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-2
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avatar+819 

The graph of the line $y = 3x + a$ intersects the graph of the parabola $y = x^{2} + x$ in two points. If the distance between these points is $\sqrt{13}$, what is the value of $a$? Express your answer as a common fraction.

 

 Sep 9, 2023
 #1
avatar+128408 
+1

We can solve for the x intersection points thusly

 

3x + a = x^2 + x

x^2 - 2x - a  = 0

x^2 - 2x  = a          complete the square on x

 

x^2 - 2x + 1 =  a + 1

( x - 1)^2 = a + 1

( x - 1)  = sqrt (a + 1)       or       (x  -1) = -sqrt (a + 1)

x= 1 + sqrt (a + 1)           or  x  =  1 - sqrt (a + 1)

 

The y values of the intersections are

3[1 + sqrt (a + 1)] + a    and     3 [ 1 - sqrt (a + 1) ] + a

 

Simplifying and using the square of the distance formula  we have

 

[ 2sqrt (a + 1) ] ^2  + [ 6sqrt (a + 1)]^2   =13

 

4 (a + 1) + 36 (a + 1)  = 13

 

40 ( a + 1)  = 13

 

40a + 40  = 13

 

40a  =  -27

 

a = -27 /40

 

BTW   ....the graph looks more like this

 

 

cool cool cool

 Sep 10, 2023
edited by CPhill  Sep 10, 2023

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