Mathematics - Quadratic Equation Question with Solution | TestHub

Let the roots be $\alpha$ and ${\alpha }^2$. Since the coefficients of the quadratic are real, if one root is complex, the other must be its conjugate. If $z$ is a complex root and $z^2=\bar{z}$, then $|z{|}^2=|\bar{z}|=|z|$, implying $|z|=1$. Also, $z^3=z\cdot z^2=z\cdot \bar{z}=|z{|}^2=1$. The complex cube roots of unity are $\omega$ and ${\omega }^2$. If the roots are $\omega$ and ${\omega }^2$, their sum is $-1$ and product is $1$. From Vieta's formulas for the given equation, $k-2=-1⟹k=1$ and $k^2-3k+1=1$. Substituting $k=1$ into the second equation gives $1^2-3(1)+1=-1\ne 1$. Thus, there are no complex roots satisfying the condition. The roots must be real. Using Vieta's formulas for real roots $\alpha$ and ${\alpha }^2$: 1) $\alpha +{\alpha }^2=k-2$ 2) ${\alpha }^3=k^2-3k+1$ From (1), $k={\alpha }^2+\alpha +2$. Substitute this into (2):${\alpha }^3=({\alpha }^2+\alpha +2{)}^2-3({\alpha }^2+\alpha +2)+1$ ${\alpha }^3=({\alpha }^4+2{\alpha }^3+5{\alpha }^2+4\alpha +4)-(3{\alpha }^2+3\alpha +6)+1$ ${\alpha }^3={\alpha }^4+2{\alpha }^3+2{\alpha }^2+\alpha -1$ ${\alpha }^4+{\alpha }^3+2{\alpha }^2+\alpha -1=0$ By inspection, $\alpha =-1$ is a root: $1-1+2-1-1=0$. So $(\alpha +1)$ is a factor. Dividing the quartic by $(\alpha +1)$ yields $(\alpha +1)({\alpha }^3+2\alpha -1)=0$. Case 1: $\alpha =-1$. From (1), $k-2=-1+(-1{)}^2=0⟹k=2$.

For $k=2$, the quadratic equation is $x^2-(2-2)x+(2^2-3(2)+1)=0⟹x^2-1=0$. The roots are $x=\pm 1$. These are real. One root is $1$, the other is $-1$. Since $1=(-1{)}^2$, the condition is satisfied. So $k=2$ is a valid value. Case 2: ${\alpha }^3+2\alpha -1=0$.

Let $f(\alpha )={\alpha }^3+2\alpha -1$. $f(0)=-1$ and $f(1)=2$. Since $f^{\prime }(\alpha )=3{\alpha }^2+2>0$, $f(\alpha )$ is strictly increasing, so there is exactly one real root, say ${\alpha }_0$, and $0<{\alpha }_0<1$. For the original quadratic to have real roots, its discriminant $D\ge 0$.$D=(k-2{)}^2-4(k^2-3k+1)=k^2-4k+4-4k^2+12k-4=-3k^2+8k=k(8-3k)$. So $k(8-3k)\ge 0⟹0\le k\le 8/3$. For the root ${\alpha }_0\in (0,1)$, the corresponding value of $k$ is $k_0={\alpha }_0^2+{\alpha }_0+2$. Since $0<{\alpha }_0<1$, we have $0<{\alpha }_0^2<1$. Thus, $0<{\alpha }_0^2+{\alpha }_0<2$, which implies $2<k_0<4$. Combining this with the discriminant condition $0\le k_0\le 8/3$, we need $k_0\in (2,8/3]$. We derived the quartic equation in $k$: $(k-2)(k^3-4k^2-k-4)=0$. Let $g(k)=k^3-4k^2-k-4$. We need to check if $g(k)=0$ has any roots in $(2,8/3]$.$g(2)=8-4(4)-2-4=-14$.$g(8/3)=(8/3{)}^3-4(8/3{)}^2-(8/3)-4=512/27-256/9-8/3-4=(512-768-72-108)/27=-436/27$.

Both $g(2)$ and $g(8/3)$ are negative. Also, $g^{\prime }(k)=3k^2-8k-1$. The roots of $g^{\prime }(k)=0$ are $k=\frac{8\pm \sqrt{64+12}}{6}=\frac{4\pm \sqrt{19}}{3}$. The local minimum is at $k_2=\frac{4+\sqrt{19}}{3}\approx 2.78$. Since $8/3\approx 2.67<k_2$, $g(k)$ is decreasing on $(2,8/3]$. As $g(2)<0$ and $g(8/3)<0$, $g(k)$ remains negative throughout this interval. Thus, $g(k)=0$ has no roots in $(2,8/3]$. Therefore, $k=2$ is the only possible real value of $k$. The product of all possible real values of $k$ is $2$.