Question

If all the roots of equation $x^4-(2k+1)x^2+k^2-3k=0$ are real then k can be

• A. -2
• B. -1
• C. 4
• D. 7

Answer 1

Answer: (C), (D)

4, 7

Answer 2

Let the given equation be $x^4-(2k+1)x^2+k^2-3k=0$. This is a biquadratic equation. Let $y = x^2$. Substituting $y$ into the equation, we get a quadratic equation in terms of $y$: $y^2-(2k+1)y+k^2-3k=0$.

For all roots of the original equation ($x$) to be real, $x^2$ must be real and non-negative. This means both roots of the quadratic equation in $y$ (let's call them $y_1$ and $y_2$) must be real and non-negative. For $y_1, y_2 \ge 0$, the following conditions must be satisfied:

1. Discriminant ($D$) must be non-negative (for real roots of $y$): $D = (-(2k+1))^2 - 4(1)(k^2-3k)$ $D = (2k+1)^2 - 4(k^2-3k)$ $D = (4k^2+4k+1) - (4k^2-12k)$ $D = 4k^2+4k+1-4k^2+12k$ $D = 16k+1$ For real roots, $D \ge 0 \implies 16k+1 \ge 0 \implies 16k \ge -1 \implies k \ge -\frac{1}{16}$.

2. Sum of roots ($y_1+y_2$) must be non-negative: For the quadratic equation $ay^2+by+c=0$, the sum of roots is $-b/a$. Here, $y_1+y_2 = -(-(2k+1))/1 = 2k+1$. For non-negative roots, $y_1+y_2 \ge 0 \implies 2k+1 \ge 0 \implies 2k \ge -1 \implies k \ge -\frac{1}{2}$.

3. Product of roots ($y_1y_2$) must be non-negative: For the quadratic equation $ay^2+by+c=0$, the product of roots is $c/a$. Here, $y_1y_2 = (k^2-3k)/1 = k^2-3k$. For non-negative roots, $y_1y_2 \ge 0 \implies k^2-3k \ge 0 \implies k(k-3) \ge 0$. This inequality holds when $k \le 0$ or $k \ge 3$.

Now, we need to find the values of $k$ that satisfy all three conditions simultaneously:

• Condition 1: $k \ge -\frac{1}{16}$
• Condition 2: $k \ge -\frac{1}{2}$
• Condition 3: $k \in (-\infty, 0] \cup [3, \infty)$

Combining Condition 1 and Condition 2: $k \ge -\frac{1}{16}$ and $k \ge -\frac{1}{2}$. The stricter of these two is $k \ge -\frac{1}{16}$.

Intersect $k \ge -\frac{1}{16}$ with Condition 3 ($k \le 0$ or $k \ge 3$):

• Case 1: $k \ge -\frac{1}{16}$ AND $k \le 0$. This implies $-\frac{1}{16} \le k \le 0$.

• Case 2: $k \ge -\frac{1}{16}$ AND $k \ge 3$. This implies $k \ge 3$.

Combining these two cases, the set of all possible values for $k$ is $[-\frac{1}{16}, 0] \cup [3, \infty)$.

Check the given options against this range: (A) -2: Not in the range, as $-2 < -\frac{1}{16}$. (B) -1: Not in the range, as $-1 < -\frac{1}{16}$. (C) 4: Is in the range, as $4 \ge 3$. (D) 7: Is in the range, as $7 \ge 3$.

Both options (C) and (D) satisfy the condition for $k$.