How do you find the roots, real and imaginary, of #y= 2(x+5)^2+(x-4)^2 # using the quadratic formula?

Question

Julia Adams · Accepted Answer

Zeros are #-2-3sqrt2i# and #-2+3sqrt2i#

Quadratic formula gives the roots of #y=ax^2+bx+c# as #x=(-b+-sqrt(b^2-4ac))/(2a)#

As #y=2(x+5)^2+(x-4)^2#

= #2(x^2+10x+25)+x^2-8x+16#

= #2x^2+20x+50+x^2-8x+16#

= #3x^2+12x+66#

Hence zeros are given by

#x=(-12+-sqrt(12^2-4×3×66))/(2×3)#

= #(-12+-sqrt(144-792))/6#

= #(-12+-sqrt(-648))/6#

= #(-12+-18sqrt2i)/6#

= #-2+-3sqrt2i#

Hence, zeros are #-2-3sqrt2i# and #-2+3sqrt2i#

Hazel Anglin · Answer

undefined To find the roots of $y = 2(x+5)^2 + (x-4)^2$ using the quadratic formula, we first expand the expression:

$y = 2(x+5)^2 + (x-4)^2$  
$y = 2(x^2 + 10x + 25) + (x^2 - 8x + 16)$  
$y = 2x^2 + 20x + 50 + x^2 - 8x + 16$  
$y = 3x^2 + 12x + 66$

Now, we set $y$ to zero and rewrite the equation in the form $ax^2 + bx + c = 0$:

$0 = 3x^2 + 12x + 66$

The quadratic formula is:

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

For this equation, $a = 3$, $b = 12$, and $c = 66$. Plugging these values into the formula, we get:

$x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 3 \cdot 66}}{2 \cdot 3}$  
$x = \frac{-12 \pm \sqrt{144 - 792}}{6}$  
$x = \frac{-12 \pm \sqrt{-648}}{6}$  
$x = \frac{-12 \pm \sqrt{648}i}{6}$  
$x = \frac{-12 \pm 18i}{6}$  
$x = -2 \pm 3i$

Therefore, the roots of $y = 2(x+5)^2 + (x-4)^2$ are $x = -2 + 3i$ and $x = -2 - 3i$.