Yes, it should be 2 or 3, not -2 or -3

Answer 10

thats it - well done!

Answer 11

now just use:\[x=y^{\frac{3}{2}}\]to find the values for x.

Answer 12

why does \[x = y ^{3/2}\]?

Answer 13

remember we started by substituting:\[y=x^{\frac{2}{3}}\]

Answer 14

so if we cube both sides, we get:\[y^3=x^2\]

Answer 15

then take square roots of both sides to get:\[y^{\frac{3}{2}}=x\]

Answer 16

understand?

Answer 17

here's another way of looking at it...

Answer 18

\[y=x^{\frac{2}{3}}\]therefore:\[y^{\frac{3}{2}}=(x^{\frac{2}{3}})^{\frac{3}{2}}=x^{\frac{2}{3}\times\frac{3}{2}}=x\]

Answer 19

I think I see. So I should put \[2^{3/2} and 3^{3/2}\] the first or second equation?

Answer 20

you found y=2 or y=3 are two solutions to the equation involving y.
so, if we take the first solution (i.e. y=2), we get:\[x=y^{\frac{3}{2}}=\pm\sqrt{y^3}=\pm\sqrt{2^3}=\pm2\sqrt{2}\]now do the same process for y=3.

Answer 21

\[\sqrt{y ^{3}}=\pm \sqrt{3^{3}} =\pm3\sqrt{3}\]

Answer 22

correct - so you end up with four solutions for x.

Answer 23

Wow, thank you

Answer 24

yw :)