# 3:03 PM Fri Jan 25139bartleby.combartleby subjectsSearchTextbooksSciencePhysicsCollege Physics Volume 1 (Chs. 1-16),10th EditionChapter 0, Problem 15Ptwo solutions.Use the quadratic formula x =In general quadratic equation has two solutions, which may be real or complex numbers.If b2 - 4ac, then two roots are equal and real numbers.If b2 〉 4ac, that is b2-4ac is positive, then two roots are unequal and real numbers.If b2く4ac, that is b2-4ac is negative, then two roots are unequal complex numbers and cannotrepresent physical quantities and quadratic equation has mathematical solution but no physicalsolutionConclusion:4.9t2 + 2t -20- 0 is a quadratic equationHere, b- > 4ac, then two roots are unequal and real numbers.Substitute in quadratic formula2(4.9)9.8-2+6/1-2-6/11Or9.89.81.83 or - 2.23Chapter 0, Problem 14PChapter 0, Problem 16PWas this solution helpful?

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I need help understanding how they got 6 square root of 11. help_outlineImage Transcriptionclose3:03 PM Fri Jan 25 139 bartleby.com bartleby subjects Search Textbooks Science Physics College Physics Volume 1 (Chs. 1-16), 10th Edition Chapter 0, Problem 15P two solutions. Use the quadratic formula x = In general quadratic equation has two solutions, which may be real or complex numbers. If b2 - 4ac, then two roots are equal and real numbers. If b2 〉 4ac, that is b2-4ac is positive, then two roots are unequal and real numbers. If b2く4ac, that is b2-4ac is negative, then two roots are unequal complex numbers and cannot represent physical quantities and quadratic equation has mathematical solution but no physical solution Conclusion: 4.9t2 + 2t -20- 0 is a quadratic equation Here, b- > 4ac, then two roots are unequal and real numbers. Substitute in quadratic formula 2(4.9) 9.8 -2+6/1-2-6/11 Or 9.8 9.8 1.83 or - 2.23 Chapter 0, Problem 14P Chapter 0, Problem 16P Was this solution helpful? fullscreen
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Step 1

We have

Step 2

Now open the bracket and simplify, we get

Step 3

Now simplify the square root...

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