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Chapter 2: Problem 77

Use the Quadratic Formula to solve the quadratic equation. $$\frac{3}{2} x^{2}-6 x+9=0$$

Short Answer

Expert verified
The roots of the quadratic equation are not real.

Step by step solution

01

Identifying coefficients

In the given equation \(\frac{3}{2} x^{2}-6 x+9=0\), using the standard form of a quadratic equation \(ax^{2} + bx + c = 0\), the coefficients can identified as \(a = \frac{3}{2}\), \(b = -6\), and \(c = 9\).
02

Substituting coefficients into the Quadratic Formula

Substitute the values of a, b, and c into the Quadratic formula to get: \(x = \frac{-(-6) \pm \sqrt{(-6)^{2} - 4 * (\frac{3}{2}) * 9}}{2 * (\frac{3}{2})}\), which simplifies to \(x = \frac{6 \pm \sqrt{36 - 54}}{3}\).
03

Solving the square root and fraction

Solve inside the square root first, \(36 - 54 = -18\). \(\sqrt{-18}\) is not a real number. Thus, the roots of the quadratic equation are not real and no x can satisfy this equation.

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Most popular questions from this chapter

Write the polynomial as the product of linear factors and list all the zeros of the function. $$h(x)=x^{2}-2 x+17$$

Find all the zeros of the function. When there is an extended list of possible rational zeros, use a graphing utility to graph the function in order to disregard any of the possible rational zeros that are obviously not zeros of the function. $$g(x)=x^{5}-8 x^{4}+28 x^{3}-56 x^{2}+64 x-32$$

Determine (if possible) the zeros of the function \(g\) when the function \(f\) has zeros at \(x=r_{1}, x=r_{2},\) and \(x=r_{3}\) $$g(x)=3 f(x)$$

Find all real zeros of the function. $$f(z)=12 z^{3}-4 z^{2}-27 z+9$$

Write the polynomial as the product of linear factors and list all the zeros of the function. $$f(x)=x^{2}+36$$
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