SOLUTION: Please help me with this question. I have to solve by whichever method is easier. ( factoring, completing the square or the quadratic formula) A) what is the discriminant of the

Algebra ->  Finance -> SOLUTION: Please help me with this question. I have to solve by whichever method is easier. ( factoring, completing the square or the quadratic formula) A) what is the discriminant of the      Log On


   

Question 896352: Please help me with this question. I have to solve by whichever method is easier. ( factoring, completing the square or the quadratic formula)
A) what is the discriminant of the equation 4x^2+8x+k=0
B) for what value of k will the equation have a double root?
C)for what values of k will the equation have two real roots?
D) for what values of k will the equation have imaginary roots?
E) name three values of k for which the given equation has rational roots

Found 2 solutions by richwmiller, MathTherapy:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a=4 b=8 c=k
A) sqrt( 8^2-4*4*k ))

solve for k
B) 0 = sqrt( 8^2-4*4*k ) if d=0 there is one root (double root)
when k=4
C) 0< sqrt( 8^2-4*4*k ) if d is positive there are two real roots
when k<4
D) 0> sqrt( 8^2-4*4*k )if d is negative then there are imaginary roots
when k>4
E) k=1 k=2 k=3
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


4%2Ax%5E2%2B8%2Ax%2B4 Start with the given expression.



4%28x%5E2%2B2x%2B1%29 Factor out the GCF 4.



Now let's try to factor the inner expression x%5E2%2B2x%2B1



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Looking at the expression x%5E2%2B2x%2B1, we can see that the first coefficient is 1, the second coefficient is 2, and the last term is 1.



Now multiply the first coefficient 1 by the last term 1 to get %281%29%281%29=1.



Now the question is: what two whole numbers multiply to 1 (the previous product) and add to the second coefficient 2?



To find these two numbers, we need to list all of the factors of 1 (the previous product).



Factors of 1:

1

-1



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to 1.

1*1 = 1
(-1)*(-1) = 1


Now let's add up each pair of factors to see if one pair adds to the middle coefficient 2:



First NumberSecond NumberSum
111+1=2
-1-1-1+(-1)=-2




From the table, we can see that the two numbers 1 and 1 add to 2 (the middle coefficient).



So the two numbers 1 and 1 both multiply to 1 and add to 2



Now replace the middle term 2x with x%2Bx. Remember, 1 and 1 add to 2. So this shows us that x%2Bx=2x.



x%5E2%2Bhighlight%28x%2Bx%29%2B1 Replace the second term 2x with x%2Bx.



%28x%5E2%2Bx%29%2B%28x%2B1%29 Group the terms into two pairs.



x%28x%2B1%29%2B%28x%2B1%29 Factor out the GCF x from the first group.



x%28x%2B1%29%2B1%28x%2B1%29 Factor out 1 from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



%28x%2B1%29%28x%2B1%29 Combine like terms. Or factor out the common term x%2B1



%28x%2B1%29%5E2 Condense the terms.



--------------------------------------------------



So 4%28x%5E2%2B2x%2B1%29 then factors further to 4%28x%2B1%29%5E2



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Answer:



So 4%2Ax%5E2%2B8%2Ax%2B4 completely factors to 4%28x%2B1%29%5E2.



In other words, 4%2Ax%5E2%2B8%2Ax%2B4=4%28x%2B1%29%5E2.



Note: you can check the answer by expanding 4%28x%2B1%29%5E2 to get 4%2Ax%5E2%2B8%2Ax%2B4 or by graphing the original expression and the answer (the two graphs should be identical).


Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression 4x%5E2%2B8x%2B3, we can see that the first coefficient is 4, the second coefficient is 8, and the last term is 3.



Now multiply the first coefficient 4 by the last term 3 to get %284%29%283%29=12.



Now the question is: what two whole numbers multiply to 12 (the previous product) and add to the second coefficient 8?



To find these two numbers, we need to list all of the factors of 12 (the previous product).



Factors of 12:

1,2,3,4,6,12

-1,-2,-3,-4,-6,-12



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to 12.

1*12 = 12
2*6 = 12
3*4 = 12
(-1)*(-12) = 12
(-2)*(-6) = 12
(-3)*(-4) = 12


Now let's add up each pair of factors to see if one pair adds to the middle coefficient 8:



First NumberSecond NumberSum
1121+12=13
262+6=8
343+4=7
-1-12-1+(-12)=-13
-2-6-2+(-6)=-8
-3-4-3+(-4)=-7




From the table, we can see that the two numbers 2 and 6 add to 8 (the middle coefficient).



So the two numbers 2 and 6 both multiply to 12 and add to 8



Now replace the middle term 8x with 2x%2B6x. Remember, 2 and 6 add to 8. So this shows us that 2x%2B6x=8x.



4x%5E2%2Bhighlight%282x%2B6x%29%2B3 Replace the second term 8x with 2x%2B6x.



%284x%5E2%2B2x%29%2B%286x%2B3%29 Group the terms into two pairs.



2x%282x%2B1%29%2B%286x%2B3%29 Factor out the GCF 2x from the first group.



2x%282x%2B1%29%2B3%282x%2B1%29 Factor out 3 from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



%282x%2B3%29%282x%2B1%29 Combine like terms. Or factor out the common term 2x%2B1



===============================================================



Answer:



So 4%2Ax%5E2%2B8%2Ax%2B3 factors to %282x%2B3%29%282x%2B1%29.



