Question

Find zeroes of the following

$x {}^{2} \: - 3x \: + 2$

Answer 1

your answer is..
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
x=2

x-1=0
x=1

Answer 2

HERE IS YOUR ANSWER...⬇⬇




$\boxed{SOUTION}$ ➡

$x {}^{2} - 3x + 2 \\ \\ = x {}^{2} - (2 + 1)x + 2 \\ \\ = x {}^{2} - 2x - x + 2 \\ \\ = x(x - 2) - 1(x -2 ) \\ \\ = (x - 1)(x - 2)$

To find the zeros,

Let,

$x {}^{2} - 3x + 2 = 0 \\ \\ = > (x - 1)(x - 2) = 0$

Therefore,

$x - 1 = 0 \\ \\ = > x = 1 \\ \\ \\ or \\ \\ \\ x - 2 = 0 \\ \\ = > x = 2$

Hence, the zeros are 1 and 2




$\boxed{ANSWER}$ ➡ 1 , 2

✝✝...HOPE IT HELPS YOU...✝✝