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

edited by Guest Oct 14, 2017

#1**+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

#1**+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**+2 **

Thanks, X^{2}.....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^{2 } found......!!!!!

CPhill
Oct 14, 2017

#3**+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