Question

Plsssss helppppp 100 points

Find the zero of the polynomial

`{x}^(2) - (3x)/(2) - 7`
​

Answer 1

We can find the zeros of the given polynomial by setting it equal to zero and solving for x.

`{\texttt{{x}^(2) - (3x)/(2) - 7 = 0}}`

To solve for x, we can use the quadratic formula:

`{\texttt{x = (-b \pm √(b^2-4ac))/(2a)}}`

where a, b, and c are the coefficients of the quadratic equation.

In this case, a = 1, b = -3/2, and c = -7. Substituting these values in the quadratic formula, we get:

`{\texttt{x = (-(-3/2) \pm √((-3/2)^2-4(1)(-7)))/(2(1))}}`

Simplifying the expression inside the square root, we get:

`{\texttt{x = (3/2 \pm √(9/4+28))/(2)}}`

`{\texttt{x = (3)/(4) \pm \sqrt{(121)/(16)}}}`

`{\texttt{x = (3)/(4) \pm (11)/(4)}}`

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Therefore, the zeros of the polynomial are:

`{\texttt{x_1 = (3)/(4) + (11)/(4) = 3}}`

`{\texttt{x_2 = (3)/(4) - (11)/(4) = -(8)/(4) = -2}}`

Hence, the zeros of the polynomial are 3 and -2.