# Math Help - Polynomial Equations help

1. ## Polynomial Equations help

Find the sum of the solutions of the equation

(3x - 8/x)^2 - 7(3x - 8/x)^2 +10 = 0

2. Hi there sri340

Notice there are some like terms here

$\left( 3x-\frac{8}{x}\right)^2-7\left( 3x-\frac{8}{x}\right)^2+10 = 0$

$-6\left( 3x-\frac{8}{x}\right)^2+10 = 0$

$6\left( 3x-\frac{8}{x}\right)^2 = 10$

Now expand the brackets and apply the quadratic fomula to solve for $x$

3. Hello, sri340!

I suspect there is a typo . . .

Find the sum of the solutions of the equation: . $\left(3x - \frac{8}{x}\right)^2 - 7\left(3x - \frac{8}{x}\right) +10 \:=\: 0$

Expand: . $9x^2 - 48 + \frac{64}{x^2} - 21x + \frac{56}{x} + 10 \:=\:0 \quad\Rightarrow\quad 9x^2 -21x - 38 + \frac{56}{x} + \frac{64}{x^2} \:=\:0$

Multiply by $x^2\!:\quad 9x^4 - 21x^3 - 38x^2 + 56x + 64 \:=\:0$

Divide by 9: . $x^4 - \frac{7}{3}x^3 - \frac{38}{9}x^2 + \frac{56}{9}x + \frac{64}{9} \:=\:0$

The sum of the roots is the negative of the $x^3$ coefficient: . $\frac{7}{3}$

4. Allow your equation to read a^2-7a+10=0.
Then (a-5)(a-2)=0.
a=5 or 2.

"a" = 3x-8/x, so 5x=3x^2-8 and 2x=3x^2-8.
Hence, 3x^2-5x-8=0 and 3x^2-2x-8=0
have solutions (3x-8)(x+1)=0 and (3x+4)(x-2)=0.

x= 8/3, -1, -4/3 and 2.

These sum to 7/3.