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John's paint crew knows from experience that its 18-ft ladder is particulary stable when the distance from the ground to the top of the ladder is 5 ft more than the distance from the building to the base of the ladder as shown in the figure. In this position, how far up the building does the ladder reach? Aug 31, 2017

#1
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The wall and the ground form a right angle...so we can use the Pythagorean theorem to find  x .

x2 + (x + 5)2   =  182

Multiply out  (x + 5)2  .

x2 +  (x + 5)(x + 5)  =  324

x2 + x2 + 10x + 25  =  324

Combine  x2  and  x2  , and subtract  324  from both sides.

2x2 + 10x - 299  =  0

Now we can use the quadratic formula to solve for  x  .

$$x = {-10 \pm \sqrt{10^2-4(2)(-299)} \over 2(2)} \\~\\ x={-10 \pm \sqrt{100+2392} \over 4} \\~\\ x={-10 \pm \sqrt{2492} \over 4} \\~\\ x={-10 \pm 2\sqrt{623} \over 4} \\~\\ x={-5 + \sqrt{623} \over 2}\,\approx\,9.98 \qquad \text{ or }\qquad x={-5 - \sqrt{623} \over 2}\, \approx\, -14.98$$

Since  -14.98 causes a negative length for the side on the ground..... x must be ≈ 9.98

So, the length of the side on the wall  =  $${-5 + \sqrt{623} \over 2}+5\,\approx\,14.98$$   feet

Aug 31, 2017

#1
+2

The wall and the ground form a right angle...so we can use the Pythagorean theorem to find  x .

x2 + (x + 5)2   =  182

Multiply out  (x + 5)2  .

x2 +  (x + 5)(x + 5)  =  324

x2 + x2 + 10x + 25  =  324

Combine  x2  and  x2  , and subtract  324  from both sides.

2x2 + 10x - 299  =  0

Now we can use the quadratic formula to solve for  x  .

$$x = {-10 \pm \sqrt{10^2-4(2)(-299)} \over 2(2)} \\~\\ x={-10 \pm \sqrt{100+2392} \over 4} \\~\\ x={-10 \pm \sqrt{2492} \over 4} \\~\\ x={-10 \pm 2\sqrt{623} \over 4} \\~\\ x={-5 + \sqrt{623} \over 2}\,\approx\,9.98 \qquad \text{ or }\qquad x={-5 - \sqrt{623} \over 2}\, \approx\, -14.98$$

Since  -14.98 causes a negative length for the side on the ground..... x must be ≈ 9.98

So, the length of the side on the wall  =  $${-5 + \sqrt{623} \over 2}+5\,\approx\,14.98$$   feet

hectictar Aug 31, 2017