In other words, 4%2Ax%5E2%2B8%2Ax%2B3=%282x%2B3%29%282x%2B1%29.



Note: you can check the answer by expanding %282x%2B3%29%282x%2B1%29 to get 4%2Ax%5E2%2B8%2Ax%2B3 or by graphing the original expression and the answer (the two graphs should be identical).


Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve 4%2Ax%5E2%2B8%2Ax%2B2=0 ( notice a=4, b=8, and c=2)





x+=+%28-8+%2B-+sqrt%28+%288%29%5E2-4%2A4%2A2+%29%29%2F%282%2A4%29 Plug in a=4, b=8, and c=2




x+=+%28-8+%2B-+sqrt%28+64-4%2A4%2A2+%29%29%2F%282%2A4%29 Square 8 to get 64




x+=+%28-8+%2B-+sqrt%28+64%2B-32+%29%29%2F%282%2A4%29 Multiply -4%2A2%2A4 to get -32




x+=+%28-8+%2B-+sqrt%28+32+%29%29%2F%282%2A4%29 Combine like terms in the radicand (everything under the square root)




x+=+%28-8+%2B-+4%2Asqrt%282%29%29%2F%282%2A4%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%28-8+%2B-+4%2Asqrt%282%29%29%2F8 Multiply 2 and 4 to get 8


So now the expression breaks down into two parts


x+=+%28-8+%2B+4%2Asqrt%282%29%29%2F8 or x+=+%28-8+-+4%2Asqrt%282%29%29%2F8



Now break up the fraction



x=-8%2F8%2B4%2Asqrt%282%29%2F8 or x=-8%2F8-4%2Asqrt%282%29%2F8



Simplify



x=-1%2Bsqrt%282%29%2F2 or x=-1-sqrt%282%29%2F2



So the solutions are:

x=-1%2Bsqrt%282%29%2F2 or x=-1-sqrt%282%29%2F2



Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for x:


Starting with the general quadratic


ax%5E2%2Bbx%2Bc=0


the general solution using the quadratic equation is:


x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve 4%2Ax%5E2%2B8%2Ax%2B1=0 ( notice a=4, b=8, and c=1)





x+=+%28-8+%2B-+sqrt%28+%288%29%5E2-4%2A4%2A1+%29%29%2F%282%2A4%29 Plug in a=4, b=8, and c=1




x+=+%28-8+%2B-+sqrt%28+64-4%2A4%2A1+%29%29%2F%282%2A4%29 Square 8 to get 64




x+=+%28-8+%2B-+sqrt%28+64%2B-16+%29%29%2F%282%2A4%29 Multiply -4%2A1%2A4 to get -16




x+=+%28-8+%2B-+sqrt%28+48+%29%29%2F%282%2A4%29 Combine like terms in the radicand (everything under the square root)




x+=+%28-8+%2B-+4%2Asqrt%283%29%29%2F%282%2A4%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




x+=+%28-8+%2B-+4%2Asqrt%283%29%29%2F8 Multiply 2 and 4 to get 8


So now the expression breaks down into two parts


x+=+%28-8+%2B+4%2Asqrt%283%29%29%2F8 or x+=+%28-8+-+4%2Asqrt%283%29%29%2F8



Now break up the fraction



x=-8%2F8%2B4%2Asqrt%283%29%2F8 or x=-8%2F8-4%2Asqrt%283%29%2F8



Simplify



x=-1%2Bsqrt%283%29%2F2 or x=-1-sqrt%283%29%2F2



So the solutions are:

x=-1%2Bsqrt%283%29%2F2 or x=-1-sqrt%283%29%2F2

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me with this question. I have to solve by whichever method is easier. ( factoring, completing the square or the quadratic formula)
A) what is the discriminant of the equation 4x^2+8x+k=0
B) for what value of k will the equation have a double root?
C)for what values of k will the equation have two real roots?
D) for what values of k will the equation have imaginary roots?
E) name three values of k for which the given equation has rational roots


Equation: 4x%5E2+%2B+8x+%2B+k+=+0+

A)	Discriminant: 8%5E2+-+4%284%29%28k%29, or highlight_green%2864+-+16k%29 

B)	64 - 16k = 0___64 = 16k___k = 4. Double roots exist when k has a value of 4.

C)	64 – 16k ≥ 0____- 16k ≥ - 64____k  ≤ %28-+64%29%2F-+16, or k ≤ 4. Two real-roots exist when k has values ≤ 4.

D)	64 – 16k < 0____- 16k < - 64____k > %28-+64%29%2F-+16, or k > 4. Imaginary roots exist when k has values > 4.

E)	64 – 16k ≥ 0, AND A PERFECT SQUARE. 64 – 16k ≥ 0____- 16k ≥ - 64____k ≤ %28-+64%29%2F-+16, or k ≤ 4.

        Rational roots exist when the k-values are ≤ 4, and DISCRIMINANT is a PERFECT SQUARE. Thus, three

        values of k that are ≤ 4, and which produces RATIONAL ROOTS VALUES (PERFECT SQUARES) are: highlight_green%28system+%2855%2F16%2C+3_and%2C%0D%0A0%29%29

You can do a check!!