1. ## Help solving difficult simultaneous quadratic equation

I'm stuck on a problem in an exercise book. It gives the answer, but I just can't get the same answer. Here's the problem:

Find the co-ordinates of the point of intersection of the given straight line and curve:

i. 2x - 5y =6
ii. 2xy - 4x² - 3y = 1

Following an example earlier in the book, I went this way:

i. 2x = 6 + 5y
i. 4x² = (6 + 5y)² = (6 + 5y)(6 + 5y) = 36 + 60y + 25y²

i into ii:

ii. y(6 + 5y) -(36 + 60y + 25y²) - 3y = 1
ii. 6y + 5y² -36 - 60y - 25y² -3y = 1
ii. -20y² - 57y - 37 = 0

When I put that last line into the quadratic formula, I only get one root. The given answer has two. I'm clearly doing something wrong. Would someone please help?

2. ## Re: Help solving difficult simultaneous quadratic equation

Originally Posted by markling
I'm stuck on a problem in an exercise book. It gives the answer, but I just can't get the same answer. Here's the problem:

Find the co-ordinates of the point of intersection of the given straight line and curve:

i. 2x - 5y =6
ii. 2xy - 4x² - 3y = 1

Following an example earlier in the book, I went this way:

i. 2x = 6 + 5y
i. 4x² = (6 + 5y)² = (6 + 5y)(6 + 5y) = 36 + 60y + 25y²

i into ii:

ii. y(6 + 5y) -(36 + 60y + 25y²) - 3y = 1
ii. 6y + 5y² -36 - 60y - 25y² -3y = 1
ii. -20y² - 57y - 37 = 0

When I put that last line into the quadratic formula, I only get one root. <--- I get: y=-1 or $y = -\frac{37}{20}$ !
... and consequently $x = \frac12~\vee~x=-\frac{13}8$

3. ## Re: Help solving difficult simultaneous quadratic equation

Agh. Such a silly mistake. Thanks very much, earboth. I can now see what I was doing wrong. I've been leaving my equations in whatever order they come out, and then they've been getting garbled when I put them into the quadratic formula. So on paper I didn't have -20y² - 57y - 37 = 0, I had -57y -20y² - 37 = 0. And then I was putting a in as b and b in as a. Thanks again.

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### 2x 3y=13 2xy 5yÂ²-4xÂ²=41

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