Quadratic equations: solving x^2 + 8x + k = 0, given that 2 is a root

[mech649]

so i know how to solve quadratic equations, but i get a littler confused when they are given to me with solutions like this.

\(\displaystyle \mbox{5. If }\, 2\, \mbox{ is one of the solu}\mbox{tions of the equa}\mbox{tion }\, x^2\, +\, 8x\, +\, k\, =\, 0,\)

. . . . .\(\displaystyle \mbox{find }\, k\, \mbox{ and the other solution.}\)

\(\displaystyle \mbox{6. If }\, -3\, \mbox{ is one of the solu}\mbox{tions of the equa}\mbox{tion }\, x^2\, -\, kx\, +\, 3\, =\, 0,\)

. . . . .\(\displaystyle \mbox{find }\, k\, \mbox{ and the other solution.}\)

\(\displaystyle \mbox{7. Find }\, k\, \mbox{ such that }\, -1\, +\, \sqrt{\strut 5\,}\, \mbox{ is a solution of } x^2\, +\, 2x\, +\, k\, =\, 0.\)

what exactly am i suppose to do with the solutions that they give?

 

[Johulus]

so i know how to solve quadratic equations, but i get a littler confused when they are given to me with solutions like this.

what exactly am i suppose to do with the solutions that they give?

So, for the fifth task: \(\displaystyle x^2-8x+k=0 \, \rightarrow \, k=-x^2+8x \). And your task states that: \(\displaystyle x_1=2 \). Now, try to work it out on your own from here. It is really obvious. One solution is actually 'x'. So, how to get k if: \(\displaystyle k=-x^2+8x \).

 

[stapel]

\(\displaystyle \mbox{5. If }\, 2\, \mbox{ is one of the solu}\mbox{tions of the equa}\mbox{tion }\, x^2\, +\, 8x\, +\, k\, =\, 0,\)

. . . . .\(\displaystyle \mbox{find }\, k\, \mbox{ and the other solution.}\)

\(\displaystyle \mbox{6. If }\, -3\, \mbox{ is one of the solu}\mbox{tions of the equa}\mbox{tion }\, x^2\, -\, kx\, +\, 3\, =\, 0,\)

. . . . .\(\displaystyle \mbox{find }\, k\, \mbox{ and the other solution.}\)

\(\displaystyle \mbox{7. Find }\, k\, \mbox{ such that }\, -1\, +\, \sqrt{\strut 5\,}\, \mbox{ is a solution of } x^2\, +\, 2x\, +\, k\, =\, 0.\)

what exactly am i suppose to do with the solutions that they give?

When they give you a quadratic (that doesn't factor nicely) to solve, what do you do? You apply the Quadratic Formula (here). That Formula spits out two values, right?

. . . . .\(\displaystyle x\, =\, \dfrac{-b\, \pm\, \sqrt{\strut b^2\, -\, 4ac\,}}{2a}\)

You plug in the values for "a", "b", and "c", and get the values of x.

In your case, you've been given a solution, along with two of the constants' values. So you still have three pieces of information. Now work backwards. For instance:

\(\displaystyle \mbox{Example: If }\, 3\, \mbox{ is one of the solu}\mbox{tions of the equa}\mbox{tion }\, x^2\, +\, x\, +\, k\, =\, 0,\)

. . . . .\(\displaystyle \mbox{find }\, k\, \mbox{ and the other solution.}\)

So I"ve got a = 1, b = 1, c = k, and one of the solutions is x = 3. Then:

. . . . .\(\displaystyle x\, =\, \dfrac{-b\, \pm\, \sqrt{\strut b^2\, -\, 4ac\,}}{2a}\)

. . . . .\(\displaystyle 3\, =\, \dfrac{-1\, \pm\, \sqrt{\strut 1^2\, -\, 4(1)(k)\,}}{2(1)}\, =\, \dfrac{-1\, \pm\, \sqrt{\strut 1\, -\, 4k}}{2}\)

This gives me:

. . . . .\(\displaystyle 3\, \cdot 2\, =\, -1\, \pm\, \sqrt{\strut 1\, -\, 4k}\)

. . . . .\(\displaystyle 6\, +\, 1\, =\, \pm\, \sqrt{1\, -\, 4k}\)

. . . . .\(\displaystyle 7^2\, =\, 1\, -\, 4k\)

. . . . .\(\displaystyle 49\, -\, 1\, =\, -4k\)

. . . . .\(\displaystyle 48\, =\, -4k\)

. . . . .\(\displaystyle -12\, =\, k\)

Then the original quadratic equation was actually x^2 + x - 12 = 0. This does actually factor, so it's easy to find the value of the other root.

