How do you find the roots, real and imaginary, of #y=x(x-1)-(3x-1)^2 # using the quadratic formula?

Question

How do you find the roots, real and imaginary, of #y=x(x-1)-(3x-1)^2      # using the quadratic formula?

Eli Assouline · Accepted Answer

Expand and simplify to find the equation in standard form, then apply the quadratic formula to obtain the roots: #x=(5+-isqrt(7))/16#.

The quadratic formula can be used with quadratic equations in standard form, #y=ax^2+bx+c#. First expand and simplify the given equation to rearrange it into standard form. #y=x(x-1)-(3x-1)^2# #y=x^2-x-(9x^2-6x+1)# #y=x^2-x-9x^2+6x-1# #y=-8x^2+5x-1# This equation is now in standard form, where #a=-8#, #b=5#, and #c=-1#. To solve for the roots of the equation, set #y=0#: #0 = -8x^2+5x-1#, Apply the quadratic formula: #x = (-b+-sqrt(b^2-4ac))/(2a)# #x = (-(5)+-sqrt((5)^2-4(-8)(-1)))/(2(-8))# #x = (-5+-sqrt(-7))/-16# #x=(5+-sqrt(-7))/16# #x=(5+-sqrt(-1)sqrt(7))/16# #x=(5+-isqrt(7))/16#

Jace Anderson · Answer

undefined To find the roots of $ y = x(x - 1) - (3x - 1)^2 $ using the quadratic formula, first, rewrite the equation in the standard form $ ax^2 + bx + c = 0 $. Then, identify the values of $ a $, $ b $, and $ c $. After that, apply the quadratic formula $ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $ to determine the roots.

Expanding the given equation yields $ y = x^2 - x - (9x^2 - 6x + 1) $. Simplifying further, we get $ y = -8x^2 + 5x - 1 $. Comparing this to the standard form, we have $ a = -8 $, $ b = 5 $, and $ c = -1 $. Substituting these values into the quadratic formula, we get:

$$ x = \frac{{-5 \pm \sqrt{{5^2 - 4(-8)(-1)}}}}{{2(-8)}} $$

$$ x = \frac{{-5 \pm \sqrt{{25 - 32}}}}{{-16}} $$

$$ x = \frac{{-5 \pm \sqrt{{-7}}}}{{-16}} $$

Since the discriminant $ b^2 - 4ac = -7 $ is negative, the roots are complex. Therefore, the roots are:

$$ x = \frac{{-5 + i\sqrt{7}}}{{-16}} $$ and $$ x = \frac{{-5 - i\sqrt{7}}}{{-16}} $$