Question
Solve the following quadratic equation.
$(2 x+3)^2=25$
$(2 x+3)^2=25$
✓
Answer
$4 x^2+12 x +9-25=0 \Rightarrow 4 x ^2+12 x -16=0$
$ \Rightarrow x ^2+3 x -4=0 \text { compare with } ax ^2+ bx + c =0 $
$\Rightarrow a =1, b=3 \text { and } c=-4 $
$\therefore b ^2-4 ac =3^2-4(1)(-4) $
$=9+16 $
$ =25$
$ x =\frac{- b \pm \sqrt{b^2-4 a c}}{2 a }$
$ \Rightarrow x =\frac{-3 \pm \sqrt{25}}{2 \times 1} $
$ \Rightarrow x =\frac{-3 \pm 5}{2} $
$ \Rightarrow x =\frac{-3+5}{2} \text { or } x =\frac{-3-5}{2} $
$\Rightarrow x =\frac{2}{2} \text { or } x =\frac{-8}{2} $
$ \Rightarrow x =1 \text { or } x =-4 $
$ \Rightarrow x ^2+3 x -4=0 \text { compare with } ax ^2+ bx + c =0 $
$\Rightarrow a =1, b=3 \text { and } c=-4 $
$\therefore b ^2-4 ac =3^2-4(1)(-4) $
$=9+16 $
$ =25$
$ x =\frac{- b \pm \sqrt{b^2-4 a c}}{2 a }$
$ \Rightarrow x =\frac{-3 \pm \sqrt{25}}{2 \times 1} $
$ \Rightarrow x =\frac{-3 \pm 5}{2} $
$ \Rightarrow x =\frac{-3+5}{2} \text { or } x =\frac{-3-5}{2} $
$\Rightarrow x =\frac{2}{2} \text { or } x =\frac{-8}{2} $
$ \Rightarrow x =1 \text { or } x =-4 $
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