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Vika [28.1K]
3 years ago
10

Complete the square to find the minimum value of the expression 4x2 + 8x + 23.

Mathematics
1 answer:
Tcecarenko [31]3 years ago
3 0
So you need to come up with a perfect square that works for the x coefficients.
like.. (2x + 2)^2
(2x+2)(2x+2) = 4x^2 + 8x + 4
Compare this to the equation given. Our perfect square has +4 instead of +23. The difference is: 23 - 4 = 19

I'm going to assume the given equation equals zero..

So, If we add subtract 19 from both sides of the equation we get the perfect square.

4x^2 + 8x + 23 - 19 = 0 - 19
4x^2 + 8x + 4 = - 19
complete the square and move 19 over..
(2x+2)^2 + 19 = 0
factor the 2 out becomes 2^2 = 4
ANSWER: 4(x+1)^2 + 19 = 0

for a short cut, the standard equation
ax^2 + bx + c = 0 becomes a(x - h)^2 + k = 0
Where "a, b, c" are the same and ..
h = -b/(2a)
k = c - b^2/(4a)

Vertex = (h, k)
this will be a minimum point when "a" is positive upward facing parabola and a maximum point when "a" is negative downward facing parabola.


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z=\frac{6.6-6}{\frac{1.7}{\sqrt{40}}}=2.232    

Since is a one sided right tailed test the p value would be:  

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Step-by-step explanation:

Data given and notation  

\bar X=6.6 represent the sample mean

\sigma=1.7 represent the population standard deviation

n=40 sample size  

\mu_o =6 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is higher than 6minutes or no, the system of hypothesis would be:  

Null hypothesis:\mu \leq 6  

Alternative hypothesis:\mu > 6  

If we analyze the size for the sample is > 30 and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

z=\frac{6.6-6}{\frac{1.7}{\sqrt{40}}}=2.232    

P-value

Since is a one sided right tailed test the p value would be:  

p_v =P(z>2.232)=0.0128  

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