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{ = }(-b\pm√(b^2-4ac))/(2a) \\ x\text{ = }(-4\pm√(\left(4\right)^2-4*1*-8))/(2*1) \\ x\text{ = }\frac{-4\text{ +}√(16-(-32))}{2}\text{ and x = }\frac{-4\text{ - }√(16-(-32))}{2} \\ x\text{ = }(-4+√(48))/(2)\text{ and x = }\frac{-4\text{ - }√(48)}{2} \end{gathered}`

Simplifying further,

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

Thus the roots of the given quadratic equation are,

`x\text{ = 2\lparen-1+}√(3))\text{ and x = 2\lparen-1-}√(3))`