Help with number 50 please. Thanks.
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Answer:
[tex] d = 7 + 3\sqrt{3} [/tex] and
[tex] d = 7 - 3\sqrt{3} [/tex]
Step-by-step explanation:
To solve the equation, [tex] d^2 - 14d - 22 = 0 [/tex], using the quadratic formula,
Recall: quadratic formula = [tex] \frac{-b ± \sqrt{b^2 - 4ac}}{2a} [/tex]
Where,
a = 1
b = -14
c = 22
Plug in your values into the formula and solve:
[tex] \frac{-(-14) ± \sqrt{(-14)^2 - 4(1)(22)}}{2(1)} [/tex]
[tex] \frac{14 ± \sqrt{196 - 88}}{2} [/tex]
[tex] \frac{14 ± \sqrt{108}}{2} [/tex]
[tex] d = \frac{14 + \sqrt{108}}{2} [/tex]
[tex] d = \frac{14 + 6\sqrt{3}}{2} [/tex]
[tex] d = (\frac{2(7 + 3\sqrt{3})}{2} [/tex]
[tex] d = 7 + 3\sqrt{3} [/tex]
And
[tex] d = \frac{14 - \sqrt{108}}{2} [/tex]
[tex] d = \frac{14 - 6\sqrt{3}}{2} [/tex]
[tex] d = (\frac{2(7 - 3\sqrt{3})}{2} [/tex]
[tex] d = 7 - 3\sqrt{3} [/tex]