Question

If $x^8−(k−1)x^4+5=0$, then least possible integral value of ' k ' so that equation has maximum number of real roots is :

Answer 1

6

Answer 2

Let the given equation be $x^8 - (k-1)x^4 + 5 = 0$. This is a polynomial equation of degree 8, so it can have at most 8 real roots. The question asks for the least possible integral value of 'k' such that the equation has the maximum number of real roots.

Let $y = x^4$. Since $x$ is a real number, $y = x^4 \ge 0$. Substituting $y$ into the equation, we get a quadratic equation in $y$: $y^2 - (k-1)y + 5 = 0$.

The number of real roots $x$ depends on the number and nature of the roots $y$ of this quadratic equation. For a real value $y$:

• If $y > 0$, the equation $x^4 = y$ has two distinct real roots: $x = \pm y^{1/4}$.
• If $y = 0$, the equation $x^4 = 0$ has one real root: $x = 0$.
• If $y < 0$, the equation $x^4 = y$ has no real roots.

Let the roots of the quadratic $y^2 - (k-1)y + 5 = 0$ be $y_1$ and $y_2$. To maximize the number of real roots for $x$, we need to maximize the number of positive roots for $y$.

Case 1: The quadratic equation in $y$ has two distinct positive real roots ($y_1 > 0, y_2 > 0, y_1 \ne y_2$). For $y_1 > 0$, $x^4 = y_1$ gives two distinct real roots $\pm y_1^{1/4}$. For $y_2 > 0$, $x^4 = y_2$ gives two distinct real roots $\pm y_2^{1/4}$. Since $y_1 \ne y_2$ and both are positive, $y_1^{1/4} \ne y_2^{1/4}$. The four roots for $x$ are $y_1^{1/4}, -y_1^{1/4}, y_2^{1/4}, -y_2^{1/4}$. These are 4 distinct real roots. The total number of real roots for $x$ is 4.

Case 2: The quadratic equation in $y$ has one positive real root with multiplicity 2 ($y_1 = y_2 = y_0 > 0$). For $y_0 > 0$, $x^4 = y_0$ gives two distinct real roots $\pm y_0^{1/4}$. The total number of real roots for $x$ is 2.

Case 3: The quadratic equation in $y$ has one positive real root and one zero root ($y_1 > 0, y_2 = 0$). If $y=0$ is a root of $y^2 - (k-1)y + 5 = 0$, then $0^2 - (k-1)(0) + 5 = 0$, which implies $5=0$. This is impossible. So, the quadratic equation cannot have a root $y=0$.

Case 4: The quadratic equation in $y$ has one positive real root and one negative real root ($y_1 > 0, y_2 < 0$). For $y_1 > 0$, $x^4 = y_1$ gives two distinct real roots $\pm y_1^{1/4}$. For $y_2 < 0$, $x^4 = y_2$ has no real roots. The total number of real roots for $x$ is 2.

Case 5: The quadratic equation in $y$ has roots that are zero or negative. If $y_1=0$ or $y_2=0$, impossible (as shown in Case 3). If $y_1 < 0$ or $y_2 < 0$, there are no real roots for $x$ from these $y$ values. If both roots are negative or complex, there are no positive real roots for $y$, so there are no real roots for $x$.

From this analysis, the maximum number of real roots for $x$ is 4, which occurs when the quadratic in $y$ has two distinct positive real roots.

Let's verify this by considering the number of roots of $x^4=c$. If $c>0$, $x^4=c$ has two real roots. If $c=0$, $x^4=0$ has one real root. If $c<0$, $x^4=c$ has zero real roots.

Let the roots of $y^2 - (k-1)y + 5 = 0$ be $y_1, y_2$. The roots of the original equation are the real solutions to $x^4=y_1$ and $x^4=y_2$. The total number of real roots is $N(x^4=y_1) + N(x^4=y_2)$, where $N(x^4=c)$ is the number of real roots of $x^4=c$.

$N(x^4=c) = 2$ if $c>0$ $N(x^4=c) = 1$ if $c=0$ $N(x^4=c) = 0$ if $c<0$

To maximize the number of real roots for $x$, we need $y_1$ and $y_2$ to be positive. If $y_1 > 0$ and $y_2 > 0$, the number of real roots is $2+2=4$. If $y_1 > 0$ and $y_2 = 0$, impossible. If $y_1 > 0$ and $y_2 < 0$, the number of real roots is $2+0=2$. If $y_1 = 0$ and $y_2 = 0$, impossible. If $y_1 = 0$ and $y_2 < 0$, impossible. If $y_1 < 0$ and $y_2 < 0$, the number of real roots is $0+0=0$.

The maximum number of real roots is 4, achieved when the quadratic in $y$ has two distinct positive real roots.

For the quadratic equation $ay^2+by+c=0$ to have two distinct positive real roots, the following conditions must be met:

1. The discriminant must be positive: $D > 0$.
2. The sum of the roots must be positive: $y_1 + y_2 > 0$.
3. The product of the roots must be positive: $y_1 y_2 > 0$.

For $y^2 - (k-1)y + 5 = 0$:

1. $D = (-(k-1))^2 - 4(1)(5) = (k-1)^2 - 20$. For distinct real roots, $D > 0 \implies (k-1)^2 - 20 > 0 \implies (k-1)^2 > 20$. This means $k-1 > \sqrt{20}$ or $k-1 < -\sqrt{20}$. $k-1 > 2\sqrt{5} \approx 2 \times 2.236 = 4.472 \implies k > 5.472$. $k-1 < -2\sqrt{5} \approx -4.472 \implies k < -3.472$.

