Question

For what value of x both the polynomials
x^2 - 3x + 2 and x^2'-6x+5 become zero?​

Answer 1

Solution:

x^2 - 3x + 2 = 0

=> x^2 - 2x - x + 2 = 0

=> x (x - 2) - 1 (x - 2) = 0

=> (x - 2)(x - 1) = 0

=> (x - 2) = 0 or, (x - 1) = 0

=> x = 2 or x = 1 ..........(1)

Also,

x^2 - 6x + 5 = 0

=> x^2 - 5x - x + 5 = 0

=> x(x - 5) - 1(x - 5) = 0

=> (x - 5)(x - 1) = 0

=> (x - 5) = 0 or (x - 1) = 0

=> x = 5 or x = 1 ...........(2)

From (1) and (2) above, it can be said that x^2 - 3x + 2 and x^2 - 6x + 5 will both become zero at x = 1.

Therefore, x = 1.

[ANSWER: 1]

Answer 2

Solution:-

Equation |,

x² - 3x + 2

=> x² -( 2 + 1)x + 2

=>x² -2x -x +2

=> x( x -2) -1(x-2)

=> (x-2)(x-1)

=> x = 2. and. x=1.

Equation ||,

x² -6x +5

=> x² - ( 5+1)x + 5

=> x² -5x -x +5

=> x( x-5) -1(x-5)

=> (x-5)(x-1)

=> x =5. and x =1.