Quadratic - Completing the square

[thomas49th]

Homework Statement

Need to solve
2x² + 5x + 3 = 0

Homework Equations

at² + bt + c = 0
so a = 2, b = 5, c = 3

The Attempt at a Solution

a x c = 6
a + c = 5
so ned to find somthing that times to make 6 and add to make 5. That is 2, 3

2x² + 2x + 3x + 3

2x (x + 1) + 3(x + 1)

(2x + 3)(x + 1) = 0
so x is either

x = 3/2 = -1.5 or x = -1

Are those answers correct?

If so how would I go about solving this question

A solution to the equation $$x^{5} - 4x^{3} = 455$$ lies between x = 3.6 and x = 3.8. Find the solution to 2 decimal places.

So should change it = 0 by -455 so i get

$$x^{5} - 4x^{3} -455 = 0$$

now where do i go?

Thx

 

[drpizza]

Your answers are correct. However, your thread title says "completing the square" - you've solved by factoring (which is easier for this particular problem.
Personally, I teach completing the square this way:
2x^2 + 5x + 3 = 0
Get rid of the 3; it's just in the way.
2x^2 + 5x = -3
now factor out any coefficients on the x^2.
2(x^2 + 5/2x ) = -3
Then, add (inside the parenthesis) half of the middle term squared.
Since you have to add the same amount to both sides, add to the right side as well; note: whatever you add inside the parenthesis will be multiplied by 2. Thus:
2(x^2 +5/2 x + 25/16)= -3 + 25/8
Factor (it'll always be half the middle term again)
2(x+5/4)^2 = 1/8
divide by 2
(x+5/4)^2 = 1/16
take the +/- square root
x+5/4 = +/- 1/4
subtract 5/4, and you have your same answers.

For your second problem, there are several ways to solve it, to get it correct to 2 decimal places... what methods have you learned? Show me how you would apply one of those methods and I or someone else can show you where you're making a mistake (or let you know if you're on the right track.)

 

[thomas49th]

$$x^{5} - 4x^{3} -455 = 0$$
$$x^{5} - 4x^{3} -455 =(x^{2} - 4x)^{3} - 64 - 455$$

is that right?

 

[HallsofIvy]

I have no idea what you are doing. But (x²)³ is x⁶ not x⁵ so that can't be right!

 

[thomas49th]

I think I've got it

x³(x²-4) can be factorised even fruther to
x³(x+2)(x-2)

that means the equation can be x = -2, x = 2 or x=0

 

[cristo]

I think I've got it

x³(x²-4) can be factorised even further to
x³(x+2)(x-2)

that means the equation can be x = -2, x = 2 or x=0

This would be true if the equation was $$x^{5} - 4x^{3} = 0$$, but your equation is $$x^{5} - 4x^{3} = 455$$. Since you're looking for an approximate value of the roots, you are not looking to solve this by factorisation.

drpizza hints at this in post #2.

For your second problem, there are several ways to solve it, to get it correct to 2 decimal places... what methods have you learned? Show me how you would apply one of those methods and I or someone else can show you where you're making a mistake (or let you know if you're on the right track.)