The solution of the equation a (x + m) &# 178; + B = 0 of X is X1 = - 2, X2 = 1, then the solution of the equation a (x + m + 2) &# 178; + B = 0 is X1 = - 2, X2 = 1

The solution of the equation a (x + m) &# 178; + B = 0 of X is X1 = - 2, X2 = 1, then the solution of the equation a (x + m + 2) &# 178; + B = 0 is X1 = - 2, X2 = 1

Take x + 2 as a whole
Using known x
a[(x+2)+m]²+b=0
Bring in
have to
x1=-4,x2=-1

Given that X1 and X2 are two real roots of the equation x2 + 3x + 1 = 0, then X12 + 8x2 + 20=______ ．

From the known, we can get X1 + x2 = - 3, x1 · x2 = 1, and ∵ X12 + 3x1 + 1 = 0, that is, X12 = - 3x1-1, ∵ X12 + 8x2 + 20 = - 3x1 + 8x2 + 19 (set as a), which is combined with X1 + x2 = - 3, we can get X1 = - A + 511, X2 = a − 2811, substituting into x1 · x2 = 1, we can get - A + 511 · a − 2811 = 1, sorting, we can get a2-23a-19 = 0, we can get a = 23 ± 11 & nbsp; 52

It is known that there are two unequal real roots of the equation x square + 2 (a + 1) x - (b-2) square = 0 about X. find the 2010 degree of A

The topic should have two equal real roots, otherwise I can't do it
Δ＝4（a+1）²－4×[-(b-2)²]＝4（a+1）²+4（b-2）²＝0
We get a = - 1
The power of a is 1

It is known that the two real roots of the equation x ^ 2 + K ^ 2x + M = 0 about X are a, B,
The two real roots of the equation y ^ 2 - 5ky + 2m + 4 = 0 are a + 3, B + 3. Find the value of K, M

By Weida theorem
a+b=-k^2
ab=m
Veda's theorem on the equation of Y
(a+3)(b+3)=5k,a+b+6=5k
a+b=-k^2
So - K ^ 2 + 6 = 5K
k^2+5k-6=0
(k+6)(k-1)=0
k=-6,k=1
(a+3)(b+3)=2m+4
ab+3(a+b)+9=2m+4
That is m + 3 (a + b) + 9 = 2m + 4
m=3(a+b)+5
If k = - 6, a + B = - K ^ 2 = - 36, M = - 103
If k = 1, a + B = - K ^ 2 = - 1, M = 2
When k = 1, M = 2, the discriminant of x ^ 2 + K ^ 2x + M = 0 is less than 0, there is no solution, and it is rounding off
So k = - 6, M = - 103