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3 years ago

Solve for x² in x²-3x+2=0

Mathematics

2 answers:

Naddika [18.5K]3 years ago

6 0

$x^2-3x+2=0\\x^2-x-2x+2=0\\x(x-1)-2(x-1)=0\\(x-2)(x-1)=0\\x=2 \vee x=1\\\\x^2=4 \vee x^2=1$

mafiozo [28]3 years ago

6 0

Answer:

Step-by-step explanation:

First we try to factor x²-3x+2.

We have to look for two numbers that multiply to 2 and add -3.

The two numbers are -1 and -2.

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

x-1 = 0 -> x = 1

x-2 = -> x = 2

Now we find x^2.

(1)^2 = 1

(2)^2 =4

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Solve for x² in x²-3x+2=0​

Solve for x² in x²-3x+2=0