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Svetradugi [14.3K]
3 years ago
12

What is the roots of the quadratic equation? 6x^2+5x-4=0

Mathematics
2 answers:
Rama09 [41]3 years ago
4 0

x = \frac{1}{2} or x = - \frac{4}{3}

consider the factors of the product 6 × - 4 = - 24 which sum to the coefficient of the x- term ( + 5)

the factors are + 8 and - 3 ( split the middle term using these factors

6x² - 3x + 8x - 4 = 0 ( factor by grouping )

3x(2x - 1) + 4(2x - 1 ) ( take out common factor of (2x - 1) )

= (2x - 1)(3x + 4) = 0

equate each factor to zero and solve for x

2x - 1 = 0 ⇒ x = \frac{1}{2}

3x + 4 = 0 ⇒ x = - \frac{4}{3}


mariarad [96]3 years ago
3 0

6x² + 5x - 4 = 0        

Multiply the first and last term (6x² * -4) to get -24x².  Now find two factors of -24x² whose sum is the middle term (5x).  -<em>3x + 8x   </em>Replace 5x with -3x + 8x, then factor and solve.

6x² - 3x    + 8x - 4 = 0

3x(<u>2x - 1</u>)  +4(<u>2x - 1</u>) = 0

(3x + 4) (<u>2x - 1</u>) = 0

3x + 4 = 0   or   2x - 1 = 0

x =  -\frac{4}{3}          or    x = \frac{1}{2}

<em>Note: you can also use the quadratic formula to find the foots.</em>

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Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25  

And rounded up we have that n=421

Step-by-step explanation:

We know that the sample proportion have the following distribution:

\hat p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by \alpha=1-0.90=0.1 and \alpha/2 =0.05. And the critical value would be given by:

z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.04 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

We assume that a prior estimation for p would be \hat p =0.5 since we don't have any other info provided. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.04}{1.64})^2}=420.25  

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