Respuesta :
I hope this helps you
1.x^2-8x-3=0
disctirminant =(-8)^2-4.1. (-3)
disctirminant =64+12
disctirminant =76
x1,2= -b(+,-)square root of disctirminant /2a
x1=-(-8)+square root of 76/2.1
x1=8+square root of 76/2
x2=8-square root of 76/2
Answer:
[tex]x-4=\sqrt{19}\Rightarrow x= 4+\sqrt{19}\\\\\ x-4=-\sqrt{19}\Rightarrow x= 4-\sqrt{19}[/tex]
is the solution set for the given equation
Step-by-step explanation:
Given : Equation [tex]x^2-8x=3[/tex]
We have to find the solution set for the given equation using completing the square method.
Consider the given equation [tex]x^2-8x=3[/tex]
Write the given equation in form of [tex]x^2+2ax+a^2=\left(x+a\right)^2[/tex]
Comparing , we have,
x = x
2ax = -8x
a = - 4
Adding [tex]a^2=16[/tex] both side, we have,
[tex]x^2-8x+\left(-4\right)^2=3+\left(-4\right)^2[/tex]
Simplify, we have,
[tex]\left(x-4\right)^2=19[/tex]
[tex]\mathrm{For\:}f^2\left(x\right)=a\mathrm{\:the\:solutions\:are\:}f\left(x\right)=\sqrt{a},\:-\sqrt{a}[/tex]
thus,
[tex](x-4)=\pm\sqrt{19}[/tex]
Therefore,
[tex]x-4=\sqrt{19}\Rightarrow x= 4+\sqrt{19}\\\\\ x-4=-\sqrt{19}\Rightarrow x= 4-\sqrt{19}[/tex]
is the solution set for the given equation
