4 x ^ 2 + 9 = kx
4 x ^ 2 - k x + 9 = 0
Quadratic equation has two real roots, if Discrininant Ä = 0
Ä = b ^ 2 - 4 * a * c
In this case :
a = 4
b = - k
c = 9
Ä = b ^ 2 - 4 * a * c
Ä = ( - k ) ^ 2 - 4 * ( 4 ) * 9
Ä = k ^ 2 - 144 = 0
k ^ 2 - 144 = 0
k ^ 2 = 144
k = sqrt ( 144 )
k = ± 12
For k = - 12
4 x ^ 2 + 9 = k x
not equal becouse left side of equation are positive ( 4 x ^ 2 + 9 ) are allways great of 0 ) ,and right side of equation are negative.
So solution are k = 12
If 4x^2+9=kx, what value of k will produce equal roots?
the answer is 12 but i don't understand how you get it, please help!
2 answers
Quadratic equation has two real equals roots, if Discrininant Ä = 0
Ä = greek letter Delta
Ä = greek letter Delta