Patrick D. answered • 05/18/17
Tutor
5
(10)
Patrick the Math Doctor
4 and 1/3 = 13/3
Let R be the smaller root.
Then the bigger root is R + 13/3
Substituting the bigger root into the equation:
3 ( R + 13/3)^2 + B(R +13/3) + 10 = 0
3 ( R^2 + 26/3 R + 169/9) + B ( R + 13/3) + 10 = 0
3*R^2 + 26*R + 169/3 + B ( R + 13/3) + 10 = 0
Multiplying everything by 3 and distributing it:
9 * r^2 + 78*R + 169 + B ( 3*R + 13) + 30 = 0
9 * r^2 + 78*R + 199 + B ( 3*R + 13) = 0
Solving for B:
B = - [ 9 * r^2 + 78*R + 199 ]/(3*R + 13) <--- label this equation ALPHA
provided that R is NOT EQUAL to -13/3
Plugging the smaller root R into the equation is simply:
3*R^2 + BR + 10 = 0
Solving for B :
B = - [ 3*R^2 + 10]/R, <--- label this equation BETA
provided that R is NOT zer0
Equations ALPHA and BETA are EQUAL because they both equal B:
- [ 9 * r^2 + 78*R + 199 ]/(3*R + 13) = - [ 3*R^2 + 10]/R
The negative signs cancel by multiplying both sides by -1.
Cross Multiplication:
[ 9 * r^2 + 78*R + 199 ] R = (3*R + 13)[ 3*R^2 + 10]
Distributing R on the left side and FOILing out the right side:
9 * r^3 + 78*R^2 + 199*R = 9 * r^3 + 39*R^2 + 30*R + 130
Moving everything to the left side, the 9*r^3 term cancels:
39*R^2 + 169*R - 130 = 0
13( 3*R^2 + 13*R - 10) = 0
13(3*R - 2 )(R + 5 ) = 0 <-- completely factored
The 13 divides out and cancels. By zero product property:
3*R - 2 = 0 ---> R = 2/3
R + 5 = 0 --> R = -5
R is the smaller root, so for R = 2/3, the larger root is 2/3+13/3 = 15/3 = 5
So the roots are X=2/3 and X=5.
Then the factors are (3X - 2)=0 and (X-5)=0
FOILing these two factors gives the quadratic 3x^2 -17X + 10 = 0
Likewise, for R=-5, the larger root is -5 + 13/3 = -15/3 + 13/3 = -2/3
So the roots are X = -5 and X = -2/3
Then the factors are (X+5)=0 and 3x + 2 = 0
FOILING these two factors also gives the quadratic 3x^2 + 17x + 10 = 0
comparing these to the original quadratic,
B = 17 or B = -17
Let R be the smaller root.
Then the bigger root is R + 13/3
Substituting the bigger root into the equation:
3 ( R + 13/3)^2 + B(R +13/3) + 10 = 0
3 ( R^2 + 26/3 R + 169/9) + B ( R + 13/3) + 10 = 0
3*R^2 + 26*R + 169/3 + B ( R + 13/3) + 10 = 0
Multiplying everything by 3 and distributing it:
9 * r^2 + 78*R + 169 + B ( 3*R + 13) + 30 = 0
9 * r^2 + 78*R + 199 + B ( 3*R + 13) = 0
Solving for B:
B = - [ 9 * r^2 + 78*R + 199 ]/(3*R + 13) <--- label this equation ALPHA
provided that R is NOT EQUAL to -13/3
Plugging the smaller root R into the equation is simply:
3*R^2 + BR + 10 = 0
Solving for B :
B = - [ 3*R^2 + 10]/R, <--- label this equation BETA
provided that R is NOT zer0
Equations ALPHA and BETA are EQUAL because they both equal B:
- [ 9 * r^2 + 78*R + 199 ]/(3*R + 13) = - [ 3*R^2 + 10]/R
The negative signs cancel by multiplying both sides by -1.
Cross Multiplication:
[ 9 * r^2 + 78*R + 199 ] R = (3*R + 13)[ 3*R^2 + 10]
Distributing R on the left side and FOILing out the right side:
9 * r^3 + 78*R^2 + 199*R = 9 * r^3 + 39*R^2 + 30*R + 130
Moving everything to the left side, the 9*r^3 term cancels:
39*R^2 + 169*R - 130 = 0
13( 3*R^2 + 13*R - 10) = 0
13(3*R - 2 )(R + 5 ) = 0 <-- completely factored
The 13 divides out and cancels. By zero product property:
3*R - 2 = 0 ---> R = 2/3
R + 5 = 0 --> R = -5
R is the smaller root, so for R = 2/3, the larger root is 2/3+13/3 = 15/3 = 5
So the roots are X=2/3 and X=5.
Then the factors are (3X - 2)=0 and (X-5)=0
FOILing these two factors gives the quadratic 3x^2 -17X + 10 = 0
Likewise, for R=-5, the larger root is -5 + 13/3 = -15/3 + 13/3 = -2/3
So the roots are X = -5 and X = -2/3
Then the factors are (X+5)=0 and 3x + 2 = 0
FOILING these two factors also gives the quadratic 3x^2 + 17x + 10 = 0
comparing these to the original quadratic,
B = 17 or B = -17
