Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 months ago

# How many real roots does the equation x^2 + 3|x| + 2 = 0 have?

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• 9 months ago

x² + 3|x| + 2 = 0 → you know that a square is always positive or zero.

x² ≥ 0 → you know that an absolute value is always positive or zero.

x² + 3|x| ≥ 0 → if you add 2 to the previous value, you can say that:

x² + 3|x| + 2 > 0 → and you can conclude that

x² + 3|x| + 2 = 0 ← is no possible

• Ian H
Lv 7
9 months ago

The equation x^2 + 3x + 2 has real roots x = -1 and x = -2

The graph of y = x^2 + 3|x| + 2 = 0 does not cross the x axis,

so, x^2 + 3|x| + 2 = 0 has no real roots, as is also confirmed by Pope

by remarking that for real values the positives on the left can not equal the negative on the right with x^2 + 3|x| = -2

The equation does have two complex roots.

• David
Lv 7
9 months ago

If you mean x^2 +3x + 2 = 0 then by solving the quadratic equation x = -1 or x = -2 meaning it has two real roots and when factored it is (x+1)(x+2) = 0

• sepia
Lv 7
9 months ago

x^2 + 3|x| + 2 = 0

x^2 + 3x + 2 = 0  or x^2 - 3x + 2 = 0

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

x^2 + 3|x| + 2 = 0 has 4 real roots.

• Pope
Lv 7
9 months ago

x² + 3|x| + 2 = 0

x² + 3|x| = -2

For all real x, the left side of the equation is non-negative, while the right side is negative. Therefore, no real x satisfies the equation.

• TomV
Lv 7
9 months ago

Assume x < 0

x² - 3x + 2 = 0

x² - 2x - x + 2 = 0

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

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

x = 1, 2 : invalid solution. Violation of the assumption.

Assume x > 0

x² + 3x + 2 = 0

(x+2)(x+1) = 0

x = -1, -2 : invalid solution. Violation of the assumption.

Ans: The equation has no real roots.