How do you find the zeroes of #g(x)=2x^2-5x-3 #?

Answer 1
#g(x) = 2x^2 -5x-3#can be factored as #color(white)("XXXX")##g(x)=(2x+1)(x-3)#
The zeroes of #g(x)# are the values of #x# for which #g(x)=0#
If #g(x) = (2x+1)(x-3) = 0# then either #color(white)("XXXX")##2x+1 = 0 rArr x= -1/2# or #color(white)("XXXX")##x-3 = 0 rArr x = 3#
So the zeros of #g(x)# are #(-1/2)# and #3#
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Answer 2

To find the zeroes of the function g(x) = 2x^2 - 5x - 3, you can use the quadratic formula, which is given by:

x = [-b ± √(b^2 - 4ac)] / (2a)

where a = 2, b = -5, and c = -3. Substituting these values into the formula, you get:

x = [5 ± √((-5)^2 - 4(2)(-3))] / (2*2)

Simplify the expression inside the square root:

x = [5 ± √(25 + 24)] / 4

x = [5 ± √49] / 4

x = [5 ± 7] / 4

This gives you two possible solutions:

  1. x = (5 + 7) / 4 = 12 / 4 = 3
  2. x = (5 - 7) / 4 = -2 / 4 = -0.5

So, the zeroes of the function g(x) = 2x^2 - 5x - 3 are x = 3 and x = -0.5.

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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.

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.

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.

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.

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