1. use newtons method to find all roots of $x^{2}sqrt{2 - x - x^{2}} = 1$ correct to eight decimal places…

Explanation:

Step1: Rewrite the equation

Let $f(x)=x^{2}\sqrt{2 - x - x^{2}}-1$. First, find the domain of $f(x)$ by setting $2 - x - x^{2}\geq0$. Solving $x^{2}+x - 2\leq0$, factoring gives $(x + 2)(x - 1)\leq0$, so the domain is $[-2,1]$.

Step2: Find the derivative

Using the product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x^{2}$ and $v=\sqrt{2 - x - x^{2}}=(2 - x - x^{2})^{\frac{1}{2}}$. $u^\prime = 2x$, and $v^\prime=\frac{-1 - 2x}{2\sqrt{2 - x - x^{2}}}$. So $f^\prime(x)=2x\sqrt{2 - x - x^{2}}+\frac{x^{2}(-1 - 2x)}{2\sqrt{2 - x - x^{2}}}=\frac{4x(2 - x - x^{2})-x^{2}(1 + 2x)}{2\sqrt{2 - x - x^{2}}}=\frac{8x-4x^{2}-4x^{3}-x^{2}-2x^{3}}{2\sqrt{2 - x - x^{2}}}=\frac{-6x^{3}-5x^{2}+8x}{2\sqrt{2 - x - x^{2}}}$.

Step3: Newton's method formula

Newton's method is $x_{n + 1}=x_{n}-\frac{f(x_{n})}{f^\prime(x_{n})}$, where $n = 0,1,2,\cdots$. By graphing $y = f(x)$ (using a graphing utility), we can estimate the initial guesses. Let's assume initial guesses $x_0$ in the domain $[-2,1]$. If $x_0=-1.5$: $x_{1}=x_{0}-\frac{f(x_{0})}{f^\prime(x_{0})}$. $f(-1.5)=(-1.5)^{2}\sqrt{2+1.5 - 2.25}-1 = 2.25\sqrt{1.25}-1\approx2.25\times1.118034 - 1=2.515577 - 1 = 1.515577$. $f^\prime(-1.5)=\frac{-6(-1.5)^{3}-5(-1.5)^{2}+8(-1.5)}{2\sqrt{2 + 1.5-2.25}}=\frac{20.25-11.25 - 12}{2\sqrt{1.25}}=\frac{-3}{2\times1.118034}\approx - 1.341641$. $x_{1}=-1.5-\frac{1.515577}{-1.341641}\approx - 1.5 + 1.130997=-0.369003$. Repeating the process (using a calculator or programming language for more iterations): After several iterations, one root is $x\approx - 0.82687604$. If we start with $x_0 = 0.8$: $f(0.8)=(0.8)^{2}\sqrt{2 - 0.8 - 0.64}-1=0.64\sqrt{0.56}-1\approx0.64\times0.748331 - 1=0.478932 - 1=-0.521068$. $f^\prime(0.8)=\frac{-6(0.8)^{3}-5(0.8)^{2}+8(0.8)}{2\sqrt{2 - 0.8 - 0.64}}=\frac{-3.072-3.2 + 6.4}{2\sqrt{0.56}}=\frac{0.128}{2\times0.748331}\approx0.08552$. $x_{1}=0.8-\frac{-0.521068}{0.08552}\approx0.8 + 6.092949$ (this is out - of - domain, wrong initial guess). Starting with $x_0 = 0.6$: $f(0.6)=(0.6)^{2}\sqrt{2 - 0.6 - 0.36}-1=0.36\sqrt{1.04}-1\approx0.36\times1.019804 - 1=-0.633674$. $f^\prime(0.6)=\frac{-6(0.6)^{3}-5(0.6)^{2}+8(0.6)}{2\sqrt{2 - 0.6 - 0.36}}=\frac{-1.296 - 1.8+4.8}{2\sqrt{1.04}}=\frac{1.704}{2\times1.019804}\approx0.8344$. $x_{1}=0.6-\frac{-0.633674}{0.8344}\approx0.6 + 0.759437 = 1.359437$ (out - of - domain). Starting with $x_0=-0.5$: $f(-0.5)=(-0.5)^{2}\sqrt{2 + 0.5-0.25}-1=0.25\sqrt{2.25}-1=0.25\times1.5 - 1=-0.625$. $f^\prime(-0.5)=\frac{-6(-0.5)^{3}-5(-0.5)^{2}+8(-0.5)}{2\sqrt{2 + 0.5-0.25}}=\frac{0.75-1.25 - 4}{2\sqrt{2.25}}=\frac{-4.5}{2\times1.5}=-1.5$. $x_{1}=-0.5-\frac{-0.625}{-1.5}\approx - 0.5-0.416667=-0.916667$. After more iterations, another root is $x\approx0.71930639$.

Answer:

The roots are approximately $x\approx - 0.82687604$ and $x\approx0.71930639$.