Question

use a graphing calculator to solve the equation by graphing. if necessary, round to the nearest hundredth. $x^3 - 9x^2 + 27x = 20$

Explanation:

Step 1: Rearrange the equation

First, we rewrite the equation \(x^{3}-9x^{2}+27x = 20\) as \(y=x^{3}-9x^{2}+27x - 20\). We need to find the \(x\)-intercepts of this function because the solution to the original equation occurs where \(y = 0\).

Step 2: Analyze the function (or use a graphing calculator)

We can analyze the function \(y=x^{3}-9x^{2}+27x - 20\). Let's try to find some roots by testing values.

• When \(x = 1\): \(y=1^{3}-9(1)^{2}+27(1)-20=1 - 9 + 27 - 20=-1\)
• When \(x = 2\): \(y=2^{3}-9(2)^{2}+27(2)-20=8 - 36 + 54 - 20=6\)
• When \(x = 5\): \(y=5^{3}-9(5)^{2}+27(5)-20=125 - 225 + 135 - 20=15\)
• When \(x = 4\): \(y=4^{3}-9(4)^{2}+27(4)-20=64 - 144 + 108 - 20=8\)
• When \(x = 1\) gives \(y=-1\), \(x = 2\) gives \(y = 6\), so by Intermediate Value Theorem, there is a root between \(x = 1\) and \(x = 2\). Let's use a graphing calculator (or more precise calculation). Alternatively, we can rewrite the left - hand side of the original equation. Notice that \(x^{3}-9x^{2}+27x=x(x^{2}-9x + 27)\), and also \(x^{3}-9x^{2}+27x=(x - 3)^{3}+0\)? Wait, \((x - 3)^{3}=x^{3}-9x^{2}+27x - 27\), so \(x^{3}-9x^{2}+27x=(x - 3)^{3}+27\). Then the equation becomes \((x - 3)^{3}+27=20\), so \((x - 3)^{3}=- 7\), then \(x-3=\sqrt[3]{-7}\approx - 1.913\), so \(x=3 - 1.913 = 1.087\approx1.09\). Wait, but let's check with the function \(y=x^{3}-9x^{2}+27x - 20\). Let's use a graphing calculator approach. If we graph \(y=x^{3}-9x^{2}+27x - 20\), we can see that the function has one real root (since the derivative \(y^\prime=3x^{2}-18x + 27=3(x^{2}-6x + 9)=3(x - 3)^{2}\geq0\), the function is non - decreasing (with a point of inflection at \(x = 3\) where the derivative is zero but the function is still non - decreasing)). Let's solve \(x^{3}-9x^{2}+27x-20 = 0\). We can also use the rational root theorem. The possible rational roots are factors of \(20\) divided by factors of \(1\), so \(\pm1,\pm2,\pm4,\pm5,\pm10,\pm20\). We saw that \(x = 1\): \(1-9 + 27-20=-1\), \(x = 2\): \(8-36 + 54-20=6\), \(x = 4\): \(64-144 + 108-20=8\), \(x = 5\): \(125-225 + 135-20=15\), \(x=\frac{20}{1}=20\) (too big). Wait, maybe my earlier approach with the cube was wrong. Wait, \((x - 3)^{3}=x^{3}-9x^{2}+27x - 27\), so \(x^{3}-9x^{2}+27x=(x - 3)^{3}+27\). Then the equation \(x^{3}-9x^{2}+27x=20\) is \((x - 3)^{3}+27 = 20\), so \((x - 3)^{3}=-7\), so \(x=3-\sqrt[3]{7}\approx3 - 1.913 = 1.087\approx1.09\). Let's check \(x = 1.09\): \((1.09)^{3}-9(1.09)^{2}+27(1.09)\). \(1.09^{3}\approx1.09\times1.09\times1.09\approx1.09\times1.1881\approx1.295\), \(9(1.09)^{2}=9\times1.1881 = 10.6929\), \(27(1.09)=29.43\). Then \(1.295-10.6929 + 29.43\approx1.295+29.43 - 10.6929\approx30.725 - 10.6929\approx20.0321\approx20\). So the solution is \(x\approx1.09\) (and also, since the function is non - decreasing, is there only one real root? Wait, the derivative \(y^\prime=3(x - 3)^{2}\geq0\), so the function is always increasing (with a horizontal tangent at \(x = 3\)). So when \(x = 3\), \(y=(3)^{3}-9(3)^{2}+27(3)-20=27-81 + 81-20=7\). When \(x\) approaches \(-\infty\), \(y=x^{3}-9x^{2}+27x - 20\) approaches \(-\infty\), and as \(x\) increases, \(y\) increases. We saw that at \(x = 1\), \(y=-1\), at \(x = 2\), \(y = 6\), so there is one real root between \(x = 1\) and \(x = 2\), and since the function is increasing, only one real root. So using the graphing calculator (or the algebraic method), we find that \(x\approx1.09\) (rounded to the nearest hundredth).

Answer:

\(x\approx1.09\)