Truth of quadratic equations statements

Question

Which of the following statements regarding quadratic equations are true? (A) Solution set for the equation $6x^2 - 5x = 4$ is $\{-\frac{1}{2}, \frac{4}{3}\}$ (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 $K = \frac{27}{256}$

Accepted Answer

B. (A) $6x^2 - 5x - 4 = 0 \implies (2x+1)(3x-4)=0 \implies x = -1/2, 4/3$. True. (B) $9x^2 + 6x + 1 = (3x+1)^2 = 0$. Discriminant $D = 36 - 36 = 0$. Roots are equal, real, and rational ($-1/3, -1/3$). True. (C) Let roots be $r$ and $3r$. Sum of roots = $4r = 3/4 \implies r = 3/16$. Product of roots = $3r^2 = K/4 \implies K = 12 r^2 = 12 \left(\frac{3}{16}ight)^2 = 12 	imes \frac{9}{256} = \frac{27}{64} 
eq \frac{27}{256}$. False. Hence, only (A) and (B) are true.

Answer

(A),(B) and (C)

Answer

(A) and (B) only

Answer

(B) and (C) only

Answer

(C) and (A) only

Truth of quadratic equations statements

Explanation

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