How do you solve the system $4 x^{2} - 56 x + 9 y^{2} + 160 = 0$ $4x^2-56x+9y^2+160=0$ and $4 x^{2} + y^{2} - 64 = 0$ $4x^2+y^2-64=0$ ?

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How do you solve the system $4 x^{2} - 56 x + 9 y^{2} + 160 = 0$ $4x^2-56x+9y^2+160=0$ and $4 x^{2} + y^{2} - 64 = 0$ $4x^2+y^2-64=0$ ?

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Answer 1 · 2016-07-09T19:18:09

$y^{2} = 64 - 4 x^{2}$ $y^2 = 64 - 4x^2$

$y = \pm \sqrt{64 - 4 x^{2}}$ $y =+- sqrt(64 - 4x^2)$

$4 x^{2} - 56 x + 9 {(\sqrt{64 - 4 x^{2}})}^{2} + 160 = 0$ $4x^2 - 56x + 9(sqrt(64 - 4x^2))^2 + 160 = 0$

$4 x^{2} - 56 x + 9 (64 - 4 x^{2}) + 160 = 0$ $4x^2 - 56x + 9(64 - 4x^2) + 160 = 0$

$4 x^{2} - 56 x + 576 - 36 x^{2} + 160 = 0$ $4x^2 - 56x + 576 - 36x^2 + 160 = 0$

$0 = 32 x^{2} + 56 x - 736$ $0 = 32x^2 + 56x - 736$

$0 = 8 (4 x^{2} + 7 x - 92)$ $0 = 8(4x^2 + 7x - 92)$

$0 = 8 (4 x^{2} - 16 x + 23 x - 92)$ $0 = 8(4x^2 - 16x + 23x - 92)$

$0 = 8 (4 x (x - 4) + 23 (x - 4))$ $0 = 8(4x(x - 4) + 23(x - 4))$

$0 = 8 (4 x + 23) (x - 4)$ $0 = 8(4x + 23)(x - 4)$

$x = - \frac{23}{4} and 4$ $x = -23/4 and 4$

Case 1:

$4 {(- \frac{23}{4})}^{2} + y^{2} - 64 = 0$ $4(-23/4)^2 + y^2 - 64 = 0$

$4 (\frac{529}{16}) + y^{2} - 64 = 0$ $4(529/16) + y^2 - 64 = 0$

$y^{2} = 64 - \frac{529}{4}$ $y^2 = 64 - 529/4$

$y = \emptyset$ $y = O/$

Case 2:

$4 {(4)}^{2} + y^{2} - 64 = 0$ $4(4)^2+ y^2 - 64 = 0$

$4 (16) + y^{2} - 64 = 0$ $4(16) + y^2 - 64 = 0$

$y^{2} = 64 - 64$ $y^2 = 64 - 64$

$y^{2} = 0$ $y^2 = 0$

$y = 0$ $y = 0$

The only real solution is $x = 4$ $x = 4$ , $y = 0$ $y = 0$ .

Thus, the solution set is ${- 4, 0}$ ${-4, 0}$ .

Hopefully this helps!

Answer 2 · 2016-07-09T20:04:40

$y = 0$ $y = 0$ and $x = 4$ $x = 4$

Explanation:

Making $x_{2} = x^{2}, y_{2} = y^{2}$ $x_2=x^2,y_2=y^2$ and solving for $x_{2}, y_{2}$ $x_2,y_2$ the system

${ (4 x_2+9y_2=56x-160), (4x_2+y_2=64) :}$

we obtain

${ (x_2 = 1/4 (92 - 7 x)), (y_2 = 7 x-28) :}$

Now we solve

$x^2 = 1/4 (92 - 7 x)$

obtaining $x = {-23/4,4}$

and

$y = pm sqrt(7x-28)$

or

$y = pm sqrt(7 cdot 4-28) = 0$ excluding complex solutions.

we have

$y = 0$ and $x = 4$

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How do you solve the system $4 x^{2} - 56 x + 9 y^{2} + 160 = 0$ $4x^2-56x+9y^2+160=0$ and $4 x^{2} + y^{2} - 64 = 0$ $4x^2+y^2-64=0$ ?

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How do you solve the system 4x^2-56x+9y^2+160=04x2−56x+9y2+160=04x^2-56x+9y^2+160=0 and 4x^2+y^2-64=04x2+y2−64=04x^2+y^2-64=0?

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How do you solve the system $4 x^{2} - 56 x + 9 y^{2} + 160 = 0$ $4x^2-56x+9y^2+160=0$ and $4 x^{2} + y^{2} - 64 = 0$ $4x^2+y^2-64=0$ ?