SOLUTION: What is the correct substitution in the quadratic formula for the solution of the following
8x^2+3a^2=10ax
A. x=-10a+-(100a^2-4*3a^2*10)^(1/2)
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Question 145098: What is the correct substitution in the quadratic formula for the solution of the following
8x^2+3a^2=10ax
A. x=-10a+-(100a^2-4*3a^2*10)^(1/2)
-----------------------------
2*8
B. x= -3a^2+-(9a^2-4*8+10a)^(1/2)
-------------------------------
2*8
c. x=10+-(100-4*8*3)^(1/2)
---------------------
2*8
d. x=10a+-(100a^2-4*8*3a^2)
----------------------
2*8
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
What is the correct substitution in the quadratic formula for the solution of the following
8x^2+3a^2=10ax-
Rearrange to get:
8x^2 - 10ax + 3a^2
Quadratic with a = 8 ; b = -10a ; c = 3a^2
-------------------------------------------
Using the quadratic formula you would get:
x = [-b +- sqrt(b^2 - 4ac)]/(2a)
-------
Substitute the values of a, b, c to get:
x = [-10a +- sqrt((10a)^2 - 4(8)(3a^2))]/(16)
x = [-10a +- sqrt(100a^2 - 96a^2)]/16
etc.
==============================================
A. x=-10a+-(100a^2-4*3a^2*10)^(1/2)
-----------------------------
2*8
B. x= -3a^2+-(9a^2-4*8+10a)^(1/2)
-------------------------------
2*8
c. x=10+-(100-4*8*3)^(1/2)
---------------------
2*8
d. x=10a+-(100a^2-4*8*3a^2)
Cheers,
Stan H.
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