Question

(x^2-3x)^2+3(x^2-3x)+2=0
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Answer 1

Answer:

The roots of the given quadratic equation are x = 2 or x = 1.

Step-by-step-explanation:

We have given a quadratic equation.

We have to find the roots of the quadratic equation.

The given quadratic equation is ( x² - 3x )² + 3 ( x² - 3x ) + 2 = 0.

Now,

( x² - 3x )² + 3 ( x² - 3x ) + 2 = 0

⇒ ( x² - 3x )² + 2 ( x² - 3x ) + ( x² - 3x ) + 2 = 0

⇒ ( x² - 3x ) [ ( x² - 3x ) + 2 ] + 1 [ ( x² - 3x ) + 2 ] = 0

⇒ [ ( x² - 3x ) + 2 ] ( x² - 3x + 1 ) = 0

⇒ ( x² - 3x + 2 ) = 0 OR ( x² - 3x + 1 ) = 0

⇒ x² - 3x + 2 = 0 OR x² - 3x + 1 = 0

⇒ x² - 2x - x + 2 = 0 OR x² - 2x - x + 1 = 0

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

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

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

⇒ ( x - 2 ) = 0 OR ( x - 1 ) = 0

⇒ x - 2 = 0 OR x - 1 = 0

⇒ x = 2 OR x = 1

∴ The roots of the given quadratic equation are x = 2 or x = 1.