How do you find the roots, real and imaginary, of $y= 2(x+1)^2-(x-4)^2$ using the quadratic formula?

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Answer 1 · 2017-09-04T10:49:07

$x = -6 + 5sqrt(2) " or " x = -6 - 5sqrt(2)$

Need to expand out the brackets:

$y = 2(x^2 + 2x + 1) - (x^2 - 8x + 16)$

$y = x^2+12x - 14$

The roots of a quadratic of the form

$ax^2 + bx + c$

given by

$x = (-b +- sqrt(b^2 - 4ac))/(2a)$ \

In this case $a = 1, b = 12, c = -14$

$x = (-12 +- sqrt(144 -4(1)(-14)))/2 = (-12+-sqrt(144+56))/2$

$x = (-12+-sqrt(200))/2$

$x = (-12 +- sqrt(100)sqrt(2))/2 = (-12+-10sqrt(2))/2$

$therefore x = -6 + 5sqrt(2) " or " x = -6 - 5sqrt(2)$

How do you find the roots, real and imaginary, of y= 2(x+1)^2-(x-4)^2 y= 2(x+1)^2-(x-4)^2 using the quadratic formula?