① If two of the equations x ^ 2 + 2 (a + 3) x + 2a-3 = 0, one is greater than 3 and the other is less than 3, find the value range of the real number a ② The equation x ^ 2 + X + 1-A = 0 has two real roots with different signs, and the value range of a is obtained I didn't understand how to solve the problem in class

① If two of the equations x ^ 2 + 2 (a + 3) x + 2a-3 = 0, one is greater than 3 and the other is less than 3, find the value range of the real number a ② The equation x ^ 2 + X + 1-A = 0 has two real roots with different signs, and the value range of a is obtained I didn't understand how to solve the problem in class

There are two unequal real roots of ∵ equation ∵ and ∵ one is greater than 3, the other is less than 3, and the coefficient of quadratic term of equation is greater than 0, that is, the function f (x) = x & sup2; + 2 (a + 3) x + 2a-3 has an opening upward and an intersection with X axis

Solve the following equation by factorization!
(1)1/9(x+1)^2-2/3(x+1)+1=0
(2)9(x-2)^2-(x+1)^2=0
(3)x^2+8x+7=0

(1)1/9(x+1)^2-2/3(x+1)+1=[1/3(1+x)]^2-2/3(x+1)+1=[1/(1+x)-1)^2=0 1/(1+x)-1=0 1/(1+x)=1 1+x=1 x=0(2)9(x-2)^2-(x+1)^2=[3(x-2)]^2- (x+1)^2=(3x-6-x-1 )(3x-6+x+1)=0 3x-6-x-1=0 x=7/3 3x-6+x+1==0 x=5/4 (3)x^...

Some factorization equation problems, master quickly
If the value of the algebraic expression 3x ^ 2 + 3x is 6, then the value of X is 6____
Solving equation 1. (x-3) ^ 2 + 2x (x-3) = 0; 2. (x-1) (x + 3) = 12 by factorization
Thank you

3x ^ 2 + 3x = 6 divide both sides by 3x ^ 2 + x = 2x ^ 2 + X-2 = 0 (x-1) (x + 2) = 0x1 = 1, X2 = - 21. (x-3) ^ 2 + 2x (x-3) = 0; (x-3) (x-3 + 2x) = 0 (that is, take x-3 as a whole) (x-3) (3x-3) = 0x1 = 3, X2 = 12. (x-1) (x + 3) = 12 x ^ 2 + 2x-3 = 12x ^ 2 + 2x-15 = 0 (x-3) (x + 5) = 0x1 = 3, X2 = - 5

A junior high school factorization equation problem!
a^2-6a+b^2+2b+10=0
^A ^ 2 is the square of A
Must have the detailed process and the written instruction, must not surpass the beginning 1 knowledge scope!
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a^2-6a+b^2+2b+10=0
Divide 10 into 9 + 1 and write the equation as
a^2-6a+9+b^2+2b+1=0
Note that a ^ 2-6a + 9 and B ^ 2 + 2B + 1 are complete square terms, and the equation is written as
(a-3)^2+(b+1)^2=0
Because the square term cannot be negative, and the sum of two non negative numbers equals 0, they must be equal to 0 respectively
So we have (A-3) ^ 2 = 0 and (B + 1) ^ 2 = 0
So a = 3, B = - 1
If you don't know how to use the full square term, I'll add the problem, and I'll tell you more about it