How do you find the roots, real and imaginary, of #y=-14 x^2 +18x +16-(x-3)^2 # using the quadratic formula?

Question

Abigail Allen · Accepted Answer

#x in {4/5-sqrt(249)/15, 4/5+sqrt(249)/15}#

The quadratic formula gives us the solutions to a quadratic equation of the form #ax^2+bx+c=0#. Thus, to use it, we must work the equation into that form.

#-14x^2+18x+16-(x-3)^2#

#= -14x^2+18x+16-(x^2-6x+9)#

#= -14x^2+18x+16-x^2+6x-9#

#=-15x^2+24x+7#

Thus, we can now look for the roots as the solutions to #ax^2+bx+c=0#, with #a=-15, b=24, c=7#

Plugging our values into the formula, we get

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

#=(-24+-sqrt(24^2-4(-15)(7)))/(2(-15))#

#=-(-24+-sqrt(576+420))/30#

#=(24+-sqrt(996))/30#

#=4/5+-sqrt(249)/15#

Thus, the roots of the given equation are #{4/5-sqrt(249)/15, 4/5+sqrt(249)/15}#

Julia Adams · Answer

undefined To find the roots of the quadratic equation $y = -14x^2 + 18x + 16 - (x - 3)^2$ using the quadratic formula, follow these steps:

1. First, rewrite the equation in standard form: $y = -14x^2 + 18x + 16 - (x^2 - 6x + 9)$.
2. Combine like terms to simplify the equation: $y = -14x^2 + 18x + 16 - x^2 + 6x - 9$.
3. Rearrange terms to standard form: $y = -15x^2 + 24x + 7$.
4. Now, identify the coefficients $a$, $b$, and $c$: $a = -15$, $b = 24$, and $c = 7$.
5. Apply the quadratic formula: $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$.
6. Substitute the values of $a$, $b$, and $c$ into the quadratic formula: $x = \frac{{-24 \pm \sqrt{{24^2 - 4(-15)(7)}}}}{{2(-15)}}$.
7. Simplify under the square root: $x = \frac{{-24 \pm \sqrt{{576 + 420}}}}{{-30}}$.
8. Further simplify: $x = \frac{{-24 \pm \sqrt{{996}}}}{{-30}}$.
9. Calculate the square root of 996: $\sqrt{996} ≈ 31.56$.
10. Plug in the value of the square root: $x = \frac{{-24 \pm 31.56}}{{-30}}$.
11. Calculate both roots: $x_1 ≈ \frac{{-24 + 31.56}}{{-30}}$ and $x_2 ≈ \frac{{-24 - 31.56}}{{-30}}$.
12. Simplify both roots: $x_1 ≈ \frac{{7.56}}{{-30}}$ and $x_2 ≈ \frac{{-55.56}}{{-30}}$.
13. Finalize the roots: $x_1 ≈ -0.252$ and $x_2 ≈ 1.852$.

Therefore, the real roots are approximately $x_1 \approx -0.252$ and $x_2 \approx 1.852$, and there are no imaginary roots.