+0  
 
0
59
3
avatar

If a quadratic has a minimum at (-6,-14) and a root at x=17, what is the other root?

Guest Oct 14, 2017
edited by Guest  Oct 14, 2017
edited by Guest  Oct 14, 2017

Best Answer 

 #1
avatar+1373 
+2

At first, I was unsure how to go about this problem. However, I think I have found a pretty easy way to find the "solution."

 

First of all, let's consider some properties about quadratics when graphing. One thing we know is that there exists an axis of symmetry. This means that if the quadratic were reflected over the axis of symmetry, then the same graph would result. Since this is true, this means that the roots are equidistant from that axis of symmetry. If this does not make any sense, then maybe this rough sketch of a parabola will help you out!

 

 

Let's find the distance between (-6,0) and (17,0). All you have to do is subtract the x-coordinates. It happens to be 23. This means that the other root is 23 units in the other direction. -6-23=-29. This means that (-29,0) is where the other zero is located.

TheXSquaredFactor  Oct 14, 2017
Sort: 

3+0 Answers

 #1
avatar+1373 
+2
Best Answer

At first, I was unsure how to go about this problem. However, I think I have found a pretty easy way to find the "solution."

 

First of all, let's consider some properties about quadratics when graphing. One thing we know is that there exists an axis of symmetry. This means that if the quadratic were reflected over the axis of symmetry, then the same graph would result. Since this is true, this means that the roots are equidistant from that axis of symmetry. If this does not make any sense, then maybe this rough sketch of a parabola will help you out!

 

 

Let's find the distance between (-6,0) and (17,0). All you have to do is subtract the x-coordinates. It happens to be 23. This means that the other root is 23 units in the other direction. -6-23=-29. This means that (-29,0) is where the other zero is located.

TheXSquaredFactor  Oct 14, 2017
 #2
avatar+78719 
+2

Thanks, X2.....your method is easiest.....

 

Here's an algebraic solution.....

 

The x coordinate of the vertex  is    -b/ [ 2a]....so we have

 

-b/ [2a]  = -6

b = 12a

 

And we know that

 

a(-6)^2  + 12a(-6)  + c   =  -14

36a  -72a + c   = -14

c = 36a - 14

 

 

And since 17  is a root.......we have that

 

a(17)^2  + 12a (17)  +  [ 36a - 14 ] =  0

289 a + 204a  + 36a  = 14

529a  =  14

a  = 14/529

b = 12a  = 168/529

c = 36 a - 14  = -6902/529

 

So.......our function is

 

f (x)  =  (14/529)x^2  +  (168/529)x - 6902/529

 

Set this to 0   and multiply through by 529 and we have that

 

14x^2   + 168x  - 6902  =  0      divide through by 14

 

x^2  + 12x - 493  = 0   and this factors as

 

(x - 17) (x + 29)  = 0

 

Setting the two linear factors to 0 and solving for x gives the two roots  x = 17  and x = -29

 

Just as X found......!!!!!

 

 

cool cool cool

CPhill  Oct 14, 2017
 #3
avatar+5249 
+1

Here is yet another way.

 

A quadratic equation can be in the form:  y - k  =  a(x - h)2  , where  (h, k)  is the vertex.

 

They tell us the vertex is  (-6, -14) , so we can say our equation is

 

y - -14  =  a(x - -6)2

 

y + 14  =  a(x + 6)2

 

Since  17  is a root, we know that when  x = 17 ,  y = 0 .

So plug in  17  for  x  and  0  for  y  and solve for  a .

 

0 + 14  =  a(17 + 6)2

 

14  =  a(529)

 

a  =  14 / 529

 

So now we can say our equation is

 

y + 14  =  (14 / 529)(x + 6)2          Plug in  0  for  y  to find the roots.

 

14  =  (14 / 529)(x + 6)2

 

1  =   (1 / 529)(x + 6)2

 

529  =  (x + 6)2

 

x  =  ± 23 - 6

 

x  =  -29     or     x  =  17

hectictar  Oct 15, 2017

7 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details