Question

Match the equation on the left with interval(s) in which its real roots lie.

Answer 1

A) Let $f\left(x\right)=(x-1)(x-3)(x-5)+\frac{2}{3}(x-2)(x-4)(x-6)$
Then from $f\left(1\right)<0,f\left(2\right)>0,f\left(3\right)>0,f\left(4\right)<0,f\left(5\right)<0,f\left(6\right)>0$
Thus roots of $f\left(x\right)=0$ lie in $[-7,7]$ and $[0,6]$
B) Let $\Delta \left(x\right)=\left|\begin{array}{ccc}(x+1)(x+2)& x+2& 1\\ (x+2)(x+3)& x+3& 1\\ (x+3)(x+4)& x+4& 1\end{array}\right|$
Applying $C_1\to C_1-(x+1)C_2,C_2\to C_2-(x+1)C_3$, we get
$\Delta \left(x\right)=-x-7$
And for $\Delta \left(x\right)=0$ roots lies between $[-7,7]$ and $[-7,0]$
C) Denote the determinant by .$\Delta \left(x\right)$.Multiply $C_1$ be $x$ and apply $C_1\to C_1+4C_2+2C_3$ to obtain
$\Delta \left(x\right)=\frac{1}{x}(x^2+20)\left|\begin{array}{ccc}1& 2& 6\\ 1& 4& 2-x\\ 1& 4+x& 2\end{array}\right|$
Apply $R_2\to R_3-R_2,R_2\to R_2-R_1$ to obtain
$\Delta \left(x\right)=\frac{1}{x}(x^2+20)\left|\begin{array}{ccc}1& 2& 6\\ 0& 2& -(x+4)\\ 0& -x& x\end{array}\right|$
$(x^2+20)(x+6)$
$\Delta \left(x\right)=0$has only one real root, viz. -6. It lies in $[-7,7]$ and $[-7,0]$
D) Denote the detenninant by $\Delta \left(x\right)$
Applying $C_1\to C_1+C_2+C_3$ we obtain
$\Delta \left(x\right)=(7-2x)\left|\begin{array}{ccc}1& 4-x& 2\\ 1& 2& 4-x\\ 1& 4-x& 2-x\end{array}\right|$
Use $R_2\to R_2-R_1,R_3\to R_3-R_1$
$\Delta \left(x\right)=(7-2x)\left|\begin{array}{ccc}1& 4-x& 2\\ 0& x-2& 2-x\\ 0& 0& -x\end{array}\right|$
$=(7-2x)(x-2)(-x)$
Thus, real roots of $\Delta \left(x\right)=0$ are $0,2$ and $7/2$ which lie in $[-7,7]$,$[0,4]$ and $[0,6]$