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Chapter 2: Problem 101

Solve each quadratic equation by the method of your choice. $$0=-2(x-3)^{2}+8$$

Short Answer

Expert verified
The solutions to the equation are \( x = -5 \) and \( x = -1 \)

Step by step solution

01

Expand the quadratic equation

Begin by expanding the equation \( -2(x-3)^2+8 \). The expanded equation becomes \( -2(x^2-6x+9)+8 = -2x^2+12x-18+8 = -2x^2+12x-10 \).
02

Rewrite the equation in standard form

Shift the equation so that it equals zero on one side. This forms the quadratic equation \( 0 = -2x^2+12x-10 \).
03

Solve the quadratic equation

Solve the quadratic equation using the quadratic formula, which is \( x = [-b 卤 sqrt(b^2-4ac)] / (2a) \). In this equation, \( a = -2\), \( b = 12 \), and \( c = -10 \). Plugging these values into the quadratic formula gives \( x = [-(-12) 卤 sqrt((-12)^2-4*-2*-10)] / (2*-2) = [12 卤 sqrt(144-80)] / -4 = [12 卤 sqrt(64)] / -4 = [12 卤 8] / -4 \). So the solutions are \( x = -5 \) and \( x = -1 \)

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Most popular questions from this chapter

Complete the square and write the equation in standard form. Then give the center and radius of each circle and graph the equation. $$x^{2}+y^{2}+3 x-2 y-1=0$$

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