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Which of the following statements regarding quadratic equations are true?
(A) Solution set for the equation $$6x^{2}-5x=4$$ is $$\left\{-\frac{1}{2},\frac{4}{3}\right\}$$
(B) The nature of the roots of the equation $$9x^{2}+6x+1=0$$ is equal, real and rational.
(C) If one root of $$4x^{2}-3x+K=0$$ is 3 times the other, then $$K=\frac{27}{256}$$
We have :
$$6x^2-5x-4\ =0\ $$
=$$6x^2-8x+3x-4\ =0\ $$
(2x+1)(3x-4)=0
so solution set is {-1/2 ,4/3}
$$9x^2+6x+1\ =0$$
D = 36-36 =0
So we can say roots are real and equal
$$4x^2-3x+k$$=0
let roots be p and 3p
we get 4p =3/4
p =3/16
Now k/4 = 3p^2
we get k = 12p^2
we get k = 27/64
So A and B are true and C is false
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