Given: 13X ^ 2-12xy + 4Y ^ 2-4x + 1 = 0, find the value of (3Y) ^ - 4x

Given: 13X ^ 2-12xy + 4Y ^ 2-4x + 1 = 0, find the value of (3Y) ^ - 4x

13X ^ 2-12xy + 4Y ^ 2-4x + 1 = 0 (2Y) ^ 2-2 * 2Y * 3x + (3x) ^ 2 + (2x) ^ 2-2 * 2x + 1 = 0 (13X ^ 2 = 9x ^ 2 + 4x ^ 2) (2y-3x) ^ 2 + (2x-1) ^ 2 = 0, so only (2y-3x) ^ 2 = 0; (2x-1) ^ 2 = 0 get: 2Y = 3x; 2x-1 = 0x = 1 / 2; y = 3 / 4. Substitute: (3Y) ^ - 4x = (3 * 3 / 4) ^ - 4 * 1 / 2 = 9 / 4 ^ - 2 = 16 /

Generalization of x ^ 2-2x + 1 = 0 4x ^ 2 + 13X + 3 = 0
Explore the law, generalize and write the formula
x^2-2x+1=0
(x-1)(x-1)=0
x=1
x^2-3x+2=0
(x-1)(x-2)=0
x=1 x=2
3x^2+x-2=0
3(x-2/3)(x+1)=0
x=2/3,x=-1
2x^2+5x+2=0
2(x+1/2)(x+2)=0
x=-1/2,x=-2
4x^2+13x+3=0
4(x+1/4)(x+3)=0
x=-1/4,x=-3
Above. Conclusion generalization, write the formula

ax^2+(ab+ac)x+abc=0
→a(x+b)(x+c)=0

4X / 9 + 0.5 = 13X / 9-1.3

4X / 9 + 0.5 = 13X / 9-1.3
13X of 9-4X of 9 = 1.8
x=1.8

Solution equation: (1-x ^ 2) ^ 2-4x (1-x-x ^ 2) = 0
Speed!

(1-x^2)^2-4x(1-x-x^2)
=(1-x^2)^2-4x(1-x^2)+4x^2
=(1-x^2+2x)^2=0
(1-x)^2=0
x=1