How to set up problems like this one?

[allegansveritatem]

I know that -b/a is the sum of quadratic solutions and c/a is the product of solutions and I wonder if that is the key to solving this problem. Other than that I HAVE NO IDEA how to even begin. The solution, according to my instructions, must be arrived at by the quadratic formula.
Here is the problem:

 

[tkhunny]

I know that -b/a is the sum of quadratic solutions and c/a is the product of solutions and I wonder if that is the key to solving this problem. Other than that I HAVE NO IDEA how to even begin. The solution, according to my instructions, must be arrived at by the quadratic formula.
Here is the problem:

View attachment 10955

1) Decide what is meant by "numbers". I suspect it is INTEGERS. Maybe not.
2) If we need ONLY integers, there are often only a few possibilities.

What two INTEGERS, when multiplied, produce +2?

+1, +2
-1, -2

That's it.

Which of these pairs sums to +3?

Done.

 

[HallsofIvy]

More generally, call the two numbers "x" and "y". The two numbers add to 3: x+ y= 3 so y= 3- x. The product of the two numbers is xy= x(3- x)= 3x- x^2= 2. So we have the quadratic equation x^2- 3x+ 2= (x- 2)(x- 1)= 0. The "two" solutions are x= 2, y= 1, and x= 1, y= 2.

 

[Steven G]

I know that -b/a is the sum of quadratic solutions and c/a is the product of solutions and I wonder if that is the key to solving this problem. Other than that I HAVE NO IDEA how to even begin. The solution, according to my instructions, must be arrived at by the quadratic formula.
Here is the problem:

View attachment 10955

So -b/a = 2 and c/a = -1 where a and b come from ax² + bx + c = 0. Choose a value for a and then compute b and c

 

[Steven G]

More generally, call the two numbers "x" and "y". The two numbers add to 3: x+ y= 3 so y= 3- x. The product of the two numbers is xy= x(3- x)= 3x- x^2= 2. So we have the quadratic equation x^2- 3x+ 2= (x- 2)(x- 1)= 0. The "two" solutions are x= 2, y= 1, and x= 1, y= 2.

What am I missing? The last two posts say that the sum is 3. If I am not insane, the OP says the sum is 2 and the product is -1. Has the problem been changed?

 

[pka]

I know that -b/a is the sum of quadratic solutions and c/a is the product of solutions and I wonder if that is the key to solving this problem. Other than that I HAVE NO IDEA how to even begin. The solution, according to my instructions, must be arrived at by the quadratic formula.
Here is the problem: View attachment 10955

Yes that is the correct idea.
\(\displaystyle \begin{align*}x^2-2x-1&=0\\x&=\dfrac{2\pm\sqrt{4-4(1)(-1)}}{2}\\&=1\pm\sqrt2 \end{align*}\)

$\displaystyle (1+\sqrt2)+(1-\sqrt2)=2\\(1+\sqrt2)(1-\sqrt2)=-1$

 

[HallsofIvy]

What am I missing? The last two posts say that the sum is 3. If I am not insane, the OP says the sum is 2 and the product is -1. Has the problem been changed?

No, you are not the one who is insane.

 

[Harry_the_cat]

I know that -b/a is the sum of quadratic solutions and c/a is the product of solutions and I wonder if that is the key to solving this problem. Other than that I HAVE NO IDEA how to even begin. The solution, according to my instructions, must be arrived at by the quadratic formula.
Here is the problem:

View attachment 10955

Let the numbers be m and n

m+n =2
m*n = -1

Solve simultaneously:

Sub m=2-n (from first equation) into m*n=-1, get a quadratic equation and solve.

 

[Steven G]

I think that the easiest method is: We know that -b/a = 2 and c/a = -1, ie b=-2a and c=-a. So pick a value for a and solve for b and c. For example if a=1, then b=-2 and c=-1. Then 1x²-2x-1=0 Done

 

[Denis]

u = 1 + √2, v = 1 - √2

u + v = 1 + √2 + 1 - √2 = 2

uv = (1 + √2)*(1 - √2) = -1

 

[allegansveritatem]

1) Decide what is meant by "numbers". I suspect it is INTEGERS. Maybe not.
2) If we need ONLY integers, there are often only a few possibilities.

What two INTEGERS, when multiplied, produce +2?

+1, +2
-1, -2

That's it.

Which of these pairs sums to +3?

Done.

I had a photo of the solution but can't find it now. But, I think there was a radical in at least one of the solutions. It was something like 1 or -1+/-square root of 2. I will check this later this evening and get back...but I think there was a radical in there somewhere. Maybe not.

