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?

Question

1744 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-04T10:49:07

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

Need to expand out the brackets:

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

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

The roots of a quadratic of the form

$a x^{2} + b x + c$ $ax^2 + bx + c$

given by

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x = (-b +- sqrt(b^2 - 4ac))/(2a)$ \

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

$x = \frac{- 12 \pm \sqrt{144 - 4 (1) (- 14)}}{2} = \frac{- 12 \pm \sqrt{144 + 56}}{2}$ $x = (-12 +- sqrt(144 -4(1)(-14)))/2 = (-12+-sqrt(144+56))/2$

$x = \frac{- 12 \pm \sqrt{200}}{2}$ $x = (-12+-sqrt(200))/2$

$x = \frac{- 12 \pm \sqrt{100} \sqrt{2}}{2} = \frac{- 12 \pm 10 \sqrt{2}}{2}$ $x = (-12 +- sqrt(100)sqrt(2))/2 = (-12+-10sqrt(2))/2$

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

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)2y= 2(x+1)^2-(x-4)^2 using the quadratic formula?