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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

#1
+5205
+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
Sort:

#1
+5205
+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
#2
+78575
+1

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

CPhill  Jul 17, 2017
#3
+5205
+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...

hectictar  Jul 17, 2017

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