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

Question 111423: I am having trouble with all five of these quadratic equations PLEASE HELP!
1.
2.
3.
4.
5.
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) (Show Source): You can put this solution on YOUR website!
#1

Start with the given equation

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 and we get:

graph of (red) and (green)

Here we can see that the two equations intersect at x values of and , so this verifies our answer.

Solved by pluggable solver: Quadratic Formula

Let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve (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 notice , , and )

Plug in a=27, b=0, and c=-49

Square 0 to get 0

Multiply to get

Combine like terms in the radicand (everything under the square root)

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

Multiply 2 and 27 to get 54

So now the expression breaks down into two parts

or

Now break up the fraction

or

Simplify

or

So the solutions are:

or

#3
Start with the given equation

Add to both sides

Factor the left side (note: if you need help with factoring, check out this solver)

Now set each factor equal to zero:

or

or Now solve for x in each case

So our solutions are or

Notice if we graph we get

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

#4

Start with the given expression

Factor out the GCF

Factor the inner expression

Now set each factor equal to zero:

, or

, , or Now solve for x in each case

Notice we get the solution twice

So our solutions are or

Notice if we graph we get

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

Solved by pluggable solver: Quadratic Formula

Let's use the quadratic formula to solve for x:

Starting with the general quadratic

the general solution using the quadratic equation is:

So lets solve (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 notice , , and )

Plug in a=1, b=0, and c=69

Square 0 to get 0

Multiply to get

Combine like terms in the radicand (everything under the square root)

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

Multiply 2 and 1 to get 2

After simplifying, the quadratic has roots of

or

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