Let x_1,x_2,x_3,x_4 be the solution of the equation 4x^4+8x^3-17x^2-12x+9=0 and (4+x_2^1)(…

Question

Let ${x}_{1},{x}_{2},{x}_{3},{x}_{4}$ be the solution of the equation $4{x}^{4}+8{x}^{3}-17{x}^{2}-12x+9=0$ and $(4+{x}_{2}^{1})(4+{x}_{2}^{2})(4+{x}_{2}^{3})(4+{x}_{2}^{4})=\frac{125}{16}m$. Then the value of $m$ is \_\_\_\_\_\_\_\_\_.

Answer: 221

Accepted Answer

First, observe that for a degree-4 polynomial

$P(x)=4{x}^{4}+8{x}^{3}-17{x}^{2}-12x+9$

with roots ${x}_{1},\ldots ,{x}_{4}$, we have

$\prod _{4}^{i=1}({x}_{2}^{i}+4)=\prod _{4}^{i=1}[({x}_{i}-2i)({x}_{i}+2i)]=\frac{P(2i)\, P(-2i)}{{4}^{2}}\, .$

Compute $P(2i)$:

$P(2i)=4(2i{)}^{4}+8(2i{)}^{3}-17(2i{)}^{2}-12(2i)+9=64-64i+68-24i+9=141-88i.$

By conjugation,

$P(-2i)=141+88i.$

Hence

$\prod _{4}^{i=1}({x}_{2}^{i}+4)=\frac{(141-88i)(141+88i)}{16}=\frac{{141}^{2}+{88}^{2}}{16}=\frac{19881+7744}{16}=\frac{27625}{16}.$

We are told

$\prod _{4}^{i=1}({x}_{2}^{i}+4)=\frac{125}{16}\, m=\frac{27625}{16},$

so

$125\, m=27625\, \implies \, m=221.$

Answer: $221.$

Solutions · 1