Explore BrainMass

# Assistance again covering quadratic equations.

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

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 ± sqrt(√b2 - 4ac)/2a Discriminant in this problem is "sqrt(√b2- 4ac)."
We may now proceed by plugging in the values above to the quadratic formula replacing the variable x with w.
w = -0 ± sqrt(√02 - 4(5)(-3)/2(5) Simplify
w = ± sqrt(√60)/10 Final result in radical form
w = sqrt(√2.2.15)/10 Factor the square root
w = 2√15/10
w = 2√15/10 Simplify common factors
w = √15/5 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 ± symbol. Being told that there are other solutions. Can you show them to me? Unsure on the process.
w = ± sqrt(√60)/10 Final result in radical form
w = 2√15/10
w = 2√15/10 Simplify common factors
w = √15/5 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?

© BrainMass Inc. brainmass.com September 27, 2022, 9:35 am ad1c9bdddf

## 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 ± sqrt(√b2 - 4ac)/2a Discriminant in this problem is "sqrt(√b2- 4ac)."
We may now proceed by plugging in the values above to the quadratic formula replacing the variable x with w.
w = -0 ± sqrt(√02 - 4(5)(-3)/2(5) Simplify
w = ± sqrt(√60)/10 Final result in radical form
w = sqrt(√2.2.15)/10 Factor the square root (we could not get ride of +/- signs since they mean there are two solutions)
w = 2√15/10
w = 2√15/10 Simplify common factors
w = √15/5 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 ± symbol. Being told that there are other solutions. Can you show them to me? Unsure on the process.
w = ± sqrt(√60)/10 Final result in radical form
w = 2√15/10
w = 2√15/10 Simplify common factors
w = √15/5 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: (-b±sqrt(b^2-4ac))/2a=(-6±sqrt(4))/(2*1)=(-6±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: (-b±sqrt(b^2-4ac))/(2a)=(-0±sqrt(60))/(2*5)=(±2sqrt(15))/10=±sqrt(15)/5

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

© BrainMass Inc. brainmass.com September 27, 2022, 9:35 am ad1c9bdddf>