Can you solve Korean SAT math Question? – korean math Question

Hello, I’m Teacher Jojo.

I’m teaching math in Korea.

The Question we’ll attempt today is a practice question for preparing for the Korean SAT.

It’s designed for 1st-year high school students.

To solve this problem, you’ll need to have knowledge about cubic polynomials, real roots, and imaginary roots.

If you’re familiar with these concepts, give it a try!

[Korean SAT math Question]

Can you solve Korean SAT math Question? -korean math Question

I will explain the Question in English.

Let $a, b,$ and $c$ be three real numbers. $P(x)$ satisfies the following conditions:

(a) The cubic equation $P(x) = 0$ has one real root and two distinct imaginary roots. The product of the two distinct imaginary roots is 5.

(b) The cubic equation $P(3x-1) = 0$ has one real root at $0$ and two distinct imaginary roots. The sum of the two distinct imaginary roots is $2$ .

What is the value of $a + b + c\;$ ?

…

Let’s think about it carefully.

The solution is written below.

If you’ve solved it, you can check the solution provided below.

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[Solution]

First, let’s denote the three roots of the cubic equation as $\alpha, \beta, and \; \gamma$ .

$\alpha$ is the real root, and we assume $\beta$ and $\gamma$ are two distinct imaginary roots.

According to condition (a), the product of $\beta$ and $\gamma$ is 5.

${\color{red}\beta\gamma = 5}$

Next,

It’s important to use condition (b) effectively.

You need to consider the roots of $P(3x-1)=0$ .

Let’s denote these roots as $\alpha, \beta, \gamma.$

We know that the roots of $P(x)=0$ are $\alpha, \beta, \gamma.$

Therefore, for x to make $P(3x-1)=0$ true,

we have $3x-1 = \alpha, 3x-1 = \beta, 3x-1 = \gamma$ .

Understanding this is crucial to solving the problem.

In conclusion, we can say that the roots of $P(3x-1)=0$ are $\cfrac{\alpha+1}{3}, \cfrac{\beta+1}{3}, \cfrac{\gamma+1}{3}$ .

Here, $\cfrac{\alpha+1}{3}$ is the real root, while $\cfrac{\beta+1}{3}, \cfrac{\gamma+1}{3}$ are imaginary roots.

Condition (b) stated that one root is $0$ , which means $\cfrac{\alpha+1}{3}=0$ .

It was also mentioned that the sum of the two distinct imaginary roots is 2,

so $\cfrac{\beta+1}{3}+\cfrac{\gamma+1}{3}=2$ .

${\color{blue}\cfrac{\alpha+1}{3}=0, \; \cfrac{\beta+1}{3}+\cfrac{\gamma+1}{3}=2}$

When we calculate and simplify,

${\color{red}\alpha=-1, \beta+\gamma=4}$

Now, we’re almost done.

If we denote the three roots of the cubic equation as $\alpha, \beta, \gamma,$ we can use these three roots to form a cubic equation with a leading coefficient of 1.

$P(x)=x^3+ax^2+bx+c$

$P(x)=x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma$

${\color{red}\alpha=-1,\; \beta+\gamma=4,\;\beta\gamma=5}$

$\alpha+\beta+\gamma=3$

$\alpha\beta+\beta\gamma+\gamma\alpha = \alpha(\beta+\gamma)+\beta\gamma=(-1)\times4+5=1$

$\alpha\beta\gamma=(-1)\times5=-5$

$P(x)=x^3-3x^2+x+5$

$a=-3, b=1, c=5$

${\color{red}a+b+c = 3}$

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How did you find today’s Question?

Did you manage to solve it and get the correct answer?

If you did, great job! It means you have a good understanding of cubic equations.

If you didn’t get it right, that’s okay too. You can study more and try again next time.

That’s it for today.

Also, I have a YouTube channel. My YouTube channel is

https://www.youtube.com/channel/UCnJ-GLzfJdWjs04eQoxfY7g

[Korean ver]

안녕하세요. 저는 한국에서 수학을 가르치고있는 조조쌤입니다.

오늘은 수학 모의고사 문제를 가지고왔습니다.

고등학교 1학년 문제입니다.

이 문제는 삼차방정식, 실근, 허근 이런것들을 알아야 풀수있는 문제입니다.

한번 도전해보시겠습니까?

[문제]

천천히 풀어보세요.

다 풀었다면 아래 풀이와 비교해보세요.

[풀이]

우선, 삼차방정식의 세 근을 $\alpha, \beta, \gamma$ 라고 설정합니다.

$\alpha$ 는 실근, $\beta$ , $\gamma$ 는 서로 다른 두 허근.

조건(가)에 따르면, $\beta$ 와 $\gamma$ 의 곱은 5입니다.

${\color{red}\beta\gamma = 5}$

그리고

조건(나)가 중요합니다.

$P(3x-1)=0$ 이 성립하게하는 $x$ 를 생각해야합니다.

우리는 $P(x)=0$ 의 근이 $\alpha, \beta, \gamma.$ 라고 했습니다.

그러므로 $P(3x-1)=0$ 을 성립하게 하는 $x$ 를 찾을때, 이렇게 생각해야합니다.

${\color{green} 3x-1 = \alpha,\; 3x-1 = \beta,\; 3x-1 = \gamma}$

$P(3x-1) = P( \alpha ) = 0$ 라는것을 이해한다면 문제는 다 풀었습니다.

따라서 $P(3x-1)=0$ 을 만족하는 $x$ 는,

$\cfrac{\alpha+1}{3}, \cfrac{\beta+1}{3}, \cfrac{\gamma+1}{3}$ .

$\cfrac{\alpha+1}{3}$ 는 실근, $\cfrac{\beta+1}{3}, \cfrac{\gamma+1}{3}$ 는 허근입니다.

조건 (나)에서 한 실근은 $0$ 이므로,

$\cfrac{\alpha+1}{3}=0$ .

또 서로 다른 두 허근의 합은 2라고 했습니다.

$\cfrac{\beta+1}{3}+\cfrac{\gamma+1}{3}=2$ .

${\color{blue}\cfrac{\alpha+1}{3}=0, \; \cfrac{\beta+1}{3}+\cfrac{\gamma+1}{3}=2}$

계산을 하여 간단하게 하면,

${\color{red}\alpha=-1, \beta+\gamma=4}$

이제 거의 다 풀었습니다.

세 근을 알 때, 삼차방정식을 만들수가 있습니다.