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3x^(2)+4x-k=0 In the given equation, k is a constant. For what value of k does the equation have no real solutions? Choose 1 answer: (A) -(4)/(3) (B) (4)/(3) (c) -(5)/(3) (D) (5)/(3)

3x2+4x−k=0 3 x^{2}+4 x-k=0 \newlineIn the given equation, k k is a constant. For what value of k k does the equation have no real solutions?\newlineChoose 11 answer:\newline(A) −43 -\frac{4}{3} \newline(B) 43 \frac{4}{3} \newline(C) −53 -\frac{5}{3} \newline(D) 53 \frac{5}{3}

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Q. 3x2+4x−k=0 3 x^{2}+4 x-k=0 \newlineIn the given equation, k k is a constant. For what value of k k does the equation have no real solutions?\newlineChoose 11 answer:\newline(A) −43 -\frac{4}{3} \newline(B) 43 \frac{4}{3} \newline(C) −53 -\frac{5}{3} \newline(D) 53 \frac{5}{3}
  1. Identify Equation: The given equation is 3x2+4x−k=03x^2 + 4x - k = 0. Here, a=3a = 3, b=4b = 4, and c=−kc = -k. We will calculate the discriminant using these values.\newlineThe discriminant (DD) is D=b2−4ac=(4)2−4(3)(−k)D = b^2 - 4ac = (4)^2 - 4(3)(-k).
  2. Calculate Discriminant: Now we calculate the discriminant: D=16−(−12k)=16+12kD = 16 - (-12k) = 16 + 12k. For the equation to have no real solutions, the discriminant must be less than zero. Therefore, we set up the inequality 16 + 12k < 0.
  3. Set Up Inequality: We solve the inequality for kk: 12k < -16.\newlineDivide both sides by 1212 to isolate kk: k < -\frac{16}{12}.\newlineSimplify the fraction: k < -\frac{4}{3}.
  4. Solve for kk: The value of kk that makes the equation have no real solutions is any number less than −43-\frac{4}{3}. Looking at the answer choices, we see that the value that fits this condition is (C) −53-\frac{5}{3}, since −53-\frac{5}{3} is less than −43-\frac{4}{3}.

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