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Bezzdna [24]
3 years ago
8

Smallest of 3 consecutive positive intergers if the product of the smaller two intergers is 5 less than 5 times the largest inte

ger
Mathematics
1 answer:
amm18123 years ago
6 0

What a delightful little problem ! (Partly because I could see
right away how to do it, and had the answer in a few minutes,
after a lot of impressive-looking algebra on my scratch-paper.)

Three consecutive integers are . . . x,  x+1,  and  x+2

The smallest two are . . . x  and  x+1
    Their product is . . . . . x(x+1)

5 times the largest one is . . . 5(x+2)
      5 less than that is . . . . . . 5(x+2)-5

Now, the conditions of the problem say that    <u>x (x + 1) = 5 (x+2) - 5</u>
THAT's the equation we have to solve, to find 'x' .

       Eliminate parentheses:    x² + x = 5x + 10 - 5
            Combine like terms:    x² + x = 5x + 5
Subtract 5x from each side:    x² - 4x = 5
Subtract  5  from each side:   <u>x² - 4x - 5 = 0</u>

You could solve that by factoring it, or use the quadratic equation.

Factored, it says that    (x + 1) (x - 5) = 0

From which       <em>x = -1</em>
             and      <em>x = +5</em>

We only want the positive results, so our three consecutive integers are

             5,  6,  and  7 .   

To answer the question, the smallest one is    <em><u>5 </u></em>.

<u>Check</u>:

   5 x 6  ? = ?  (7 x 5) - 5

       30  ? = ?  (35) - 5

         30   =   30

             yay !   


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Answer:

The top of the ladder is now at 10 ft.

Step-by-step explanation:

At the start, we have a height H=6, a length L=10 and a base B, that has to be calculated by the Pythagorean theorem:

B^2=L^2-H^2=10^2-6^2=100-36=64\\\\B=\sqrt{64}=8

The base is moved twice the distance the height moves up.

We called this distance x, so we have:

L^2=(H+x)^2+(B-2x)^2=H^2+2Hx+x^2+B^2-4Bx+4x^2\\\\L^2=(H^2+B^2)+5x^2+(2H-4B)x\\\\L^2=L^2+5x^2+(2H-4B)x\\\\0=5x^2+(2H-4B)x\\\\5x+(2H-4B)=0\\\\x=\dfrac{4B-2H}{5}=\dfrac{4*8-2*6}{5}=\dfrac{32-12}{5}=\dfrac{20}{5}=4

The new height (H+x) is

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This means that the ladder is all against the wall (L=H').

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