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## #1 2023-11-15 03:53:48

sologuitar
Member
Registered: 2022-09-19
Posts: 467

### Real Solutions of the Quadratic Equation...B

Show that the real solutions of the equation ax^2 + bx + c = 0 are the reciprocals of the real solutions of the equation cx^2 + bx + a = 0. Assume that b^2 - 4ac is greater than or equal to 0.

Any hints? What is the first step to consider?

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## #2 2023-11-15 04:20:20

Bob Registered: 2010-06-20
Posts: 9,863

### Re: Real Solutions of the Quadratic Equation...B

I've written down the solutions using the quadratic formula, let's say x and x' for the first and y and y' for the second.

To be reciprocals x times y (or maybe y') must equal 1 and similarly x' with y' or y

Haven't tried it yet but I think it should work.

edit: tried it now. It does if you do x times y' (in other words take the plus root for one with the minus root of the other ... then loads of canceling,  leaving 4ac/4ac = 1).

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob Offline

## #3 2023-11-15 04:32:03

amnkb
Member
Registered: 2023-09-19
Posts: 175

### Re: Real Solutions of the Quadratic Equation...B

sologuitar wrote:

Show that the real solutions of the equation ax^2 + bx + c = 0 are the reciprocals of the real solutions of the equation cx^2 + bx + a = 0. Assume that b^2 - 4ac is greater than or equal to 0.

Last edited by amnkb (2023-11-16 18:23:04)

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## #4 2023-11-16 18:21:37

sologuitar
Member
Registered: 2022-09-19
Posts: 467

### Re: Real Solutions of the Quadratic Equation...B

Bob wrote:

I've written down the solutions using the quadratic formula, let's say x and x' for the first and y and y' for the second.

To be reciprocals x times y (or maybe y') must equal 1 and similarly x' with y' or y

Haven't tried it yet but I think it should work.

edit: tried it now. It does if you do x times y' (in other words take the plus root for one with the minus root of the other ... then loads of canceling,  leaving 4ac/4ac = 1).

Bob

The wording in this application is horrible.

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## #5 2023-11-16 18:24:27

sologuitar
Member
Registered: 2022-09-19
Posts: 467

### Re: Real Solutions of the Quadratic Equation...B

amnkb wrote:
sologuitar wrote:

Show that the real solutions of the equation ax^2 + bx + c = 0 are the reciprocals of the real solutions of the equation cx^2 + bx + a = 0. Assume that b^2 - 4ac is greater than or equal to 0.

Again, what data in the application led you to work out the problem as you did? You said (1/r) is a solution to the quadratic equation given. When you say "a solution", do you mean there are more solutions?

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## #6 2023-11-17 17:41:00

KerimF
Member
From: Aleppo-Syria
Registered: 2018-08-10
Posts: 53

### Re: Real Solutions of the Quadratic Equation...B

sologuitar wrote:

When you say "a solution", do you mean there are more solutions?

As you know, in general, a quadratic equation has two roots (real or complex) if (b^2 - 4ac) ≠ 0. We may say that 'r' is one of them (for known a, b and c).

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## #7 2023-11-18 04:00:19

sologuitar
Member
Registered: 2022-09-19
Posts: 467

### Re: Real Solutions of the Quadratic Equation...B

KerimF wrote:
sologuitar wrote:

When you say "a solution", do you mean there are more solutions?

As you know, in general, a quadratic equation has two roots (real or complex) if (b^2 - 4ac) ≠ 0. We may say that 'r' is one of them (for known a, b and c).

Much better this time. Less lectures and philosophy and more mathematics.

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