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X^2= 14x +11
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User Axelly
by
7.1k points

2 Answers

6 votes

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


\frac{(x-7)²}=√(60)

we get

x-7=
√(5*3*2*2)

x-7=±2
√(15)

So

x=7±2
√(15)

Either

x=7+2
√(15)

or

x=7-2
√(15)

User Dan Coughlin
by
7.6k points
7 votes

Answer:


x=7+2√(5) ,7-2√(5)

Explanation:


x^(2) =14x+11


x^(2) -14x-11=0


x\frac{-b\pm\sqrt{b^(2) -4ac} }{2(a)}


x=\frac{14\pm\sqrt{(-14)^(2) -4(1)(-11)} }{2(1)}


x=(14\pm√(240) )/(2)


x=(14\pm4√(15) )/(2)


x=7+2√(5) ,7-2√(5)

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OAmalOHopeO

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User Arnoutaertgeerts
by
8.6k points

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