104th Problem 2016

$4x^{2}+8x+8=4$ $\iff 4(x^{2}+2x+2)=4$ $\iff x^{2}+2x+2=\frac{4}{4}=1$ $\iff (x^{2}+2x+1)+1=1$ $\iff (x+1)^{2}+1=1$ $\iff (x+1)^{2}=1-1=0$ Because $(x+1)^2=0$ , so $x+1=\pm 0$ On the other hand, $-0=0$ , so $x+1=0$ $\boxed{x=-1~only}$

$4x^2 +8x + 8 = 4$

$\Rightarrow 4x^2 + 8x +4 = 0$

Solve it with the quadratic formula :

• For a quadratic equation of the form $ax^2 + bx + c = 0$ the solutions are $x_1, _2 = \dfrac { -b \pm \sqrt {b^2 - 4ac} } {2a}$

Now, applying the formula.

For $a = 4, b= 8, c = 4$

$x_1, _2 = \dfrac {-8 \pm \sqrt {8^2 - 4 \times 4 \times 4} } { 2 \times 4}$

Since, $8^2 - 4 \times 4 \times4 = 0$

$x = \dfrac{-8}{2 \times 4}$

$= \dfrac{-8}{-8}$

$= - \dfrac{8}{8}$

$= \boxed{-1}$

$4x ^ { 2 } + 8x + 4 = 0 \Rightarrow x ^ { 2 } + 2x + 1 = 0 \Rightarrow ( x + 1 ) ^ { 2 } = 0 \Rightarrow x + 1 = 0 \Rightarrow x = -1$ .

$\begin{matrix} &4x^2+8x+8=4&\\ &4x^2+8x+8-4=4-4&\\ &4x^2+8x+4=0&\\ &4(x^2+2x+1)=0&\\ &4(x^2+x+x+1)=0&\\ &4(\left(x^2+x\right)+\left(x+1\right))=0&\\ &4(x\left(x+1\right)+1\left(x+1\right))=0&\\ &4\left(x+1\right)\left(x+1\right)=0&\\ &4\left(x+1\right)^2=0&\\ 4=0&&x+1=0\\ 4\neq 0&&x+1-1=0-1\\ &&\mathbf{x=-1}\\ \end{matrix}$