What is the sum of roots of the equation (x^2 - 3x)^2 - (3x^2 - 9x) = 4?

Steps to Solve

1. Rewrite the equation Rewrite the equation to isolate everything on one side. The original equation is $$(x^2 - 3x)^2 - (3x^2 - 9x) = 4.$$ Rearranging gives $$(x^2 - 3x)^2 - (3x^2 - 9x) - 4 = 0.$$

2. Simplify the equation Next, substitute ( y = x^2 - 3x ) into the equation. This changes the left side to: $$y^2 - (3(x^2 - 3x)) - 4 = 0 \rightarrow y^2 - (3y) - 4 = 0.$$

3. Apply the quadratic formula Now, apply the quadratic formula to solve for ( y ): $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where ( a = 1, b = -3, c = -4 ). So, $$y = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)}.$$

4. Calculate the discriminant Calculate the discriminant: $$(-3)^2 - 4(1)(-4) = 9 + 16 = 25.$$ So, $$y = \frac{3 \pm 5}{2}.$$

5. Find the values of ( y ) Now, solve for ( y ):

• First root: ( y = \frac{3 + 5}{2} = 4 )
• Second root: ( y = \frac{3 - 5}{2} = -1 )

6. Revert to ( x ) and find the roots Now revert to the original variable ( x ):

• From ( y = x^2 - 3x = 4 ): $$x^2 - 3x - 4 = 0$$ Using the quadratic formula, we get: $$x = \frac{3 \pm \sqrt{(3)^2 - 4(1)(-4)}}{2(1)}$$ Calculate the roots and sum.
• From ( y = x^2 - 3x = -1 ): $$x^2 - 3x + 1 = 0$$ Using the quadratic formula, we get the roots and sum.

7. Sum the Roots The sum of roots for a quadratic ( ax^2 + bx + c ) is given by ( -\frac{b}{a} ). For ( y = 4 ): The sum is ( 3 ).

For ( y = -1 ): The sum is ( 3 ) again.

So, total sum = ( 3 + 3 = 6 ).