Complete the square f(x) = x^2 + 4x - 8 PLZ help me ;c

1 answer:

6 0

Ok, first group x terms
f(x)=(x²+4x)-8
factor out quadratic coefient (no need but that's the step)
f(x)=1(x²+4x)-8
take 1/2 of the linear coefient and square it
4/2=2, (2)²=4
add positive and negative of it insides the parenthasees
f(x)=1(x²+4x+4-4)-8
factor perfect square
f(x)=1((x+2)²-4)-8
distribute
f(x)=1(x+2)²-4-8
f(x)=1(x+2)²-12
and, now if we wanted to find the x intercepts where f(x)=0 then
0=1(x+2)²-12
12=(x+2)²
+/-2√3=x+2
-2+/-2√3=x

x=-2+2√3 or -2-2√3
that is where the x intercept are

and completed square form is
f(x)=(x+2)²-12

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Let A denote the event that the podiatrist finds the first person with an ingrown toenail.

And (1 - A) denote the event that the podiatrist does not find the ingrown toenail.

While examining seven people, the podiatrist can find the very first person to have an ingrown toenail. Similarly he can find the second patient to have the ingrown toenail. Going in this way the probability of the first person to have an ingrown toenail is given by:

= A + (1 - A) × A + (1 - A) × (1 - A) × A + (1 - A) × (1 - A) × (1 - A) × A + (1 - A) × (1 - A) × (1 - A) × (1 - A) × A + (1 - A) × (1 - A) × (1 - A) × (1 - A) × (1 - A) × A + (1 - A) × (1 - A) × (1 - A) × (1 - A) × (1 - A) × (1 - A) × A.

= $\frac{1}{3} + \frac{2}{3} \frac{1}{3} + \frac{2}{3} \frac{2}{3} \frac{1}{3} + \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{1}{3} + \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{1}{3} + \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{1}{3} + \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{2}{3} \frac{1}{3} .$

= $\frac{1}{3} + \frac{2}{3} \frac{1}{3} + (\frac{2}{3}) ^{2} \frac{1}{3} + (\frac{2}{3})^{3} \frac{1}{3} + (\frac{2}{3})^{4} \frac{1}{3} + (\frac{2}{3})^{5} \frac{1}{3} + (\frac{2}{3})^{6} \frac{1}{3}.$