Systems of Quadratics

[Xearf_987]

What I'm doing is pretty much self-explanatory.... solving systems.

#1

Question: This problem confused me

The problem:

1) 2x^2 - y^2 = 7
2) 2x - 3y = -7

My Attempt:
(I used the substitution method)

-3y = -2x -7

3y = 2x +7

y = (2/3)x + 7/3

1) 2x^2 - (2/3x + 7/3)^2 = 7

2x^2 - (4/9x + 49/9) = 7

2x^2 - 4/9x - 49/9 = 7

2x^2 - 4/9x - 112/9 = 0

This is about where I decided I might have made a mistake. What did I do wrong?


#2

Question: How do I do this problem?

The problem:

1) x^2 + 4y^2 = 68
2) xy = 8

My Attempt: (I'm pretty sure this problem requires substitution)

y = 8/x

1) x^2 + 4(8/x)^2 = 68

x^2 + 4(64/x^2) = 68

x^2 + 256/x^2 = 68

256/x^2 = -x^2 + 68

(x^2)(256/x^2) = (-x^2 + 68)(x^2)

256 = -x^4 + 68x^2

I'm not sure if this is right :?

I greatly appreciate your help!

 

[stapel]

#1) Solve the following system:

. . . . .2x² - y² = 7
. . . . .2x - 3y = -7

My Attempt:
(I used the substitution method)

For this sort of system, "substitution" is pretty much all you've got. Unfortunately, neither of the equations solves neatly, so this solution is likely to be fairly messy.

. . . . .-3y = -2x -7

. . . . .3y = 2x +7

. . . . .y = (2/3)x + 7/3

. . . . .2x^2 - (2/3x + 7/3)^2 = 7

. . . . .2x^2 - (4/9x + 49/9) = 7

No; (a + b)² = a² + 2ab + b², not a² + b².

. . . . .2x² - (2/3x + 7/3)² = 7

. . . . .2x² - [(4/9)x² + (28/9)x + (49/9)] = 7

. . . . .2x² - (4/9)x² - (28/9)x - (49/9) = 7

. . . . .18x² - 4x² - 28x - 49 = 63

See where that takes you. (It could still be messy, though.)

#2) Solve the following system:

. . . . .x² + 4y² = 68
. . . . .xy = 8

My attempt:

. . . . .y = 8/x

. . . . .x^2 + 4(8/x)^2 = 68

. . . . .x^2 + 4(64/x^2) = 68

. . . . .x^2 + 256/x^2 = 68

. . . . .256/x^2 = -x^2 + 68

. . . . .(x^2)(256/x^2) = (-x^2 + 68)(x^2)

. . . . .256 = -x^4 + 68x^2

Now put all the terms on one side, and solve the resulting quadratic:

. . . . .x⁴ - 68x² + 256 = 0

. . . . .(x²)² - 68(x²) + 256 = 0

Let a = 1, b = -68, and c = 256; plug into the Quadratic Formula to find the value(s) of x². Take the square roots to find the value(s) of x.

Eliz.

 

[soroban]

Hello, Xearf_987!

Problem #2
\(\displaystyle [1]\;x^2\,+\,4y^2\:=\:68\)
\(\displaystyle [2]\;xy\:=\:8\)

There is a back-door approach to this problem . . .

Multiply [2] by 4: \(\displaystyle \;\;\;\;4xy\;=\;32\)
. . . . . . . Add [1]: \(\displaystyle \,\;x^2\,+\,4y^2\:=\:68\)

And we get: \(\displaystyle \,x^2\,+\,4xy\,+\,4y^2\;=\;100\;\;\Rightarrow\;\;(x\,+\,2y)^2\:=\:100\)

Then: \(\displaystyle \,x\,+\,2y\:=\:\pm10\;\;\Rightarrow\;\;x\:=\:-2y\,\pm\,10\)

Substitute into [1]: \(\displaystyle \-2y\,\pm\,10)^2\,+\,4y^2\:=\:68\)

\(\displaystyle \;\;4y^2\,\pm\,40y\,+\,100\,+\,4y^2\:=\:68\;\;\Rightarrow\;\;8y^2\,\pm\,40y\,+\,32\:=\:0\)

Divide by 8: \(\displaystyle \;y^2\,\pm\,5y\,+\,4\:=\:0\)

Factor: \(\displaystyle \y\,\pm\,1)(y\,\pm\,4)\:=\:0\;\;\Rightarrow\;\;y\,=\,\pm1,\;\pm4\)

Substitute into [2]: \(\displaystyle \:x\,=\,\pm8,\;\pm2\)


There are four solutions: \(\displaystyle \,(x,y)\:=\2,4),\;(-2,-4),\;(8,1),\;(-8,-1)\)

 

[Denis]

1) 2x^2 - y^2 = 7
2) 2x - 3y = -7

Try to keep your work neater; less chances of error.

2) 3y = 2x + 7
y = (2x + 7) / 3 : that's easier to handle than (2/3)x + 7/3

Substitute in 1):
2x^2 - [(2x + 7) / 3]^2 = 7
2x^2 - [(4x^2 + 28x + 49) / 9] = 7
18x^2 - 4x^2 - 28x - 49 = 63
14x^2 - 28x - 112 = 0
x^2 - 2x - 8 = 0

See what I mean? Finish it :wink: