$\large x^{4}+x^{2}(1-2k)+k^{2}-1=0$
Find the value of constant $k$ for which the equation above has 3 distinct real solutions.
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Nice interpretation of the problem.
$x^4+(1-2k)x^2+k^2-1 = 0$ is a quadratic equation for $x^2$ . In general there are either no real root or two real root of $x^2$ . When there are two positive real roots of $x^2$ , there can be four roots of $x$ . For the equation to have three real roots of $x$ , one of the real root of $x^2$ must be $0$ . Therefore, the equation is of the form $x^2(x^2-a) = 0$ since there is no $x^3$ term. This is possible when $k^2-1=0$ $\Longrightarrow k = \pm 1$ and the equation is either:
$\begin{cases} k = 1 & \Rightarrow x^2(x^2-1) & \Rightarrow x = -1,0,1 & \text{3 real roots} \\ k = -1 & \Rightarrow x^2(x^2+3) & \Rightarrow x = 0 & \text{1 real root -- rejected} \end{cases} \\ \Rightarrow k = \boxed{1}$
Why is this true?
For the equation to have three real roots of $x$ , one of the real root of $x^2$ must be $0$ .
When there is two real roots of $x^2$ , then there are four real roots of $x$ ? $x^4 - x^2 - 2 = 0$ have real roots of $x^2=2,-1$ but only two real roots of $x$ .
I was trying to explain that the quadratic equation is of the form of $(x^2-\alpha)(x^2-\beta)$ . For real roots of $x$ , $\alpha$ and $\beta$ must be non-negative. If $\alpha, \beta < 0$ , then there are four real roots of $x$ . To have only three roots a pair of roots must be repeated. This is only possible only when either $\alpha$ or $\beta$ is $0$ , because $+0=-0$ .
Let $x^{2}=t$ then the given equation transforms into a quadratic.
$t^{2}+t(1-2k)+k^{2}-1$
Now since the biquadratic equation has 3 solutions, the quadratic expression formed will satisfy the following conditions.
$D>0 \implies k<\dfrac{5}{4}$
$\dfrac{-b}{2a} >0 \implies k>\dfrac{1}{2}$
$t=0$ is a zero of the quadratic $\implies k=±1$
Intersection of all these conditions gives $k=1$ .
There is a need to modify the question for clarity.
First of all, an equation needs to have an equality relation.
The question should read
$x^4+x^2(1-2k)+k^2-1=0$
Secondly, the question does not specify if the three solutions need to be 'real'. Without the restriction $k=-1$ would also be a valid answer.
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Clearly, the graph of the equation is symmetric about x=0; so, the no. of roots on left hand side and right side would be equal which implies that there will be even number of roots unless there is a root( which is actually a double root) at the origin. So, x=0 is a root of the equation; hence k=1,-1. But, since x is real, k=1