2. Sum of roots: $y_1 + y_2 = -(-(k-1))/1 = k-1$. For positive roots, $y_1 + y_2 > 0 \implies k-1 > 0 \implies k > 1$.

3. Product of roots: $y_1 y_2 = 5/1 = 5$. The product of roots is 5, which is positive. This condition is always satisfied.

We need to satisfy $k > 5.472$ or $k < -3.472$ (for distinct real roots) AND $k > 1$ (for positive sum of roots). Combining these conditions: If $k < -3.472$, then $k < 1$, so $k>1$ is not satisfied. If $k > 5.472$, then $k > 1$ is satisfied. So, for two distinct positive real roots for $y$, we need $k > 5.472$.

The question asks for the least possible integral value of $k$. The integers greater than 5.472 are 6, 7, 8, ... The least integer value of $k$ satisfying $k > 5.472$ is 6.

Let's check $k=6$: The quadratic in $y$ is $y^2 - (6-1)y + 5 = 0 \implies y^2 - 5y + 5 = 0$. The roots are $y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(5)}}{2(1)} = \frac{5 \pm \sqrt{25 - 20}}{2} = \frac{5 \pm \sqrt{5}}{2}$. $y_1 = \frac{5 + \sqrt{5}}{2}$ and $y_2 = \frac{5 - \sqrt{5}}{2}$. $\sqrt{5} \approx 2.236$. $y_1 \approx \frac{5 + 2.236}{2} = \frac{7.236}{2} = 3.618 > 0$. $y_2 \approx \frac{5 - 2.236}{2} = \frac{2.764}{2} = 1.382 > 0$. The roots $y_1$ and $y_2$ are distinct and positive. So, for $y_1$, $x^4 = y_1$ gives $\pm y_1^{1/4}$ (2 real roots). For $y_2$, $x^4 = y_2$ gives $\pm y_2^{1/4}$ (2 real roots). Since $y_1 \ne y_2$, $y_1^{1/4} \ne y_2^{1/4}$. The four roots $\pm y_1^{1/4}, \pm y_2^{1/4}$ are distinct. The total number of real roots for $x$ is 4.

The maximum number of real roots for this type of equation $x^{2n} + ax^n + b = 0$ is $2 \times (\text{number of distinct positive roots of } y^2+ay+b=0)$, where $y=x^n$. In our case $n=4$. The maximum number of real roots is $2 \times 2 = 4$.

The question asks for the least possible integral value of 'k' so that the equation has maximum number of real roots. The maximum number of real roots is 4. This occurs when $k \ge 6$. The least integral value is 6.

Final check: If $k=5$, $y^2 - 4y + 5 = 0$. $D = (-4)^2 - 4(5) = 16 - 20 = -4 < 0$. No real roots for $y$, so no real roots for $x$. If $k=-3$, $y^2 - (-4)y + 5 = 0 \implies y^2 + 4y + 5 = 0$. $D = 4^2 - 4(5) = 16 - 20 = -4 < 0$. No real roots for $y$, so no real roots for $x$. If $k=-4$, $y^2 - (-5)y + 5 = 0 \implies y^2 + 5y + 5 = 0$. $D = 5^2 - 4(5) = 25 - 20 = 5 > 0$. Roots are $y = \frac{-5 \pm \sqrt{5}}{2}$. Both roots are negative. No real roots for $x$.

The condition for maximum number of real roots (4) is $k > 5.472$. The least integer value is 6.

The final answer is $\boxed{6}$.

Explanation of the solution: The equation $x^8 - (k-1)x^4 + 5 = 0$ can be transformed into a quadratic equation in $y=x^4$ as $y^2 - (k-1)y + 5 = 0$. For each positive real root $y$ of this quadratic, there are two real roots for $x$ (namely $\pm y^{1/4}$). If $y=0$ is a root, there is one real root for $x$ ($x=0$). If $y<0$ is a root, there are no real roots for $x$. The quadratic $y^2 - (k-1)y + 5 = 0$ cannot have a root $y=0$ since the constant term is 5. To maximize the number of real roots for $x$, the quadratic in $y$ must have the maximum possible number of distinct positive real roots. This occurs when the quadratic has two distinct positive real roots. The conditions for this are: discriminant $D > 0$, sum of roots $> 0$, and product of roots $> 0$. For $y^2 - (k-1)y + 5 = 0$:

1. $D = (k-1)^2 - 20 > 0 \implies (k-1)^2 > 20 \implies k-1 > \sqrt{20}$ or $k-1 < -\sqrt{20}$. $k > 1 + \sqrt{20} \approx 5.472$ or $k < 1 - \sqrt{20} \approx -3.472$.
2. Sum of roots $= k-1 > 0 \implies k > 1$.
3. Product of roots $= 5 > 0$ (always true). Combining these conditions, we need $k > 5.472$. The least integer value of $k$ satisfying this condition is 6. For $k=6$, the quadratic $y^2 - 5y + 5 = 0$ has two distinct positive roots $y = (5 \pm \sqrt{5})/2$. Each positive root $y$ gives two real roots for $x$, resulting in a total of $2+2=4$ distinct real roots for $x$. This is the maximum number of real roots for this type of equation.

The final answer is $\boxed{6}$.