Question

Find the solution of the given quadratic equation:x2 + 4x + 4 = 12

Answer 1

The equation is given as,

$x^2+4x+4\text{ = 12}$

Re-arranging the equation,

$\begin{gathered} x^2+4x+4-12\text{ = 0 } \\ x^2+4x-8\text{ = 0} \\ \end{gathered}$

Comparing the equation with the standard form of quadratic equation,

$\begin{gathered} ax^2+bx+c\text{ = 0} \\ a\text{ = 1} \\ b\text{ = 4} \\ c\text{ = -8} \end{gathered}$

By applying the quadratic formula,

$\begin{gathered} x\text{ = }\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x\text{ = }\frac{-4\pm\sqrt{\left(4\right)^2-4\times1\times-8}}{2\times1} \\ x\text{ = }\frac{-4\text{ +}\sqrt{16-(-32)}}{2}\text{ and x = }\frac{-4\text{ - }\sqrt{16-(-32)}}{2} \\ x\text{ = }\frac{-4+\sqrt{48}}{2}\text{ and x = }\frac{-4\text{ - }\sqrt{48}}{2} \end{gathered}$

Simplifying further,

$\begin{gathered} x\text{ = }\frac{-4+4\sqrt{3}}{2}\text{ and x = }\frac{-4\text{ - 4}\sqrt{3}}{2} \\ x\text{ = }\frac{2(-2+2\sqrt{3})}{2}\text{ and x = }\frac{2(-2-2\sqrt{3})}{2} \\ x\text{ = -2 + 2}\sqrt{3}\text{ and x = -2 - 2}\sqrt{3} \\ x\text{ = 2\lparen-1+}\sqrt{3})\text{ and x = 2\lparen-1-}\sqrt{3}) \end{gathered}$

Thus the roots of the given quadratic equation are,

$x\text{ = 2\lparen-1+}\sqrt{3})\text{ and x = 2\lparen-1-}\sqrt{3)}$