 

[allegansveritatem]

Yes that is the correct idea.
\(\displaystyle \begin{align*}x^2-2x-1&=0\\x&=\dfrac{2\pm\sqrt{4-4(1)(-1)}}{2}\\&=1\pm\sqrt2 \end{align*}\)

$\displaystyle (1+\sqrt2)+(1-\sqrt2)=2\\(1+\sqrt2)(1-\sqrt2)=-1$

I think this is the solution I looked up in the back of the book. I took a photo of it but somehow the file has been mislaid.

How did you get that equation?

 

[pka]

How did you get that equation?

In the quadratic equation $\displaystyle x^2+Bx+C=0$ the sum of its roots is $\displaystyle -B$ and the product of its roots is $\displaystyle C$.
You posted that the sum of the roots was $\displaystyle 2$, the product was $\displaystyle -1$ so that means $\displaystyle B=-2~\&~C=-1$
$\displaystyle x^2+Bx+C=x^2-2x-1=0$

 

[Steven G]

I had a photo of the solution but can't find it now. But, I think there was a radical in at least one of the solutions. It was something like 1 or -1+/-square root of 2. I will check this later this evening and get back...but I think there was a radical in there somewhere. Maybe not.

Of course the solution you are referring to is wrong. Did you check what the sum of the two numbers was? How about the product of the two numbers? Both results failed the two given requirements.

 

[Steven G]

In the quadratic equation $\displaystyle x^2+Bx+C=0$ the sum of its roots is $\displaystyle -B$ and the product of its roots is $\displaystyle C$.
You posted that the sum of the roots was $\displaystyle 2$, the product was $\displaystyle -1$ so that means $\displaystyle B=-2~\&~C=-1$
$\displaystyle x^2+Bx+C=x^2-2x-1=0$

Yes, this is the way to do this type of problem. Simply let a=1 as I noted in my previous post.

 

[allegansveritatem]

I think that the easiest method is: We know that -b/a = 2 and c/a = -1, ie b=-2a and c=-a. So pick a value for a and solve for b and c. For example if a=1, then b=-2 and c=-1. Then 1x²-2x-1=0 Done

I think this is what I was looking for. The funny thing is this: I learned about the fact that -b/a is the sum and c/a is the product of the solutions not from my author's text but from one of the word problems he offers at the back of each section--and this particular problem was one of the last of a series. So this info was almost buried. And it was buried in the back of an earlier chapter and not the chapter that this problem we are dealing with here comes from. Not only that but there is not one example in the whole text of how to solve this kind of problem. Como?

 

[allegansveritatem]

Of course the solution you are referring to is wrong. Did you check what the sum of the two numbers was? How about the product of the two numbers? Both results failed the two given requirements.

No, I had no idea how that answer was come by and that question was hogging my mental resources.

So, the coefficient of a is a guess?

 

[JeffM]

I know that -b/a is the sum of quadratic solutions and c/a is the product of solutions and I wonder if that is the key to solving this problem. Other than that I HAVE NO IDEA how to even begin. The solution, according to my instructions, must be arrived at by the quadratic formula.
Here is the problem:

View attachment 10955

Yes, it is true that the solutions to a quadratic equation are

$\displaystyle ax^2 + bx + c \implies x = \dfrac{-\ b + \sqrt{b^2 - 4ac}}{2a} \text { or } x = \dfrac{- \ b - \sqrt{b^2 - 4ac}}{2a}.$

That formula comes up often enough that it is convenient to remember it.

And from that formula, one can easily deduce that the sum of the solutions is

$\displaystyle \dfrac{-\ b + \sqrt{b^2 - 4ac}}{2a} + \dfrac{-\ b - \sqrt{b^2 - 4ac}}{2a} =$

$\displaystyle \dfrac{-\ b - b + \sqrt{b^2 - 4ac} - \sqrt{b^2 - 4ac}}{2a} = \dfrac{-\ 2b}{2a} = -\ \dfrac{b}{a}.$

And deduce that the product of the solutions is

$\displaystyle \dfrac{-\ b + \sqrt{b^2 - 4ac}}{2a} * \dfrac{-\ b - \sqrt{b^2 - 4ac}}{2a} =$

$\displaystyle \dfrac{(-\ b)^2 - b\{\sqrt{b^2 - 4ac} - \sqrt{b^2 - 4ac}\} - \{\sqrt{b^2 - 4ac}\}^2}{2a * 2a} =$

$\displaystyle \dfrac{b^2 - b(0) - (b^2 - 4ac)}{4a^2} = \dfrac{4ac}{4a^2} = \dfrac{c}{a}.$

But those two facts are not worth memorizing.

Math is not a bunch of facts to be memorized. It is a small set of techniques that you learn how to apply in different situations. You memorized some facts and wondered if two of those facts might be helpful in solving a specific problem. That is not a useful approach. There are literally an infinite number of facts about the real numbers.

