# Solve each quadratic function. 1. y^2=5x-3 3x^2-3y^2=1 2. y=3x+1 x^2+y^2=3

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### 3 Answers

1.

Substitute the first equation into the second equation:

`3x^2-3(5x-3)=1`

Expand, and rearrange to be in the quadratic form of: `ax^2+bx+c`

`3x^2-15x+9-1=0`

`3x^2-15x+8=0`

Use the quadratic formula to solve for the values of x that satisfy the equation:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

a=3; b=-15; c=8

`x=(15+-sqrt((-15)^2-4(3)(8)))/(2(3))`

`x=(15+-sqrt(225-96))/6`

`x_1=(15+sqrt(129))/6` and `x_2=(15-sqrt(129))/6`

2.

Same as with the first question, substitute the first equation into the second equation:

`x^2+(3x+1)^2=3`

Expand and rearrange in the quadratic form used above:

`x^2+9x^2+6x+1-3=0`

`10x^2+6x-2=0`

Use the quadratic formula to solve for the possible values of x:

a=10; b=6; c=-2

`x=(-6+-sqrt(6^2-4(10)(-2)))/(2(10))`

`x=(-6+-sqrt(36+80))/20`

`x_1=(-6+sqrt(116))/20` and `x_2=(-6-sqrt(116))/20`

**Sources:**

`y=3x+1` (1)

`x^2+y^2=3` (2)

Squaring (1):

`y^2=9x^2+6x+1` (3)

subsituing in (2)

`x^2+9x^2+6x+1=3`

`10x^2+6x-2=0`

`5x^2+3x-1=0`

`Delta= 9 +20=29 >0`

Has two dfferent real solutions:

`x=(-3+-sqrt(29))/10`

`x_1=0.23851648071345040312507104915403`

`y_1=1.71554944214035120937521`

`x_2=-0.83851648071345040312507104915403`

`y_2=-1.515549442140351209375213147462`

`y^2=5x-3`

`3x^2-3y^2=1`

`5x-y^2=3` (1)

`3x^2-3y^2=1` (2)

Subtracting from (2) (1) multiplied by 3:

`3x^2-3y^2-3(5x-y^2)=1-3xx3=-8`

`3x^2-3y^2-15x-3y^2=-8`

`3x^2-15x+8=0`

`Delta = 225 - 4xx3xx8=129>0` has two real differetn solutions:

`x=(15+-sqrt(129))/6`

`x_1=4.3929694486000912036307793279975`

`y_1=4.3548647789570293405374240854713`

`x_2=0.6070305513999087963692206720025`

`y_2=0.18749068510073768954429730179736`