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Chapter 2: Problem 102

Solve each quadratic equation by the method of your choice. $$-x^{2}-2 x+1=0$$

Short Answer

Expert verified
The solutions of the equation \(-x^{2}-2x+1=0\) are \(x=1\) and \(x=-1\).

Step by step solution

01

Rewrite the Equation in Standard Form

The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). The given equation is \(-x^{2}-2x+1=0\). Multiply through the equation by -1 to transform it into a more familiar form of \(x^{2}+2x-1=0\).
02

Factor the Quadratic Equation

Find two numbers that multiply to -1 (the value of c) and add to 2 (the value of b). The numbers are -1 and 1. Therefore, the factored form of the quadratic equation is \((x - 1)(x + 1)=0\).
03

Setting Each Factor Equal to Zero

The zero product property states that if a product of factors equals zero, then at least one of the factors must be zero. Set each factor equal to zero and solve for x: \n\n\(x-1=0\) => \(x=1\) \n\n \((x+1)=0\) => \(x=-1\)

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Most popular questions from this chapter

Express the given function \(h\) as \(a\) composition of two functions \(f\) and \(g\) so that \(h(x)=(f \circ g)(x)\). $$h(x)=\sqrt[3]{x^{2}-9}$$

Find a. \((f \circ g)(x)\) b. the domain of \(f \circ g\) $$f(x)=\frac{2}{x+3}, g(x)=\frac{1}{x}$$

Find a. \((f \circ g)(x)\) b. \((g \circ f)(x)\) c. \((f \circ g)(2)\) d. \((g \circ f)(2)\) $$f(x)=\frac{1}{x}, g(x)=\frac{1}{x}$$

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}-x+2 y+1=0$$

Graph \(y_{1}=x^{2}-2 x, y_{2}=x,\) and \(y_{3}=y_{1} \div y_{2}\) in the same \([-10,10,1]\) by \([-10,10,1]\) viewing rectangle. Then use the TRACE \(]\) feature to trace along \(y_{3} .\) What happens at \(x=0 ?\) Explain why this occurs.
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