How do you use the sum of two squares formula to solve #4x^2 -8=0#?

Question

Isaiah Anthony · Accepted Answer

I am not sure that this is what you want, but...
Considering: #(a^2-b^2)=(a+b)(a-b)#
Write:
#4x^2-8=0#
As:
#(2x+sqrt(8))(2x-sqrt(8))=0# So:
#(2x+sqrt(8))=0#
#(2x-sqrt(8))=0#
#x_1=-sqrt(8)/2#
#x_2=sqrt(8)/2#

Kennedy Adams · Answer

undefined To solve the quadratic equation $4x^2 - 8 = 0$ using the sum of two squares formula, follow these steps:

1. Move the constant term to the other side of the equation to isolate the quadratic term: 
   $4x^2 = 8$

2. Divide both sides of the equation by the coefficient of the quadratic term to make the leading coefficient equal to 1:
   $x^2 = 2$

3. Recognize that $2$ can be expressed as the square of $\sqrt{2}$.

4. Apply the sum of two squares formula, which states that for any real numbers $a$ and $b$, $a^2 - b^2 = (a + b)(a - b)$.

5. Substitute $a = x$ and $b = \sqrt{2}$ into the formula: 
   $x^2 - (\sqrt{2})^2 = (x + \sqrt{2})(x - \sqrt{2})$

6. Expand the expression:
   $x^2 - 2 = (x + \sqrt{2})(x - \sqrt{2})$

7. Set each factor equal to zero and solve for $x$:
   $x + \sqrt{2} = 0$ or $x - \sqrt{2} = 0$

8. Solve for $x$ in each equation:
   $x = -\sqrt{2}$ or $x = \sqrt{2}$

Therefore, the solutions to the equation $4x^2 - 8 = 0$ using the sum of two squares formula are $x = -\sqrt{2}$ and $x = \sqrt{2}$.