Question
Solve the following equation: $4x^2 + 4 bx - (a^2 - b^2) = 0$
✓
Answer
$4 x^2+4 b x-\left(a^2-b^2\right)=0 $
$ x^2+b x-\frac{\left(a^2-b^2\right)}{4}=0$
$x ^2+\frac{ a + b }{2} \times-\frac{ a - b }{2} \times-\frac{ a ^2- b ^2}{4}=0$
$x \left\{ x +\frac{ a + b }{2}\right\}-\frac{( a - b )}{2}\left\{ x +\frac{ a + b }{2}\right\}=0$
$\left\{ x +\frac{( a + b )}{2}\right\}\left\{ x -\frac{( a - b )}{2}\right\}=0$
$\left\{ x +\frac{( a + b )}{2}\right\}=0,\left\{ x -\frac{( a - b )}{2}\right\}=0$
$x =-\frac{( a + b )}{2}, x =\frac{( a - b )}{2}$
$ x^2+b x-\frac{\left(a^2-b^2\right)}{4}=0$
$x ^2+\frac{ a + b }{2} \times-\frac{ a - b }{2} \times-\frac{ a ^2- b ^2}{4}=0$
$x \left\{ x +\frac{ a + b }{2}\right\}-\frac{( a - b )}{2}\left\{ x +\frac{ a + b }{2}\right\}=0$
$\left\{ x +\frac{( a + b )}{2}\right\}\left\{ x -\frac{( a - b )}{2}\right\}=0$
$\left\{ x +\frac{( a + b )}{2}\right\}=0,\left\{ x -\frac{( a - b )}{2}\right\}=0$
$x =-\frac{( a + b )}{2}, x =\frac{( a - b )}{2}$
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