Help me with this queston......?

Question

Show all ur working...

(a) 5x/2-9=0

(b) x^2 +12x + 3=0

Answer 1

5x/2 - 9 = 0

Transpose 9

5x / 2 - 9 + 9 = 0 + 9

5x / 2 = 9

Clear the fraction. Multiply both sides of the equation by 2

2(5x/2) = 2(9)

5x = 18

Divide both sides of the equation by 5

5x/5 = 18/5

x = 18/5

x = 3 3/5

X = 3.6

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Completing the square

x² + 12x + 3 = 0

Transpose 3

x² + 12x + 3 - 3 = 0 - 3

x² + 12x = - 3

x² + 12x +_____ = - 3 +_____

x² + 12x + 36 = - 3 + 36

(x + 6)(x + 6) = 33

(x + 6)² = 33

(√x + 6) = ± √33

x + 6 = ± √33

x + 6 = ± 5.744562647

Transpose 6

x + 6 - 6 = - 6 ± 5.74462647

x = - 6 ± 5.744562647

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Solving for +

x = - 6 + 5.744562647

x = - 0.255437353

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Solving for -

x = - 6 - 5.744562647

x = - 11.74456265

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The above equation can be solved by using the quadratic formula

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Answer 2

What he have here are simply linear and quadratic polynomials so you have to use some of that good old fashioned logic that you learned back in your Pre-algebra classes.

In the first problem you need to move the 9 over to the other side of the equation and then multiply by 2 which means 5x = 18 and therefore x = 18/5 which cannot be simplified any further.

The answer is there for x = 18/5

You may or may not want to approximate that solution on your calculator depending on the context of the problem.

For the second problem you have to try to factor it first. It does not appear to factor cleanly so you have to use the quadratic formula which is x = (-b + or - sqrt(b^2-4ac))/2a. This quadratic formula works to find all of the real roots of a quadratic equation in the form ax^2 + bx + c. Now if that works out you are all set punch the values of a, b, and c in the original equation and punch them down into the quadratic formula and you will be well on your way to solving this problem. Good luck to ya in the rest of the class and be sure to study hard on this stuff as it will likely appear on your final exam because it is very important stuff, especially as you get a little farther down the road in the math game.

I would just like to add a little note down here about the solutions that you will come up with using the quadratic formula. It depends on he context of the problem whether or not you want to use the exact numbers which will most likely include radicals. The only way that a solution found using the quadratic formula will not contain radicals is if the discriminant is a perfect square, which, in this case, it is not. If you are working on a word problem then you should definitely find a 2 to 3 decimal place approximation to the problem because your teacher will have a hard time making sense of your answer if you write it in radical form, and he will probably mark you off a few points for doing such a thing.

Answer 3

a)5x/2-9=0
5x/2=9
5x=18
x=18/5



b)x^2+12x+3=0
use formula (-b+-squareroot b^2-4ac)/2a
then,x=[-12+sr12^2-4(1)(3)]/2or
x=[-12-sr12^2-4(1)(3)]/2
x=-0.51/2 or x=-23.49/2
x=-0.255 or x=-11.75

Answer 4

(a) 5x/2 = 9
5x = 9*2
x = 18/5

(b) x = -0.255 and -11.74

Answer 5

5x/2 = 9
5x = 9x2
5x = 18
x = 18/5
x =3.6

Answer 6

5x/2-9=0
5x/2=9
5x=2(multiply)9
5x=18
x=18/5
x=3,6