The fundamental idea in algebra is that you find unknown numbers by finding what they are equal to. This problem is amazingly easy if you remember that fundamental idea because the equalities are laid on your plate.

$\displaystyle x + y = 2 \text { and } xy = -\ 1.$

You have two equations in two unknowns. There is a general technique for solving such systems of equations called substitution.

$\displaystyle y = 2 - x \implies -\ 1 = xy = x(2 - x) = 2x - x^2 \implies x^2 - 2x - 1 = 0.$

Now, as I said, such quadratic equations in one unknown come up often enough that it is convenient to memorize a formula for solving them, but the formula is not necessary. There is a general technique for solving these equations called completing the square.

$\displaystyle x^2 - 2x - 1 = 0 \implies x^2 - 2x - 1 + 2 = 2 \implies x^2 - 2x + 1 = 2 \implies$

$\displaystyle (x - 1)^2 = 2 \implies (x - 1) = \pm \sqrt{2} \implies x = 1 + \sqrt{2} \text { or } x = 1 - \sqrt{2} \implies$

$\displaystyle y = 2 - (1 + \sqrt{2}) = 1 - \sqrt{2} \text { or } y = 2 - (1 - \sqrt{2}) = 1 + \sqrt{2}.$

Algebra is about applying a very small number of ideas and techniques to a very wide range of problems. It is not about amassing a huge number of mathematical facts to apply to the small proportion of problems to which those facts are relevant.

 

[allegansveritatem]

Yes, it is true that the solutions to a quadratic equation are

$\displaystyle ax^2 + bx + c \implies x = \dfrac{-\ b + \sqrt{b^2 - 4ac}}{2a} \text { or } x = \dfrac{- \ b - \sqrt{b^2 - 4ac}}{2a}.$

That formula comes up often enough that it is convenient to remember it.

And from that formula, one can easily deduce that the sum of the solutions is

$\displaystyle \dfrac{-\ b + \sqrt{b^2 - 4ac}}{2a} + \dfrac{-\ b - \sqrt{b^2 - 4ac}}{2a} =$

$\displaystyle \dfrac{-\ b - b + \sqrt{b^2 - 4ac} - \sqrt{b^2 - 4ac}}{2a} = \dfrac{-\ 2b}{2a} = -\ \dfrac{b}{a}.$

And deduce that the product of the solutions is

$\displaystyle \dfrac{-\ b + \sqrt{b^2 - 4ac}}{2a} * \dfrac{-\ b - \sqrt{b^2 - 4ac}}{2a} =$

$\displaystyle \dfrac{(-\ b)^2 - b\{\sqrt{b^2 - 4ac} - \sqrt{b^2 - 4ac}\} - \{\sqrt{b^2 - 4ac}\}^2}{2a * 2a} =$

$\displaystyle \dfrac{b^2 - b(0) - (b^2 - 4ac)}{4a^2} = \dfrac{4ac}{4a^2} = \dfrac{c}{a}.$

But those two facts are not worth memorizing.

Math is not a bunch of facts to be memorized. It is a small set of techniques that you learn how to apply in different situations. You memorized some facts and wondered if two of those facts might be helpful in solving a specific problem. That is not a useful approach. There are literally an infinite number of facts about the real numbers.

The fundamental idea in algebra is that you find unknown numbers by finding what they are equal to. This problem is amazingly easy if you remember that fundamental idea because the equalities are laid on your plate.

$\displaystyle x + y = 2 \text { and } xy = -\ 1.$

You have two equations in two unknowns. There is a general technique for solving such systems of equations called substitution.

$\displaystyle y = 2 - x \implies -\ 1 = xy = x(2 - x) = 2x - x^2 \implies x^2 - 2x - 1 = 0.$

Now, as I said, such quadratic equations in one unknown come up often enough that it is convenient to memorize a formula for solving them, but the formula is not necessary. There is a general technique for solving these equations called completing the square.

$\displaystyle x^2 - 2x - 1 = 0 \implies x^2 - 2x - 1 + 2 = 2 \implies x^2 - 2x + 1 = 2 \implies$

$\displaystyle (x - 1)^2 = 2 \implies (x - 1) = \pm \sqrt{2} \implies x = 1 + \sqrt{2} \text { or } x = 1 - \sqrt{2} \implies$

$\displaystyle y = 2 - (1 + \sqrt{2}) = 1 - \sqrt{2} \text { or } y = 2 - (1 - \sqrt{2}) = 1 + \sqrt{2}.$

Algebra is about applying a very small number of ideas and techniques to a very wide range of problems. It is not about amassing a huge number of mathematical facts to apply to the small proportion of problems to which those facts are relevant.

Good. I will copy this and paste it in my digital notebook. Sounds true. Thanks