It is known that a b is the root of the two real numbers of the equation x x 2013 0, then the value

Solution: a2+2a+b=(a2+a)+(a+b) Because a is a root of the equation, a2+a-2013=0, find a2+a=2013

Since a,b are the two real roots of the equation, a+b=-(1 1)=-1 so the result is 2012

Find the quadratic coefficient, the primary term coefficient and the constant term of the one-element quadratic equation, and use the relationship between the root and the coefficient to find the value of the sum of the two roots a+b and the product of the two roots ab. With a as the solution of the equation, x=a is substituted into the equation to obtain the equation about a, and the value of a +a is obtained after deformation, and the second term 2a of the equation is changed to a+a, the first two items are combined, the last two items are combined, and the respective values are substituted to obtain the value

Solution: a and b are the two real roots of the equation x + x-2013=0, a+b=-1, ab=-2013;Veda's theorem

and a is the real root of the equation x +x-2013=0, substituting x=a into the equation obtains: a +a-2013=0, that is, a +a = 2013, then a +2a+b = (a +a) + (a + b) = 2013-1 = 2012

So the answer is: 2012

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a is the root of the equation x + x-2015 = 0.

a + a-2015 = 0, then a + a = 2015 a and b are the roots of the equation x + x - 2015 = 0.

From Vedder's theorem: a+b=-1

a²+2a+b=a²+a+a+b

Substituting a into the equation yields: a+a-2015=0

Relationship between root and coefficient: a+b=-1

The addition of the two formulas yields: a +2a+b-2015=-1, so a +2a+b=2014

a, b are the two real roots of the equation x 2 + x - 2015 = 0, a 2 + a = 2015, a + b = -1

a^2+2a+b

a^2+a+a+b

Knowing that a,b are the two real roots of the equation x +x-2011=0, then:

a +a-2011 = 0 i.e. a + a = 2011 and can be obtained from Veda's theorem: a + b = -1

So: a +2a+b=a +a+a+b=2011-1=2010

Since a and b are the roots of the equation x x 2013=0, then:

a²＋a－2013=0

b²＋b－2013=0

Subtract the two formulas to get:

a²－b²)＋a－b)=0

a－b)(a＋b＋1)=0

Since a≠b, then: a b 1 = 0, i.e., a b = 1a 2a b

a²＋a)＋(a＋b)

Substituting a into the rolling of the family formula vernacular: a +a-2015=0 by the relationship between the root and the coefficient: a + b = -1

The addition of the two formulas yields: a large dispersion + 2a + b - 2015 = -1 so a + 2a + b = 2014

Solution source shielding: a, b are the real roots of two unphased hail modes of equation x +x-2013=0, then a+b=-1, a and a=2013, a +2a+b=(a +a)+(a+b)=-1+2013=2012

Substitute a for x +x-2012=0

A +a-2012=0, i.e., a +a=2012 is obtained according to the relationship between roots and coefficients (Veda's theorem), a+b=-1, so a +2a+b=a +a+a+b=2012-1=2011

Hello: There are many ways to solve this problem, try it with the formula method.

Here are my steps:

Answer: Your method is very good, why can't you use the crosses to multiply?

x=a, then a +a-2013=0

a²=-a+2013

Veda's theorem a+b=-1

So the original formula =-a+2013+2a+b

a+b+2013

It can be obtained according to the title.

a²+a-2013=0

i.e. a + a = 2013

and a+b=-1

So a +2a+b=a +a+a+b