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How do you solve #2x^2+3x+1=0#?

Answer 1

Eva Abramson

The solutions are:
#x = - 1/2#

#x = - 1 #

#2x^2 + 3x + 1 = 0 # 4 The equation is of the form #color(blue)(ax^2+bx+c=0# where:

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

The source of the discriminant is:

#Delta=b^2-4*a*c#

# = (3)^2-(4 * 2 * 1)#

# = 9 - 8 = 1 #

The solutions are found using the formula #x=(-b+-sqrtDelta)/(2*a)#

#x = ((-3)+-sqrt(1))/(2*2) = (-3 +- ( 1) )/4#

#x = (- 3 + 1 ) / 4 = -2/ 4 = - 1/2#

#x = (- 3 - 1 ) / 4 = -4/4 = - 1 #

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Answer 2

Hazel Anglin

To solve the quadratic equation 2x^2 + 3x + 1 = 0, you can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

where a = 2, b = 3, and c = 1.

Plugging these values into the formula:

x = (-3 ± √(3^2 - 4 * 2 * 1)) / (2 * 2)

x = (-3 ± √(9 - 8)) / 4

x = (-3 ± √1) / 4

x = (-3 ± 1) / 4

This gives two solutions:

x₁ = (-3 + 1) / 4 = -2/4 = -1/2 x₂ = (-3 - 1) / 4 = -4/4 = -1

Therefore, the solutions to the equation 2x^2 + 3x + 1 = 0 are x = -1/2 and x = -1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.