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 ..........( Equation1)

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 ...........( Equation 2)

From ( equation1) and ( equation 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

❣Holla user❣

Here is ur ans⬇️⬇️⬇️

First Equation:-

f(x)=x^2 - 3x + 2 = 0

f(x)= x^2 - 2x - x + 2 = 0

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

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

f(x)=(x - 2) = 0

f(x)= (x - 1) = 0

f( x )= 2 or f(x) = 1

Second Equation:-

f(x)=x^2 - 6x + 5 = 0

f(x)= x^2 - 5x - x + 5 = 0

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

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

f(x)= (x - 5) = 0

f(x)=(x - 1) = 0

So

f(x)=5

f(x)=1

In both equation (x-1) is a common factor

and the other two resultant factor is

(x-5) and (x-2)

Hope it helps u^_^