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# Help would be appreciated!

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Let f(x)=(x^2+6x+9)^{50}-4x+3, and r_1,r_2,...r_100 be the roots of f(x).

Compute (r_1+3)^100+(r_2+3)^100+...+(r_100+3)^100

Apr 16, 2022

#1
+9459
+2

Note that

$$\begin{array}{rcl}f(x) &=& (x^2 + 6x + 9)^{50} - 4x + 3\\ &=&\left( (x + 3)^2\right)^{50} - 4x + 3 \\&=&(x+3)^{100}-4x+3\end{array}$$

For each of $$i=1,2,\cdots,100$$, since $$x = r_i$$ is a root of $$f(x)$$:

$$\begin{array}{rcl} f(r_i) &=& 0\\ (r_i+ 3)^{100} &=& 4r_i - 3 \end{array}$$

Now, using the equation above,

$$\begin{array}{cl} &(r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}\\ =&(4r_1 - 3) + (4r_2 - 3) + \cdots + (4r_{100} - 3)\\ =&4(r_1 + r_2 + \cdots + r_{100}) - 300 \end{array}$$

By Vieta's formula, we have:

$$r_1 + r_2 + \cdots + r_{100} = -\dfrac{\text{coefficient of }x^{99}\text{ in }f(x)}{\text{coefficient of }x^{100}\text{ in }f(x)}$$

Using the binomial theorem, we can expand $$(x + 3)^{100}$$.

$$(x + 3)^{100} -4x+3 = x^{100} + 300x^{99}+\cdots$$

Therefore,

$$r_1 + r_2 + \cdots + r_{100} = -\dfrac{300}1 = -300$$

Recall that $$(r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}=4(r_1 + r_2 + \cdots + r_{100}) - 300$$. Then:

$$\begin{array}{cl} &(r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}\\ =&4(-300) -300\\ =& -1500 \end{array}$$

Apr 16, 2022

#1
+9459
+2

Note that

$$\begin{array}{rcl}f(x) &=& (x^2 + 6x + 9)^{50} - 4x + 3\\ &=&\left( (x + 3)^2\right)^{50} - 4x + 3 \\&=&(x+3)^{100}-4x+3\end{array}$$

For each of $$i=1,2,\cdots,100$$, since $$x = r_i$$ is a root of $$f(x)$$:

$$\begin{array}{rcl} f(r_i) &=& 0\\ (r_i+ 3)^{100} &=& 4r_i - 3 \end{array}$$

Now, using the equation above,

$$\begin{array}{cl} &(r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}\\ =&(4r_1 - 3) + (4r_2 - 3) + \cdots + (4r_{100} - 3)\\ =&4(r_1 + r_2 + \cdots + r_{100}) - 300 \end{array}$$

By Vieta's formula, we have:

$$r_1 + r_2 + \cdots + r_{100} = -\dfrac{\text{coefficient of }x^{99}\text{ in }f(x)}{\text{coefficient of }x^{100}\text{ in }f(x)}$$

Using the binomial theorem, we can expand $$(x + 3)^{100}$$.

$$(x + 3)^{100} -4x+3 = x^{100} + 300x^{99}+\cdots$$

Therefore,

$$r_1 + r_2 + \cdots + r_{100} = -\dfrac{300}1 = -300$$

Recall that $$(r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}=4(r_1 + r_2 + \cdots + r_{100}) - 300$$. Then:

$$\begin{array}{cl} &(r_1 + 3)^{100} + (r_2 + 3)^{100} + \cdots + (r_{100} + 3)^{100}\\ =&4(-300) -300\\ =& -1500 \end{array}$$

MaxWong Apr 16, 2022
#2
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Good Job!

Guest Apr 21, 2022