# Algebraically find all the real solutions for the polynomial?

2x^6+28=15x^3

### 6 Answers

- PhilipLv 63 weeks ago
2x^6 +28 = 15x^3...(1).; Put x^3 = u. Then (1) ---> 2u^2+28 = 15u, ie., 2u^2-15u+28= 0, ie.,(2u-7)(u-4) = 0. Then u = (7/2) or 4 and x = (7/2)^(1/3) or 4^(1/3) = 1.518294486 or 1.587401052.;

Put a = u^(1/3). Then a^3 = [u^(1/3)]^3 = u = x^3 and x^3-a^3 = 0...(2).;

For u = (7/2), a = (7/2)^(1/3) = 1.518294486.. For u = 4, a = 2^(1/3) = 1.587401052;

Each a-value occurs in a form (2) eqn equivalent to (x-a)(x^2+ax+a^2) = 0...(3).

We already got 2 x-roots of eqn (1). 2 x-roots are located in each quadratic eqn (3).

x^2+ax+a^2 = 0 ---> 2x = -a(+/-)D, where D^2 = a^2 -4a^2 = -3a^2 = (ia*rt3)^2.;

Then x = (1/2)[-a(+/-)a*i*rt3] = (a/2)[-1(+/-)i*rt3].;

For a = (7/2)^(1/3), we have x = (1/2)(1.528294486)[-1(+/-)i*rt3].;

For a.=.......2^(1/3), we have x = (1/2)(1.587401052)[-1(+/-)i*rt3].

I got a little too excited and decided to solve for all 6 roots of (1). Some of you may

appreciate my enthusiasm, others may groan. Oh well, that's life.

- ?Lv 73 weeks ago
2x⁶ + 28 = 15x³

2x⁶ - 15x³+ 28 = 0.........[1]

Let u = x³ then [1] can be rewritten as

2u² - 15u + 28 = 0

// Solve for u using the quadratic formula

.......- (-15) ± √((-15)² - 4(2)(28))

u = ------------------------------------

................2(2)

.................15 ± √1

..........u = -----------

......................4

..........u = ⁷⁄₂, 4

// Now, back substitute for u = x³

x³ = ⁷⁄₂, 4

x = ∛⁷⁄₂ or x = ∛4

x = 1.518, or 1.587.......................ANS

- KrishnamurthyLv 73 weeks ago
2x^6 + 28 = 15x^3

2x^6 - 15x^3 + 28 = 0

(x^3 - 4) (2 x^3 - 7) = 0

Real solutions:

x = 2^(2/3)

x = (7/2)^(1/3)

- husoskiLv 73 weeks ago
Rewrite that as:

2x^6 - 15x^3 + 28 = 0

Then treat (x^3) as a variable:

2(x^3)^2 - 15(x^3) + 28 = 0

Factor by grouping, notiting that 2*28 = 56 = (-7)(-8) and (-7) + (-8) = -15:

2(x^3)^2 - 7(x^3) - 8(x^3) + 28 = 0

[2(x^3) - 7](x^3) - (4)[2(x^3) - 7] = 0

[2(x^3) - 7] [x^3 - 4] = 0

That's true when either 2x^3 = 7 or x^3 = 4.

The only real values are cuberoot(7/2) and cuberoot(4).

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- 3 weeks ago
2x^6 - 15x^3 + 28 = 0

x^3 = (15 +/- sqrt(225 - 4 * 2 * 28)) / 4

x^3 = (15 +/- sqrt(225 - 224)) / 4

x^3 = (15 +/- sqrt(1)) / 4

x^3 = (15 +/- 1) / 4

x^3 = 16/4 , 14/4

x^3 = 4 , 3.5

x = 4^(1/3) , 3.5^(1/3)