Given: Different equations
To Determine: Which would be best solved using difference of two squares
Solution
The factorization of a difference of two squares is given below
[tex]a^2-b^2=(a-b)(a+b)[/tex]
Let us examine each of the given equation
[tex]\begin{gathered} x^2+3x-10=0 \\ x^2-2x+5x-10=0 \\ x(x-2)+5(x-2)=0 \\ (x-2)(x+5)=0 \end{gathered}[/tex][tex]\begin{gathered} 3x^2+12x=8 \\ 3x^2+12x-8=0 \end{gathered}[/tex][tex]\begin{gathered} x^3+5x^2-9x-45=0 \\ x^2(x+5)-9(x+5)=0 \\ (x+5)(x^2-9)=0 \\ x^2-9=x^2-3^2=(x+3)(x-3) \\ Therefore \\ (x+5)(x+3)(x-3)=0 \end{gathered}[/tex][tex]\begin{gathered} x^2-25=0 \\ x^2-5^2=0 \\ (x+5)(x-5)=0 \end{gathered}[/tex]
From the above,
[tex]\begin{gathered} x^3+5x^2-9x-45=0 \\ AND \\ x^2-25=0 \end{gathered}[/tex]
The two equation above can be solved by difference of two square, but the equation below is the easiest solved using differnce of two square
x² - 25 = 0
