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Verdich [7]
3 years ago
9

Find thetwo consecutive integers whose product is 24​

Mathematics
1 answer:
malfutka [58]3 years ago
5 0

Answer:

There is no solution to this equation that is an integer for x.

Step-by-step explanation:

x is the first integer, and x+1 the second

x(x+1) = 24\\x^{2} + x - 24 =0\\

x = \frac{-1+\sqrt{1^{2}+4*24}}{2} =\frac{-1+\sqrt{97}}{2}\\x = \frac{-1-\sqrt{1^{2}+4*24}}{2} =\frac{-1-\sqrt{97}}{2}

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6. What are the zeroes for the<br>function<br>f(x) = 2x3 + 12x – 10x2?<br>​
agasfer [191]

Answer:

<h2>x = 0, x = 5 + √13 and x = 5 - √13.</h2>

Step-by-step explanation:

f(x) = 2x^3 + 12x – 10x^2 can and should be rewritten in descending powers of x:  

f(x) = 2x^3 – 10x^2 + 12x

This, in turn, can be factored into f(x) = x·(x² - 10x + 12).

Setting this last result = to 0 results in f(x) = x·(x² - 10x + 12).

Thus, x = 0 is one root.  Two more roots come from x² - 10x + 12 = 0.

Let's "complete the square" to solve this equation.

Rewrite x² - 10x + 12 = 0  as  x² - 10x +              12 = 0.

a) Identify the coefficient of the x term.  It is -10.

b) take half of this result:  -5

c) square this last result:  (-5)² = 25.

d) Add this 25 to both sides of x² - 10x +              12 = 0:

     x² - 10x +  25      +      12 = 0 + 25

e) rewrite x² - 10x +  25 as the square of a binomial:

       (x - 5)² = 13

f)  taking the sqrt of both sides:  x - 5 = ±√13

g) write out the zeros:  x = 5 + √13 and x = 5 - √13.

The three roots are x = 0, x = 5 + √13 and x = 5 - √13.

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