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A square is drawn such that one of its sides coincides with the line $y = 5$, and so that the endpoints of this side lie on the parabola $y = x^2 + 3x + 2$. What is the area of the square?

michaelcai  Jul 17, 2017

Best Answer 

 #1
avatar+4480 
+1

Let's find the two points on the parabola that touch the line y = 5 .

 

y  =  x2 + 3x + 2             We want to find the  x  values when  y  is equal to  5  .

 

5  =  x2 + 3x + 2             Subtract  5  from both sides of this equation.

 

0  =  x2 + 3x - 3              Use the quadratic formula to find  x  .

 

x  =  \({-3 \pm \sqrt{21} \over 2}\)

 

So we know that the points     \(({-3 + \sqrt{21} \over 2}\,,\,5)\)     and     \(({-3 - \sqrt{21} \over 2}\,,\,5)\)     are the endpoints of one side of the square.     Here's a graph that shows this: https://www.desmos.com/calculator/djyyxourv3

 

The side length of the square is the distance between these two points.

 

side length    \(={-3 + \sqrt{21} \over 2} - {-3 - \sqrt{21} \over 2} \\~\\ ={-3 + \sqrt{21} \,-\,(-3-\sqrt{21})\over 2} \\~\\ ={-3 + \sqrt{21} + 3+\sqrt{21}\over 2} \\~\\ =\frac{2\sqrt{21}}{2} \\~\\ =\sqrt{21}\)

 

 

and...

 

area of square  =  (side length)2

 

area of square  =  \(\sqrt{21}^2\)

 

area of square  =  21   square units

hectictar  Jul 17, 2017
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3+0 Answers

 #1
avatar+4480 
+1
Best Answer

Let's find the two points on the parabola that touch the line y = 5 .

 

y  =  x2 + 3x + 2             We want to find the  x  values when  y  is equal to  5  .

 

5  =  x2 + 3x + 2             Subtract  5  from both sides of this equation.

 

0  =  x2 + 3x - 3              Use the quadratic formula to find  x  .

 

x  =  \({-3 \pm \sqrt{21} \over 2}\)

 

So we know that the points     \(({-3 + \sqrt{21} \over 2}\,,\,5)\)     and     \(({-3 - \sqrt{21} \over 2}\,,\,5)\)     are the endpoints of one side of the square.     Here's a graph that shows this: https://www.desmos.com/calculator/djyyxourv3

 

The side length of the square is the distance between these two points.

 

side length    \(={-3 + \sqrt{21} \over 2} - {-3 - \sqrt{21} \over 2} \\~\\ ={-3 + \sqrt{21} \,-\,(-3-\sqrt{21})\over 2} \\~\\ ={-3 + \sqrt{21} + 3+\sqrt{21}\over 2} \\~\\ =\frac{2\sqrt{21}}{2} \\~\\ =\sqrt{21}\)

 

 

and...

 

area of square  =  (side length)2

 

area of square  =  \(\sqrt{21}^2\)

 

area of square  =  21   square units

hectictar  Jul 17, 2017
 #2
avatar+76145 
+1

 

Very impressive, hectictar.....!!!!

 

Forget that college stuff....you're already too smart for your own good....LOL!!!!!

 

 

cool cool cool

CPhill  Jul 17, 2017
 #3
avatar+4480 
+1

Haha thank you! We'll see about that college stuff...I have a feeling that I'm gonna be asking a bunch of questions on here when that time comes... surpriselaugh

hectictar  Jul 17, 2017

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