Solve -q2 + 2 = 3q

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Solution.

Given equation is -q² + 2 = 3q

Converting given equation into Standard Form of Quadratic Equation. ax² + bx + c = 0

-q² + 2 = 3q converted into -1x² - 3x + 2 = 0

Comparing it with the standard Form of Quadratic Equation ax² + bx + c = 0

a = -1, b = -3, c = 2

As we know, discriminant = b² - 4ac

Discriminant = (-3)² - 4(-1)(2)

Discriminant = 9 - (-8)

Discriminant = 17

Since discriminant > 0
Both roots are real and unequal.

Using quadratic formula

Roots(x₁, x₂) =	−b ± √ b2 − 4ac
	2a

Roots(x₁, x₂) =	−b ± √ D
	2a

x₁ =	−b + √D
	2a

=	−(-3) + √17
	2(-1)

=	3 + √17
	-2

x₂ =	−b - √D
	2a

=	−(-3) - √17
	2(-1)

=	3 - √17
	-2

Roots: x₁ = , x₂ =

Sum of roots = -b/a

Sum of roots = -(-3)/(-1)

Sum of roots = -3

Product of roots = c/a

Product of roots = (2)/(-1)

Product of roots = -2

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