SOLUTION: Help with a few problems, please. 1. Solve by completing the square: 2x^2 + x + 2 = 0 2. Solve by using the quadratic formula: 5x^2 - 2x + 3 = 0 choices: 3. Write a quadr

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Help with a few problems, please. 1. Solve by completing the square: 2x^2 + x + 2 = 0 2. Solve by using the quadratic formula: 5x^2 - 2x + 3 = 0 choices: 3. Write a quadr      Log On


   

Question 728581: Help with a few problems, please.
1. Solve by completing the square: 2x^2 + x + 2 = 0
2. Solve by using the quadratic formula: 5x^2 - 2x + 3 = 0
choices:
3. Write a quadratic equation that has integer coefficients and has solutions 4/3 and 6.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

1.
Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form


y=2+x%5E2%2B1+x%2B2 Start with the given equation



y-2=2+x%5E2%2B1+x Subtract 2 from both sides



y-2=2%28x%5E2%2B%281%2F2%29x%29 Factor out the leading coefficient 2



Take half of the x coefficient 1%2F2 to get 1%2F4 (ie %281%2F2%29%281%2F2%29=1%2F4).


Now square 1%2F4 to get 1%2F16 (ie %281%2F4%29%5E2=%281%2F4%29%281%2F4%29=1%2F16)





y-2=2%28x%5E2%2B%281%2F2%29x%2B1%2F16-1%2F16%29 Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of 1%2F16 does not change the equation




y-2=2%28%28x%2B1%2F4%29%5E2-1%2F16%29 Now factor x%5E2%2B%281%2F2%29x%2B1%2F16 to get %28x%2B1%2F4%29%5E2



y-2=2%28x%2B1%2F4%29%5E2-2%281%2F16%29 Distribute



y-2=2%28x%2B1%2F4%29%5E2-1%2F8 Multiply



y=2%28x%2B1%2F4%29%5E2-1%2F8%2B2 Now add 2 to both sides to isolate y



y=2%28x%2B1%2F4%29%5E2%2B15%2F8 Combine like terms




Now the quadratic is in vertex form y=a%28x-h%29%5E2%2Bk where a=2, h=-1%2F4, and k=15%2F8. Remember (h,k) is the vertex and "a" is the stretch/compression factor.




Check:


Notice if we graph the original equation y=2x%5E2%2B1x%2B2 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C2x%5E2%2B1x%2B2%29 Graph of y=2x%5E2%2B1x%2B2. Notice how the vertex is (-1%2F4,15%2F8).



Notice if we graph the final equation y=2%28x%2B1%2F4%29%5E2%2B15%2F8 we get:


graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C2%28x%2B1%2F4%29%5E2%2B15%2F8%29 Graph of y=2%28x%2B1%2F4%29%5E2%2B15%2F8. Notice how the vertex is also (-1%2F4,15%2F8).



So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.






2.
Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve 5%2Ax%5E2-2%2Ax%2B3=0 ( notice a=5, b=-2, and c=3)





x+=+%28--2+%2B-+sqrt%28+%28-2%29%5E2-4%2A5%2A3+%29%29%2F%282%2A5%29 Plug in a=5, b=-2, and c=3




x+=+%282+%2B-+sqrt%28+%28-2%29%5E2-4%2A5%2A3+%29%29%2F%282%2A5%29 Negate -2 to get 2




x+=+%282+%2B-+sqrt%28+4-4%2A5%2A3+%29%29%2F%282%2A5%29 Square -2 to get 4 (note: remember when you square -2, you must square the negative as well. This is because %28-2%29%5E2=-2%2A-2=4.)




x+=+%282+%2B-+sqrt%28+4%2B-60+%29%29%2F%282%2A5%29 Multiply -4%2A3%2A5 to get -60




x+=+%282+%2B-+sqrt%28+-56+%29%29%2F%282%2A5%29 Combine like terms in the radicand (everything under the square root)




x+=+%282+%2B-+2%2Ai%2Asqrt%2814%29%29%2F%282%2A5%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%282+%2B-+2%2Ai%2Asqrt%2814%29%29%2F%2810%29 Multiply 2 and 5 to get 10




After simplifying, the quadratic has roots of


x=1%2F5%2Bsqrt%2814%29%2F5%2Ai or x=1%2F5-sqrt%2814%29%2F5%2Ai




3.
%28x-x%5B1%5D%29%28x-x%5B2%5D%29....if x%5B1%5D=4%2F3 and x%5B2%5D=6, than
%28x-4%2F3%29%28x-6%29
x%5E2-%284%2F3%29x-6x%2B%284%2F3%296
x%5E2-%28%284%2F3%29%2B6%29x%2B4%2A2
x%5E2-%2822%2F3%29x%2B8