I am really struggling with this one. I'm teaching my self pre-calc out of a book and it isn't showing me how to do this. I've been all over the internet and could only find a few examples. I only know how to solve quadratic equations by converting them to vertex form and would like to stick with this method until it really sinks in. What am I doing wrong?

1.) Distance formula $\sqrt{(x-1)^2 + (-1 -1 + x)^2}=2$

2.) remove sqrt, $(x - 1)(x - 1) + (x - 2)(x - 2) = 4$

3.) multiply, $x^2 - 2x +1 + x^2 -4x +4 = 4$

4.) combine, $2x^2 -6x +5 = 4$

5.) general form, $2x^2 -6x +1$

6.) convert to vertex form (find the square), $2(x^2 - 3x + 1.5^2)-2(1.5)^2+1$

7.) Vertex form, $2(x-1.5)^2 -3.5$

8.) Solve for x, $x-1.5 = \pm\sqrt{1.75}$

9.) $x = 1.5 - 1.32$ and $x = 1.5 + 1.32$

10.) $x = 0.18$ and $2.82$

When I plug these two $x$ values back into the vertex form of the quadratic equation, I'm getting $y = 0.02$ for both $x$ values. These points are not on the line. Can someone tell me what I'm doing wrong please?

  • $\begingroup$ Assuming that you’ve solved the quadratic equation correctly, plug the values that you got for $x$ into the equation of the line. $\endgroup$ – amd Jul 19 '18 at 23:41
  • $\begingroup$ I did this but the points don't show up on the line. This is why I know I'm doing something wrong. $\endgroup$ – maybedave Jul 19 '18 at 23:46
  • $\begingroup$ In the last paragraph of your question you write that you plug these values back into the quadratic equation. (In which case, you’ve probably made some error since you should get zero.) $\endgroup$ – amd Jul 19 '18 at 23:47
  • $\begingroup$ yes and then I get points (0.18, 0.02) and (2.82, 0.02) which are not on the line of y = 1 - x $\endgroup$ – maybedave Jul 19 '18 at 23:48
  • $\begingroup$ If I understand your solution, you are finding $y$ from the quadratic $2x^2-6x+1=y$. But the quadratic at step 5 is $2x^2-6x+1=0$ not $2x^2-6x+1=y$. $\endgroup$ – David Jul 19 '18 at 23:50

Aside from truncating the answers to only a few significant digits, which can throw things off, everything is fine up until the very last step when you try to compute the corresponding $y$-values. You find those by plugging the $x$-values that you’ve computed into the equation $y=1-x$ of the line. If you do that, you can’t help but get points that lie on the line, assuming of course that you’ve solved the quadratic equation correctly. Specifically, the two points on the line are $(0.18, 0.82)$ and $(2.82, -1.82)$ or, more precisely, $\left(\frac32+\frac{\sqrt7}2,-\frac12+\frac{\sqrt7}2\right)$ and $\left(\frac32-\frac{\sqrt7}2,-\frac12-\frac{\sqrt7}2\right)$.


Another, possibly easier way to tackle this is to find the intersection points between a circle centered at $(1,-1)$ with radius $2$, and $y=1-x.$

The equation for the circle would be $$(x-1)^2+(y+1)^2=4$$ The equation of the line is $$y = 1-x$$

We can plug in $1-x$ for $y$ in the equation for the circle to get $$(x-1)^2+(2-x)^2=4$$ Expanding, $$x^2-2x+1+4-4x+x^2=4$$ or $$2x^2-6x+1=0$$ If you use the quadratic equation, you'll get $\frac32\pm\frac{\sqrt 7}2$. Plugging back into to either equation (preferably the linear equation), you'll get the coordinates as $(\frac32\pm\frac{\sqrt 7}2,-\frac12\pm\frac{\sqrt 7}2)$

  • $\begingroup$ This is exactly the same equation that the O.P. has in step 5 using an equivalent derivation. $\endgroup$ – amd Jul 19 '18 at 23:52

Since you have the general form $2x^2-6x+1=0$

Now solve for x, $$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$ $$x=\dfrac32\pm\dfrac12\sqrt{7},$$

Now try to plug in these values of $x$.


$$2(x-1.5)^2-3.5=0$$ $$2(x^2+2.25-3x)=3.5$$ $$x^2-3x+2.25=1.75$$ $$x^2-3x+0.5=0$$ $$2x^2-6x+1=0$$

  • $\begingroup$ The O.P. already has this. The problem isn’t solving the derived quadratic equation, but rather what to do with these solution. $\endgroup$ – amd Jul 19 '18 at 23:49
  • $\begingroup$ The $x$ values which the O.P found were wrong. $\endgroup$ – Key Flex Jul 19 '18 at 23:50
  • $\begingroup$ @Key Flex, yes, your values are what the book has. I guess I should just be using the quadratic equation and not using the vertex form to solve them so I guess I messed it up somewhere in that form. $\endgroup$ – maybedave Jul 19 '18 at 23:51
  • $\begingroup$ I was hoping someone could tell me what steps I did wrong though in the O.P. $\endgroup$ – maybedave Jul 19 '18 at 23:52
  • $\begingroup$ The values are correct to the two decimal places used. $\endgroup$ – amd Jul 19 '18 at 23:54

You have

$2x^2 - 6x + 1 = 0$

This looks good.

then I would say $x = \frac {3\pm\sqrt 7}{2}$ is simpler than $x = 1.5 \pm \sqrt {1.75}$

Plug each value of $x$ into $y = 1-x$ to find $y.$

$y = 1 - \frac {3+\sqrt 7}{2} = \frac {-1-\sqrt 7}{2}\\y = 1 - \frac {3-\sqrt 7}{2} = \frac {-1+\sqrt 7}{2}$


After getting those two roots $ (x_1,x_2) $ we get next by plugging in $ (1-y)$ for $x$ a second quadratic in $y:$

$$ 2y^2+2y-3=0$$

which supplies corresponding roots

$$ (y_1,y_2)= \frac{-1\pm \sqrt7}{2}.$$


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