Do the same sort of thing with your examples.

 

[Harry_the_cat]

When they give you a quadratic (that doesn't factor nicely) to solve, what do you do? You apply the Quadratic Formula (here). That Formula spits out two values, right?

. . . . .\(\displaystyle x\, =\, \dfrac{-b\, \pm\, \sqrt{\strut b^2\, -\, 4ac\,}}{2a}\)

You plug in the values for "a", "b", and "c", and get the values of x.

In your case, you've been given a solution, along with two of the constants' values. So you still have three pieces of information. Now work backwards. For instance:

\(\displaystyle \mbox{Example: If }\, 3\, \mbox{ is one of the solu}\mbox{tions of the equa}\mbox{tion }\, x^2\, +\, x\, +\, k\, =\, 0,\)

. . . . .\(\displaystyle \mbox{find }\, k\, \mbox{ and the other solution.}\)

So I"ve got a = 1, b = 1, c = k, and one of the solutions is x = 3. Then:

. . . . .\(\displaystyle x\, =\, \dfrac{-b\, \pm\, \sqrt{\strut b^2\, -\, 4ac\,}}{2a}\)

. . . . .\(\displaystyle 3\, =\, \dfrac{-1\, \pm\, \sqrt{\strut 1^2\, -\, 4(1)(k)\,}}{2(1)}\, =\, \dfrac{-1\, \pm\, \sqrt{\strut 1\, -\, 4k}}{2}\)

This gives me:

. . . . .\(\displaystyle 3\, \cdot 2\, =\, -1\, \pm\, \sqrt{\strut 1\, -\, 4k}\)

. . . . .\(\displaystyle 6\, +\, 1\, =\, \pm\, \sqrt{1\, -\, 4k}\)

. . . . .\(\displaystyle 7^2\, =\, 1\, -\, 4k\)

. . . . .\(\displaystyle 49\, -\, 1\, =\, -4k\)

. . . . .\(\displaystyle 48\, =\, -4k\)

. . . . .\(\displaystyle -12\, =\, k\)

Then the original quadratic equation was actually x^2 + x - 12 = 0. This does actually factor, so it's easy to find the value of the other root.

Do the same sort of thing with your examples.

Another (easier) approach using the example above:

If 3 is a solution to the equation \(\displaystyle x^2+x+k =0\), then \(\displaystyle 3^2+3+k=0\), so k=-12.
Then proceed to find the other root.

 

[Steven G]

so i know how to solve quadratic equations, but i get a littler confused when they are given to me with solutions like this.

\(\displaystyle \mbox{5. If }\, 2\, \mbox{ is one of the solu}\mbox{tions of the equa}\mbox{tion }\, x^2\, +\, 8x\, +\, k\, =\, 0,\)

. . . . .\(\displaystyle \mbox{find }\, k\, \mbox{ and the other solution.}\)

\(\displaystyle \mbox{6. If }\, -3\, \mbox{ is one of the solu}\mbox{tions of the equa}\mbox{tion }\, x^2\, -\, kx\, +\, 3\, =\, 0,\)

. . . . .\(\displaystyle \mbox{find }\, k\, \mbox{ and the other solution.}\)

\(\displaystyle \mbox{7. Find }\, k\, \mbox{ such that }\, -1\, +\, \sqrt{\strut 5\,}\, \mbox{ is a solution of } x^2\, +\, 2x\, +\, k\, =\, 0.\)

what exactly am i suppose to do with the solutions that they give?

Hi, I think (there I go again) that some of the posters might have missed a problem that you have. Do you know what it means that x=2 is a solution? It means that the left side of the equation equals the right side of the equation. So you plug in 2 where ever you see x and get 4+16+k=0 or 20+k=0 and so k=-20. Now plug in -20 for +k and try to find the other root.