Answer:
[tex]x=7+2\sqrt{5} ,7-2\sqrt{5}[/tex]
Step-by-step explanation:
[tex]x^{2} =14x+11[/tex]
[tex]x^{2} -14x-11=0[/tex]
[tex]x\frac{-b\pm\sqrt{b^{2} -4ac} }{2(a)}[/tex]
[tex]x=\frac{14\pm\sqrt{(-14)^{2} -4(1)(-11)} }{2(1)}[/tex]
[tex]x=\frac{14\pm\sqrt{240} }{2}[/tex]
[tex]x=\frac{14\pm4\sqrt{15} }{2}[/tex]
[tex]x=7+2\sqrt{5} ,7-2\sqrt{5}[/tex]
------------------------
OAmalOHopeO
-----------------------
Answer:
Solution given:
x²=14x+11
keeping all term on same side
x²-14x-11=0
trying to make it a perfect square
x²-2*7*x+7²-7²-11=0
By using formula of (x-y)²=x²-2xy+y²we get
(x-7)²-60=0
keeping 60 in right side
(x-7)²=60
doing square root on both side
[tex]\frac{(x-7)²}=\sqrt{60}[/tex]
we get
x-7=[tex]\sqrt{5*3*2*2}[/tex]
x-7=±2[tex]\sqrt{15}[/tex]
So
x=7±2[tex]\sqrt{15}[/tex]
Either
or
