Question

(x^2-2x)^2 - 4(x^2-2x) +3 = 0. SOLVE THIS BY USING QUADRATIC FORMULA​

Answer 1

Answer :

x= -1 , 3 , 1+√2 , 1-√2

Step by step exaplaination :

=>Here, the given Quadratic equation is

(x²-2x)²-4(x²-2x)+3 =0

The General form of quadratic equation :

ax²+bx+c =0

=> Suppose, (x²-2x)= y

=> Now,

y²-4y+3 =0

y²-3y-y+3 =0

y(y-3)-1(y-3) =0

(y-1)(y-3) =0 ...(1)

=> Putting value of y in equation(1) ..

(x²-2x-1)(x²-2x-3) =0

=>Here, x²-2x-1=0

∆=b²-4ac =(-2)²-4(1)(-1) =4+4 =8 >0

(alpha,beta) = -b±√∆ /2a = -(-2)±√8 /2(1) = 2±√8 /2

alpha=2+√8 /2 ; beta=2-√8 /2

alpha×beta = c/a = -1/1 = -1

alpha+beta = -b/a = -(-2)/1 = 2

=>We get,

x= 1+√2 , 1-√2

=>Let, p(x)=0

x²-2x-3=0

x²-3x+x-3=0

x(x-3)+1(x-3)=0

(x+1)(x-3)=0

x+1=0 => x= -1 or

x-3=0 => x= 3

=> We get,

x= -1 , 3 , 1+√2 , 1-√2