Question

81y^2-4= (x-2)(x+3)+6(x-1)^2=

Answer 1

(9y-2)(9y+2)

x²+3x-2x-6+6(x²-2x+1)= x²+x-6+6x²-12x+6= 7x²-11x= x(7x-11)

Answer 2

To solve the equation $81y^2 - 4 = (x - 2)(x + 3) + 6(x - 1)^2$, we'll begin by expanding the right-hand side and then simplify the equation:

Step 1: Expand the right-hand side: $81y^2 - 4 = (x - 2)(x + 3) + 6(x - 1)^2$

Step 2: Perform the multiplication on the right-hand side: $81y^2 - 4 = (x^2 - 2x + 3x - 6) + 6(x^2 - 2x + 1)$

Step 3: Combine like terms: $81y^2 - 4 = (x^2 + x - 6) + 6x^2 - 12x + 6$

Step 4: Combine the x^2 terms and the x terms: $81y^2 - 4 = 7x^2 - 11x + 6$

Now, we have a quadratic equation in terms of x. To further solve for x, let's set the equation equal to zero:

$7x^2 - 11x + 6 = 0$

This is a quadratic equation in the form of $ax^2 + bx + c = 0$. To solve for x, we can use factoring or the quadratic formula. Let's use factoring:

Step 5: Factor the quadratic equation: $7x^2 - 11x + 6 = (x - 2)(7x - 3) = 0$

Now, we can set each factor equal to zero and solve for x:

$x - 2 = 0 \quad \text{or} \quad 7x - 3 = 0$

Step 6: Solve for x: $x = 2 \quad \text{or} \quad x = \frac{3}{7}$

Now that we have the values of x, we can find the corresponding values of y by substituting these values back into the original equation:

For $x = 2$:

$81y^2 - 4 = (2 - 2)(2 + 3) + 6(2 - 1)^2 = 0 + 6 \times 1 = 6$

$81y^2 = 6 + 4 = 10$

$y^2 = \frac{10}{81}$

$y = \pm \frac{\sqrt{10}}{9}$

For $x = \frac{3}{7}$:

$81y^2 - 4 = \left(\frac{3}{7} - 2\right)\left(\frac{3}{7} + 3\right) + 6\left(\frac{3}{7} - 1\right)^2 = -\frac{44}{49} + \frac{36}{49} = -\frac{8}{49}$

$81y^2 = -\frac{8}{49} + 4 = \frac{8}{49}$

$y^2 = \frac{8}{49 \times 81}$

$y = \pm \frac{2\sqrt{2}}{63}$

So, the solutions to the equation are:

$x = 2$ and $y = \pm \frac{\sqrt{10}}{9}$