# How to solve this quadratic equation?

please can you help me solve this quadratic equation as i am a but stuck.

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

Relevance

3x^2 - y^2 = 3

2x - 1 = y

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

3x^2 - (4x^2 - 4x + 1) = 3

3x^2 - 4x^2 + 4x - 1 = 3

-x^2 + 4x - 4 = 0

x^2 - 4x + 4 = 0

(x - 2)^2 = 0

x = 2

and if x = 2, then y = 2(2) - 1 = 3

solution: (x , y) = (2 , 3)

edit: interpreting this geometrically:

the first equation is a hyperbola

the second equation is a line

so there is one point of intersection, and the line is a tangent of the hyperbola

check:

3(2^2) - (3^2) = 12 - 9 = 3 yep

2(2) - 1 = 3 yep

(2 , 3)

• 3x²-y²=3 (1)

2x-1=y (2)

Let's put y in function of x in (1).

3x²-(2x-1)²=3

3x²-(4x²-4x+1)=3

-x²+4x-4=0

x²-4x+4=0

(x-2)²=0

x-2=0

x=2

y=2x-1

y=2.2-1

y=3

The system of equations has got an unique solution.

x=2

y=3

• 3x^2 - y^2 = 3, 2x - 1 = y. Therefore 3x^2 - (2x - 1)^2 = 3, or 0 = x^2 - 4x + 4 = (x - 2)^2, so x = 2. Since 2x - 1 = y we have 2(2) - 1 = y, or y = 3.

• 3x² - (2x - 1)² = 3

3x² - (4x² - 4x + 1) = 3

3x² - 4x² + 4x - 1 = 3

x² - 4x + 4 = 0

(x - 2)(x - 2) = 0

x = 2

y = 3

• Anonymous
7 years ago

1. Substitute eqn. 2 for y in eqn.1

2. Expand the the new y^2 term

3. Regroup and cancel so in form ax^2+bx+c=0

4. Use the standard quadratic equation formula.

• i exploit the "new and more suitable factoring AC approach" (Google or Yahoo seek) to unravel the equation 2y^3 + 7y^2 - 30y = 0. First factor out y: y(2y^2 + 7y - 30) = 0. next, resolve the quadratic equation in parentheses. the two roots have opposite indications. Compose the aspects of ac = -60 with first numbers destructive. they are: (-a million, 60)(-2, 30)(-3, 20)(-4, 15)(-5, 12). end. This final sum is -5 +12 = 7 = b. next, replace in the equation the term (7y) by capacity of the two words (-5y) and (12y), then put in elementary factor: 2y^2 - 5y + 12y - 30 = 2y(y + 6) - 5(y + 6) = (y + 6)(2y - 5) = 0. next, resolve the binomials: y + 6 = 0 --> y = -6 2y - 5 = 0 --> y = 5/2