# Check the solution x 4 in the original equation 2 7 2

• Test Prep
• 285

This preview shows page 273 - 277 out of 285 pages.

##### We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
The document you are viewing contains questions related to this textbook.
Chapter 2 / Exercise 84
Applied Calculus
Berresford/Rockett
Expert Verified
Check the solution x= 4 in the original equation.2 72(4) = (4)Simplify the expression.16 = (4) or 4 = 4This is not true. There is one solution for the original equation, x= 12.501 Algebra Questions
##### We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
The document you are viewing contains questions related to this textbook.
Chapter 2 / Exercise 84
Applied Calculus
Berresford/Rockett
Expert Verified
475.Square both sides of the equation.x2= 32x+ 10Add (23x10) to both sides of the equation.x232x10 = 0Multiply the equation by 2 to eliminate the fraction.2x23x20 = 0Factor using the trinomial factor form.2x23x20 = (x4)(2x+ 5) = 0Letting each factor of the trinomial factors equal zero results in two possible solutions for the original equation, x= 4 and/or x= 212.Check the first possible solution in the original equation.4 = 32(4) + 10Simplify the radical expression.4 = 16 or 4 = 4The solution x= 4 checks out as a solution for the original equation.Check the second possible solution in the original equation.212= 32(212) + 10A negative number cannot be equal to a positive square root as the radical sign in the original expression calls for. Therefore, x= 212is nota solution to the original equation. The only solution for this equation is x= 4.260501 Algebra Questions
In this chapter,you will have the opportunity to practice solving equa-tions using the quadratic formula. In Chapter 17, you practiced using fac-toring to solve quadratic equations, but factoring is useful only for thoseequations that can easily be factored. The quadratic formula will allow youto find solutions for any quadratic equation that can be put in the form ax2+ bx+ c= 0, where a, b, and care numbers.The quadratic formula tells you that for any equation in the form ax2+bx+ c= 0, the solution will be x=b±2ba24ac. The solutions found using thequadratic formula are also called the roots of the equation. Some solutionswill be in the form of whole numbers or fractions. Some will be in the formof a radical. Some will be undefined, as when the radicand is equal to a neg-ative number. The ± in the quadratic equation tells you that there will betwo solutions, one when you add the radical and one when you subtract it.20Solving Equationswith the Quadratic Formula
Tips for Solving Equations with the Quadratic FormulaTransform the equation into the form ax2+ bx+ c= 0. Use the valuesfor a, b, and cin the quadratic equation to determine the solution forthe original equation.For solutions that contain a radical, be sure to simplify the radical asyou practiced in Chapter 18.When you are asked to find the solution to the nearest hundredth, youcan use a calculator to find the value of the radical.Solve the following equations using the quadratic formula. Reduce answers to theirsimplest form or to the simplest radical form. Use of a calculator is recommended.476.x2+ 2x8 = 0477.2x27x30 = 0478.6x2+ 13x28 = 0479.18x2+ 9x+ 1 = 0480.6x2+ 17x= 28481.