Explained: "How to think while solving question like this?..."

High School
Algebra
Algebra I
Quadratic function and equation

talpajocotefnf3

Answered question

2022-03-31

How to think while solving question like this? If

a x^{2} + b x + c = 0

has roots A and

A^{n}

, then

{(a c^{n})}^{\frac{1}{n + 1}} + {(a^{n} c)}^{\frac{1}{n + 1}} + b = 0

Answer & Explanation

Dixie Reed

Beginner2022-04-01Added 15 answers

Step 1
Since $\frac{c}{a} = A \times A^{n} = A^{n + 1}$ , we that
${(a c^{n})}^{\frac{1}{n + 1}} + {(a^{n} c)}^{\frac{1}{n + 1}} + b =$
$= \sqrt[n + 1]{a c^{n}} + \sqrt[n + 1]{a^{n} c} + b =$
$= c \sqrt[n + 1]{} + a \sqrt[n + 1]{} + b =$
$= c \sqrt[n + 1]{} + a \sqrt[n + 1]{A^{n + 1}} + b =$
$= d \frac{c}{A} + a A + b =$
$= d \frac{1}{A} (c + a A^{2} + b A) =$
$= d \frac{1}{A} (a A^{2} + b A + c) =$
$= 0$
indeed A is a root of $a x^{2} + b x + c$
So we have proved that
${(a c^{n})}^{\frac{1}{n + 1}} + {(a^{n} c)}^{\frac{1}{n + 1}} + b = 0; .$

Roy Brady

Beginner2022-04-02Added 19 answers

Step 1
Since we are given the roots of the quadratic equation, we can write

a \times (x - A) \times (x - A^{n}) = a x^{2} + b x + c

This tells us that

b = a \times (- A - A^{n}) \Rightarrow a \times (A + A^{n}) + b = 0

and

a \times A \times A^{n} = a \times A^{n + 1} = c

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