Math quesion, please help me!?

Question

the problem is 2x^7/2+8x^5/2=24x^3/2 the directions says solve by factoring and the hint for the problem is "first factor out a fractional root" i am not sure how to do that so please can someone help me!

Answer 1

solution...

2x^7/2 + 8x^5/2 - 24x^3/2 = 0....transfer the 24x^3/2 to the left.

2(x^7/2 + 4x^5/2 - 12x^3/2) = 0...factor the common between the three values which is 2...

2(x^7 + 4x^5 - 12x^3)^1/2 = 0...the factor inside the parenthesis is 1/2...

Answer 2

(1) I assume your expression (more) accurately reads:

2x^(7/2) + 8x^(5/2) = 24x^(3/2)

(2) If so, a factor common to all three terms is:

x^(3/2),

so factoring that out and equating the expression to zero gives:

x^(3/2)*[2x^2 + 8x - 24] = 0

(3) One of the solutions is x = 0

(4) The other (non-zero) solutions are obtained as follows:

(5) Divide by x^(3/2) and 2, obtaining

x^2 + 4x - 12 = 0,

which is factored to read: (x + 6)(x - 2) = 0

(6) Hence, the solutions are: x = 0, x = - 6 and x = + 2.

Answer 3

2x^7/2+8x^5/2=24x^3/2

2x^7/2 + 8x^5/2 - 24x^3/2 = 0

What the hint is saying is that square root of x is in all 3 terms and can be factored out.

√x (2x^7 + 8x^5 - 24x^3) = 0

2x³ can also be factored out

x³√x (x^4 + 4x^2 - 12) = 0

now you can factor

x³√x (x² + 6)(x² -  2) = 0

x³√x (x² + 6)(x + 2)(x - 2) = 0

solutions are x = 0, +2, -2, +6i, -6i

.

Answer 4

2x^7/2 + 8x^5/2 = 24x^3/2

2x^4 * x^(3/2) + 8x^2 * x^(3/2) = 24 * x^(3/2)

2x^4 * x^(3/2) + 8x^2 * x^(3/2) - 24 * x^(3/2) = 0

x^(3/2) (2x^4 + 8x^2 - 24) = 0

x^(3/2) (2x^2 + 12)(x^2 - 2) = 0

Answer 5

Subtract 24x^3/2 from both sides, then factor out what they have in common : 2 x^(3/2)

2x^(3/2) [ x^ (4/2) + 4x ^ (2/2) - 12 ] = 0 which simplifies to

2x ^ (3/2) [ x^2 + 4x - 12 ] = 0

(2x^(3/2)) (x + 6) (x - 2) = 0....use zero product property

2x^(3/2) = 0 OR x + 6 = 0 OR x - 2 = 0

Solve each equation for x

x = 0 or x = - 6 or x = 2

This was a tough problem !

Good luck to you !