For this case we have to:
Given the quadratic equation of the form:
![ax ^ 2 + bx + c = 0](https://tex.z-dn.net/?f=ax%20%5E%202%20%2B%20bx%20%2B%20c%20%3D%200)
The roots are given by:
![x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%20%7B-b%20%5Cpm%20%5Csqrt%20%7Bb%20%5E%202-4%20%28a%29%20%28c%29%7D%7D%20%7B2a%7D)
If we have: ![x ^ 2 + 14x + 17 = -96](https://tex.z-dn.net/?f=x%20%5E%202%20%2B%2014x%20%2B%2017%20%3D%20-96)
We can rewrite it in the following way:
![x ^ 2 + 14x + 17 + 96 = 0\\x ^ 2 + 14x + 113 = 0](https://tex.z-dn.net/?f=x%20%5E%202%20%2B%2014x%20%2B%2017%20%2B%2096%20%3D%200%5C%5Cx%20%5E%202%20%2B%2014x%20%2B%20113%20%3D%200)
Where:
![a = 1\\b = 14\\c = 113](https://tex.z-dn.net/?f=a%20%3D%201%5C%5Cb%20%3D%2014%5C%5Cc%20%3D%20113)
Where we have:
![x = \frac {-14 \pm \sqrt {(14) ^ 2-4 (1) (113)}} {2 (1)}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%20%7B-14%20%5Cpm%20%5Csqrt%20%7B%2814%29%20%5E%202-4%20%281%29%20%28113%29%7D%7D%20%7B2%20%281%29%7D)
![x = \frac {-14 \pm \sqrt {(196-452)}} {2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%20%7B-14%20%5Cpm%20%5Csqrt%20%7B%28196-452%29%7D%7D%20%7B2%7D)
![x = \frac {-14 \pm \sqrt {-256}} {2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%20%7B-14%20%5Cpm%20%5Csqrt%20%7B-256%7D%7D%20%7B2%7D)
By definition: ![\sqrt {-1} = i](https://tex.z-dn.net/?f=%5Csqrt%20%7B-1%7D%20%3D%20i)
![x = \frac {-14 \pm \sqrt {256} i} {2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%20%7B-14%20%5Cpm%20%5Csqrt%20%7B256%7D%20i%7D%20%7B2%7D)
![x = \frac {-14 \pm16i} {2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%20%7B-14%20%5Cpm16i%7D%20%7B2%7D)
![x = \frac {-14} {2} \pm \frac {16i} {2}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%20%7B-14%7D%20%7B2%7D%20%5Cpm%20%5Cfrac%20%7B16i%7D%20%7B2%7D)
![x = -7 \pm8i](https://tex.z-dn.net/?f=x%20%3D%20-7%20%5Cpm8i)
Thus, the roots are given by imaginary numbers:
![x_ {1} = - 7 + 8i\\x_ {2} = - 7-8i](https://tex.z-dn.net/?f=x_%20%7B1%7D%20%3D%20-%207%20%2B%208i%5C%5Cx_%20%7B2%7D%20%3D%20-%207-8i)
Answer:
![x_ {1} = - 7 + 8i\\x_ {2} = - 7-8i](https://tex.z-dn.net/?f=x_%20%7B1%7D%20%3D%20-%207%20%2B%208i%5C%5Cx_%20%7B2%7D%20%3D%20-%207-8i)
I wasn't going to click on this one, but the all-caps enthralled me and hypnotized me.
first off, let's change all the mixed fractions to "improper" and proceed from there, keep in mind that if we subtract the two bags' weight from the total, what's leftover is the 3rd bag's weight.
![\bf \stackrel{mixed}{22\frac{1}{2}}\implies \cfrac{22\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{45}{2}} \\\\\\ \stackrel{mixed}{6\frac{3}{8}}\implies \cfrac{6\cdot 8+3}{8}\implies \stackrel{improper}{\cfrac{51}{8}} \\\\\\ \stackrel{mixed}{3\frac{1}{4}}\implies \cfrac{3\cdot 4+1}{4}\implies \stackrel{improper}{\cfrac{13}{4}}\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B22%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B22%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B45%7D%7B2%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bmixed%7D%7B6%5Cfrac%7B3%7D%7B8%7D%7D%5Cimplies%20%5Ccfrac%7B6%5Ccdot%208%2B3%7D%7B8%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B51%7D%7B8%7D%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bmixed%7D%7B3%5Cfrac%7B1%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B3%5Ccdot%204%2B1%7D%7B4%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B13%7D%7B4%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)
7x-4=2
add 4 to both sides
7x=2+4
7x=6
divide 7 to both sides
x=6/7
Answer:
<em>B. a = y - k/(x-h)² </em>
Step-by-step explanation:
Given the expression y = a (x-h)² + k
First make a the subject of the formula
Subtract k from both sides
y = a (x-h)² + k
y- k = a (x-h)² + k - k
y - k = a (x-h)²
Divide through by (x-h)²
y - k/(x-h)² = a
a = y - k/(x-h)²
Hence option B is correct