Answer :
Divide by 2. You will have:
[tex]x^2-8x+15=0[/tex]
First method:
[tex]x^2-8x+15=0 \\ x^2-8x=-15 \qquad \qquad /+16 \\ \underbrace{x^2-8x+16}_{(x-4)^2}=-15+16 \\ \\ (x-4)^2=1 \qquad /\sqrt{} \\ |x-4|=1 \\ \hbox{From deefinition of value absolute:} \\ x_1-4=1 \qquad \hbox{and} \qquad x_2-4=-1 \\ \hbox{So solutions are:} \\ x_1=5 \\ x_2=3[/tex]
Second method:
[tex]x^2-8x+15=0 \\ \Delta= (-8)^2 -4 \cdot 15 \cdot 1 = 64-60=4 \\ \sqrt{\Delta}=2 \\ \hbox{So solutions are:} \\ x_1=\frac{8+2}{2}=5 \\ x_2=\frac{8-2}{2}=3[/tex]
Thirt method:
[tex]x^2-8x+15=0 \\ \hbox{See, that} \ \ -8x= -5x - 3x : \\ \underbrace{x^2-5x}_{x(x-5)} \underbrace{-3x+15}_{-3(x-5)}=0 \\ \\ x(x-5)-3(x-5)=0 \\ \hbox{Factor out the} \ \ x-5: \\ (x-3)(x-5)=0 \\ \hbox{Each factor have to be zero, so:} \\ x-3=0 \\ x-5=0 \\ \hbox{So:} \\ x_1=3 \\ x_2 = 5[/tex]
CHOOSE YOUR FAVOURITE METHOD! :)
[tex]x^2-8x+15=0[/tex]
First method:
[tex]x^2-8x+15=0 \\ x^2-8x=-15 \qquad \qquad /+16 \\ \underbrace{x^2-8x+16}_{(x-4)^2}=-15+16 \\ \\ (x-4)^2=1 \qquad /\sqrt{} \\ |x-4|=1 \\ \hbox{From deefinition of value absolute:} \\ x_1-4=1 \qquad \hbox{and} \qquad x_2-4=-1 \\ \hbox{So solutions are:} \\ x_1=5 \\ x_2=3[/tex]
Second method:
[tex]x^2-8x+15=0 \\ \Delta= (-8)^2 -4 \cdot 15 \cdot 1 = 64-60=4 \\ \sqrt{\Delta}=2 \\ \hbox{So solutions are:} \\ x_1=\frac{8+2}{2}=5 \\ x_2=\frac{8-2}{2}=3[/tex]
Thirt method:
[tex]x^2-8x+15=0 \\ \hbox{See, that} \ \ -8x= -5x - 3x : \\ \underbrace{x^2-5x}_{x(x-5)} \underbrace{-3x+15}_{-3(x-5)}=0 \\ \\ x(x-5)-3(x-5)=0 \\ \hbox{Factor out the} \ \ x-5: \\ (x-3)(x-5)=0 \\ \hbox{Each factor have to be zero, so:} \\ x-3=0 \\ x-5=0 \\ \hbox{So:} \\ x_1=3 \\ x_2 = 5[/tex]
CHOOSE YOUR FAVOURITE METHOD! :)
