PRECALCULUS:...COMMON CORE ED.-W/ACCESS
PRECALCULUS:...COMMON CORE ED.-W/ACCESS
10th Edition
ISBN: 9780134781839
Author: Demana
Publisher: PEARSON
Expert Solution & Answer
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Chapter P.5, Problem 68E
Solution

Derive the quadratic formula, Follow these steps to use completing the square to solve ax2+bx+c=0,a0 .

  1. Subtract c from both sides of the original equation and divide both sides of the resulting equation by a to obtain x2+bax=ca
  2. Add the square of one-half of the coefficient of x in (a) to both sides and simplify to obtain
  3. (x+b2a)2=b24ac4a2

  4. Extract square roots in (b) and solve for x to obtain the quadratic formula
  5. x=b±b24ac2a

Derived quadratic formula x=b±b24ac2a

Given:

The equation is, ax2+bx+c=0

Concept Used:

  • Completing the Square is a method used to solve a quadratic equation by changing the form of the equation so that the left side is a perfect square trinomial.
  • Transform the equation so that the constant term, c, is alone on the right side.
  • If a, the leading coefficient (the coefficient of the  x2 term), is not equal to 1, divide both sides by a.
  • Add the square of half the coefficient of the x -term, ba2 to both sides of the equation.
  • Factor the left side as the square of a binomial.
  • Take the square root of both sides. 

Calculation:

(a)

  ax2+bx+c=0ax2+bx+=cdividebothsidebyax2+bxa+=ca

(b)

   add ( 1 2 b a ) 2 bothside x 2 + bx a + ( 1 2 b a ) 2 = c a + ( 1 2 b a ) 2 x 2 + 2xb 2a + ( b 2a ) 2 = c a + b 4 a 2 2 x+b2a2=4ac+b24a2x+b2a2=b24ac4a2

(c)

  x+b2a=±b24ac4a2x=±b24ac2ab2ax=±b24acb2ax=b±b24ac2a

So the derived quadratic formula is x=b±b24ac2a

Chapter P Solutions

PRECALCULUS:...COMMON CORE ED.-W/ACCESS

Ch. P.1 - Prob. 1ECh. P.1 - Prob. 2ECh. P.1 - Prob. 3ECh. P.1 - Prob. 4ECh. P.1 - Prob. 5ECh. P.1 - Prob. 6ECh. P.1 - Prob. 7ECh. P.1 - Prob. 8ECh. P.1 - Prob. 9ECh. P.1 - Prob. 10ECh. P.1 - Prob. 11ECh. P.1 - Prob. 12ECh. P.1 - Prob. 13ECh. P.1 - Prob. 14ECh. P.1 - Prob. 15ECh. P.1 - Prob. 16ECh. P.1 - Prob. 17ECh. P.1 - Prob. 18ECh. P.1 - Prob. 19ECh. P.1 - Prob. 20ECh. P.1 - Prob. 21ECh. P.1 - Prob. 22ECh. P.1 - Prob. 23ECh. P.1 - Prob. 24ECh. P.1 - Prob. 25ECh. P.1 - Prob. 26ECh. P.1 - Prob. 27ECh. P.1 - Prob. 28ECh. P.1 - Prob. 29ECh. P.1 - Prob. 30ECh. P.1 - Prob. 31ECh. P.1 - Prob. 32ECh. P.1 - Prob. 33ECh. P.1 - Prob. 34ECh. P.1 - Prob. 35ECh. P.1 - Prob. 36ECh. P.1 - Prob. 37ECh. P.1 - Prob. 38ECh. P.1 - Prob. 39ECh. P.1 - Prob. 40ECh. P.1 - Prob. 41ECh. P.1 - Prob. 42ECh. P.1 - Prob. 43ECh. P.1 - Prob. 44ECh. P.1 - Prob. 45ECh. P.1 - Prob. 46ECh. P.1 - Prob. 47ECh. P.1 - In Exercises 4752, simplify the expression. Assume...Ch. 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