# Assistance again covering quadratic equations.

Problem 2. x2 + 6x +8 = 0 (Dugopolski, p. 635, 2012).

Through observation, we can see that the quadratic equation in our problem is factorable. Solution by the factoring method:

x2 +6x + 8 = 0 Original equation

(x + 2)(x+4) = 0 Factoring left hand side.

x + 2 = 0 or x + 4 = 0 Zero Factor property will find the value of x.

- 2 = - 2 or -4 = -4 Subtract variable

x = - 2 or x = - 4 Final answer

{-2, -4} Solution set braced by above answers.

On the above problem, how would I check the answers?

The factoring method is straightforward as long as the quadratic equation being solved is factorable.

Problem 78. 5w2 - 3 = 0 (Dugopolski, p. 637, 2012).

Though the problem is quite short for a quadratic equation, nevertheless, it is still a quadratic equation. Using the standard form of quadratic equation ax2 +bx + c = 0, the equation 5w2 - 3 = 0 can be represented as 5w2 +0x - 3 = 0, we know that a = 5; b = 0 and c = - 3.

Solution using the Quadratic formula:

x = - b Â± sqrt(âˆšb2 - 4ac)/2a Discriminant in this problem is "sqrt(âˆšb2- 4ac)."

We may now proceed by plugging in the values above to the quadratic formula replacing the variable x with w.

w = -0 Â± sqrt(âˆš02 - 4(5)(-3)/2(5) Simplify

w = Â± sqrt(âˆš60)/10 Final result in radical form

w = sqrt(âˆš2.2.15)/10 Factor the square root

w = 2âˆš15/10

w = 2âˆš15/10 Simplify common factors

w = âˆš15/5 Final answer

Reference:

Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY: McGraw-Hill Publishing.

Problem 78 has caused me some difficulty, after I dropped the Â± symbol. Being told that there are other solutions. Can you show them to me? Unsure on the process.

w = Â± sqrt(âˆš60)/10 Final result in radical form

w = 2âˆš15/10

w = 2âˆš15/10 Simplify common factors

w = âˆš15/5 Professor is stating there are two solutions? Final answer"

Also, how do I get the decimal equivalent for problem 78?

When it comes to solving quadratic equations, will completing the square always work?

Will using the quadratic formula always work? Can you explain using both of my problems?

https://brainmass.com/math/basic-algebra/assistance-again-covering-quadratic-equations-577981

## SOLUTION This solution is **FREE** courtesy of BrainMass!

Problem 2. x2 + 6x +8 = 0 (Dugopolski, p. 635, 2012).

Through observation, we can see that the quadratic equation in our problem is factorable. Solution by the factoring method:

x2 +6x + 8 = 0 Original equation

(x + 2)(x+4) = 0 Factoring left hand side.

x + 2 = 0 or x + 4 = 0 Zero Factor property will find the value of x.

- 2 = - 2 or -4 = -4 Subtract variable

x = - 2 or x = - 4 Final answer

{-2, -4} Solution set braced by above answers.

On the above problem, how would I check the answers?

We plug x=-2 and -4 into equation x^2+6x+8=0 to see if the equation is legimate.

For example, left side=(-2)^2+6*(-2)+8=4-12+8=0=right side. So x=-2 is one of the solutions.

With the same principle, we could plug x=-4 into the equation as well.

The factoring method is straightforward as long as the quadratic equation being solved is factorable.

Problem 78. 5w2 - 3 = 0 (Dugopolski, p. 637, 2012).

Though the problem is quite short for a quadratic equation, nevertheless, it is still a quadratic equation. Using the standard form of quadratic equation ax2 +bx + c = 0, the equation 5w2 - 3 = 0 can be represented as 5w2 +0x - 3 = 0, we know that a = 5; b = 0 and c = - 3.

Solution using the Quadratic formula:

x = - b Â± sqrt(âˆšb2 - 4ac)/2a Discriminant in this problem is "sqrt(âˆšb2- 4ac)."

We may now proceed by plugging in the values above to the quadratic formula replacing the variable x with w.

w = -0 Â± sqrt(âˆš02 - 4(5)(-3)/2(5) Simplify

w = Â± sqrt(âˆš60)/10 Final result in radical form

w = sqrt(âˆš2.2.15)/10 Factor the square root (we could not get ride of +/- signs since they mean there are two solutions)

w = 2âˆš15/10

w = 2âˆš15/10 Simplify common factors

w = âˆš15/5 Final answer

Reference:

Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY: McGraw-Hill Publishing.

Problem 78 has caused me some difficulty, after I dropped the Â± symbol. Being told that there are other solutions. Can you show them to me? Unsure on the process.

w = Â± sqrt(âˆš60)/10 Final result in radical form

w = 2âˆš15/10

w = 2âˆš15/10 Simplify common factors

w = âˆš15/5 Professor is stating there are two solutions? Final answer"

(as I indicated earlier, we could not get rid of +/- sign in the process since they mean two solutions )

Also, how do I get the decimal equivalent for problem 78?

(For the positive solution, w=sqrt(15)/5=3.8730/5=0.7746

For the negative solution, w=-sqrt(15)/5=-3.8730/5=-0.7746

Note: we could use calculator to find out the value for square root of 15)

When it comes to solving quadratic equations, will completing the square always work?

Yes.

Will using the quadratic formula always work? Can you explain using both of my problems?

The quadratic formula always work.

For problem 1, x^2+6x+8=0

a=1, b=6, c=8

determinant: b^2-4ac=6^2-4*1*8=4

solutions: (-bÂ±sqrt(b^2-4ac))/2a=(-6Â±sqrt(4))/(2*1)=(-6Â±2)/2

one solution: (-6-2)/2=-4, the other solution: (-6+2)/2=-2.

For problem 2, 5w^2-3=0

a=5, b=0, c=-3

determinant: b^2-4ac=0^2-4*(-3)*5=60

solutions: (-bÂ±sqrt(b^2-4ac))/(2a)=(-0Â±sqrt(60))/(2*5)=(Â±2sqrt(15))/10=Â±sqrt(15)/5

https://brainmass.com/math/basic-algebra/assistance-again-covering-quadratic-equations-577981