# f(x)= 2x2-3x-2 Determine the zeros of each function by factoring:

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### 3 Answers

The roots of f(x) = 2x^2-3x-2 are required.

Solve the quadratic equation 2x^2 - 3x - 2 = 0.

The roots of a quadratic equation ax^2 + bx + c = 0 are given by `(-b+-sqrt(b^2-4ac))/(2a)`

For the equation 2x^2 - 3x - 2 = 0, a = 2, b = -3 and c = -2, the roots of the equation are:

`(-(-3)+-sqrt((-3)^2-4*2*-2))/(2*2)`

`(3+-sqrt(9 + 16))/4`

`(3+-sqrt 25)/4`

`(3+- 5)/4`

8/4 and -2/4

2 and -1/2

The roots of f(x) = 2x^2-3x-2 are 2 and -1/2

The function is:

f(x) = 2x^2 - 3x - 2

We have to find out the values of x for which f(x) = 0. That is when:

2x^2 - 3x - 2 = 0

To find the values of x satisfying this equation we factorise the left hand side of the equation as follows:

2x^2 - 4x + x - 2 = 0

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

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

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

Therefor:

x = 2, and -1/2

f(x) = 2x^2-3x-2.

To determine zeros of the function, we equate the function and solve for x by factorisation:

2x^2-3x-2 = 0. Or

2x^2-4x +x-2 = 0. Or

2x(x-2) +1(x-2) = 0. Or

(x-2)(2x+1) = 0. Or

x-2=0. Or 2x+1 = 0. Or

x=2 Or 2x=-1 implies x = -1/2.

So x=2 and x=-1/2 are the values of x for which (x) becomes zero.