How do you find the roots, real and imaginary, of #y= x^2 + 8x -9- (-x-1)^2 # using the quadratic formula?

Question

Eli Assouline · Accepted Answer

x=5/3

There are no imaginary roots of this equation. To get the roots, put y=0 Now expand #(-x-1)^2# which is actually #(x+1)^2# #x^2+8x-9-(x^2+2x+1)=0# #cancel(x^2)+8x-9cancel(-x^2)-2x-1=0# #6x-10=0# #x=5/3# Additionally you can also see that this equation is actually equation of a straight line so it can have at most 1 zero (which is real).

Eva Adamson · Answer

undefined To find the roots of the equation $y = x^2 + 8x -9- (-x-1)^2$, first, rewrite the equation to simplify:

$y = x^2 + 8x -9- (x+1)^2$

$y = x^2 + 8x -9 - (x^2 + 2x + 1)$

$y = x^2 + 8x - 9 - x^2 - 2x - 1$

Combine like terms:

$y = -3x - 10$

Now, since this equation is already simplified, we can find the roots directly using the quadratic formula:

$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$

where $a = 1$, $b = -3$, and $c = -10$.

Substitute these values into the formula and solve for $x$.

$x = \frac{{-(-3) \pm \sqrt{{(-3)^2 - 4(1)(-10)}}}}{{2(1)}}$

$x = \frac{{3 \pm \sqrt{{9 + 40}}}}{2}$

$x = \frac{{3 \pm \sqrt{{49}}}}{2}$

$x = \frac{{3 \pm 7}}{2}$

This gives two possible values for $x$:

$x_1 = \frac{{3 + 7}}{2} = 5$

$x_2 = \frac{{3 - 7}}{2} = -2$

Therefore, the roots of the equation are $x = 5$ and $x = -2$.