SOLUTION: I am having trouble with all five of these quadratic equations PLEASE HELP! 1. {{{ 9(x-8)^2 = 36 }}} 2. {{{ 27x^2 - 49 = 0 }}} 3. {{{ -16x = -x^2 }}} 4. {{{ x^3 -4x^2 + 4X = 0

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: I am having trouble with all five of these quadratic equations PLEASE HELP! 1. {{{ 9(x-8)^2 = 36 }}} 2. {{{ 27x^2 - 49 = 0 }}} 3. {{{ -16x = -x^2 }}} 4. {{{ x^3 -4x^2 + 4X = 0       Log On


   

Question 111423: I am having trouble with all five of these quadratic equations PLEASE HELP!
1. +9%28x-8%29%5E2+=+36+
2. +27x%5E2+-+49+=+0+
3. +-16x+=+-x%5E2+
4. +x%5E3+-4x%5E2+%2B+4X+=+0+
5. +x%5E2+%2B+69+=+0+
Any help would be greatly appreciated. I am going to continue working on these problems. They are due today by 4:30 p.m. est.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
#1


Start with the given equation




%28x-8%29%5E2=4 Divide both sides by 9



Take the square root of both sides




Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



Add 8 to both sides to isolate x.


Break down the expression into two parts: 




   or


Now combine like terms for each expression: 

   or



-----------------------------------
Answer:
So our solution is 

  or


Notice when we graph the equations y=9%28x-8%29%5E2 and y=36 we get:

graph of y=9%28x-8%29%5E2 (red) and y=36 (green)



Here we can see that the two equations intersect at x values of x=10 and x=6, so this 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 27%2Ax%5E2-49=0 (note: since the polynomial does not have an "x" term, the 2nd coefficient is zero. In other words, b=0. So that means the polynomial really looks like 27%2Ax%5E2%2B0%2Ax-49=0 notice a=27, b=0, and c=-49)





x+=+%280+%2B-+sqrt%28+%280%29%5E2-4%2A27%2A-49+%29%29%2F%282%2A27%29 Plug in a=27, b=0, and c=-49




x+=+%280+%2B-+sqrt%28+0-4%2A27%2A-49+%29%29%2F%282%2A27%29 Square 0 to get 0




x+=+%280+%2B-+sqrt%28+0%2B5292+%29%29%2F%282%2A27%29 Multiply -4%2A-49%2A27 to get 5292




x+=+%280+%2B-+sqrt%28+5292+%29%29%2F%282%2A27%29 Combine like terms in the radicand (everything under the square root)




x+=+%280+%2B-+42%2Asqrt%283%29%29%2F%282%2A27%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%280+%2B-+42%2Asqrt%283%29%29%2F54 Multiply 2 and 27 to get 54


So now the expression breaks down into two parts


x+=+%280+%2B+42%2Asqrt%283%29%29%2F54 or x+=+%280+-+42%2Asqrt%283%29%29%2F54



Now break up the fraction



x=0%2F54%2B42%2Asqrt%283%29%2F54 or x=0%2F54-42%2Asqrt%283%29%2F54



Simplify



x=0%2B7%2Asqrt%283%29%2F9 or x=0-7%2Asqrt%283%29%2F9



So the solutions are:

x=0%2B7%2Asqrt%283%29%2F9 or x=0-7%2Asqrt%283%29%2F9








#3
+-16x+=+-x%5E2+ Start with the given equation


+x%5E2-16x+=0++Add x%5E2 to both sides


x%28x-16%29=0 Factor the left side (note: if you need help with factoring, check out this solver)



Now set each factor equal to zero:

x=0 or x-16=0

x=0 or x=16 Now solve for x in each case


So our solutions are x=0 or x=16


Notice if we graph y=x%5E2-16x we get

+graph%28500%2C500%2C-20%2C20%2C-10%2C10%2C+x%5E2-16x%29+

and we can see that the graph has roots at x=0 or x=16, so this verifies our answer.



#4

x%5E3-4x%5E2%2B4x=0 Start with the given expression


x%28x%5E2-4x%2B4%29=0 Factor out the GCF x


x%28x-2%29%28x-2%29=0 Factor the inner expression x%5E2-4x%2B4

Now set each factor equal to zero:

x=0, x-2=0 or x-2=0

x=0, x=2, or x=2 Now solve for x in each case

Notice we get the solution x=2 twice


So our solutions are x=0 or x=2


Notice if we graph y=x%5E3-4x%5E2%2B4x we get

+graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C+x%5E3-4x%5E2%2B4x%29+

and we can see that the graph has roots at x=0 and x=2 so this verifies our answer.




#5

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 x%5E2%2B69=0 (note: since the polynomial does not have an "x" term, the 2nd coefficient is zero. In other words, b=0. So that means the polynomial really looks like x%5E2%2B0%2Ax%2B69=0 notice a=1, b=0, and c=69)





x+=+%280+%2B-+sqrt%28+%280%29%5E2-4%2A1%2A69+%29%29%2F%282%2A1%29 Plug in a=1, b=0, and c=69




x+=+%280+%2B-+sqrt%28+0-4%2A1%2A69+%29%29%2F%282%2A1%29 Square 0 to get 0




x+=+%280+%2B-+sqrt%28+0%2B-276+%29%29%2F%282%2A1%29 Multiply -4%2A69%2A1 to get -276




x+=+%280+%2B-+sqrt%28+-276+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




x+=+%280+%2B-+2%2Ai%2Asqrt%2869%29%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%280+%2B-+2%2Ai%2Asqrt%2869%29%29%2F%282%29 Multiply 2 and 1 to get 2




After simplifying, the quadratic has roots of


x=0%2Bsqrt%2869%29i or x=0-sqrt%2869%29i