Precalculus: Graphical, Numerical, Algebraic Common Core 10th Edition
Precalculus: Graphical, Numerical, Algebraic Common Core 10th Edition
10th Edition
ISBN: 9780134673301
Author: Waits, Foley, Kennedy, Bock Demana
Publisher: PEARSON
Expert Solution & Answer
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Chapter P.5, Problem 71E
Solution

The equation 2x2+bx+c=0 has two solutions x1 and x2 . If x1+x2=5 and x1⋅x2=3 . Find the two solutions.

  x1=5+132

  x2=5−132

Given:

The equation is, 2x2+bx+c=0

Concept Used:

  • Use factorization method to solve.
  • To reduce the algebraic or quadratic equation into its simpler form.
  • The equations are represented as the product of factors instead of expanding the brackets.

Calculation:

Reduce the algebraic or quadratic equation and then solve as shown below,

   2 x 2 +bx+c=0 here a=2, b=b,  c=c x= −b± b 2 −4ac 2a x= −b± b 2 −4⋅2⋅c 2⋅2 x= −b± b 2 −8c 4

   x= −b+ b 2 −8c 4 , −b− b 2 −8c 4 x 1 = −b+ b 2 −8c 4   &  x 2 = −b− b 2 −8c 4   According to question−

   −b+ b 2 −8c 4 + −b− b 2 −8c 4  =5 −b+ b 2 −8c −b− b 2 −8c 4 =5 −2b 4 =5 −b 2 =5 b=−10 and  x 1 ⋅ x 2 =3

   ( −b+ b 2 −8c 4 )⋅( −b− b 2 −8c 4 )=3 −( −b+ b 2 −8c 4 )⋅( b+ b 2 −8c 4 )=3 −[ ( b 2 −8c ) 2 − (b) 2 ] 16 =3 −[ ( b 2 −8c ) 2 − (b) 2 ]=3⋅16 −( b 2 −8c− b 2 )=48 8c=48 c= 48 8 c=6

  putting the values of b and c above−as above x1=−b+b2−8c4x1=−(−10)+(−10)2−8⋅64x1=10+100−484x1=10+524x1=10+2⋅2⋅134x1=10+2134x1=5+132x2=−b−b2−8c4x2=−(−10)−(−10)2−8⋅64x2=10−100−484x2=10−524x2=10−2⋅2⋅134x2=10−2134x2=5−132

Thus, the solution of the given equation is x={+32, −32 } .

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Chapter P.5, Problem 70E
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Chapter P.6, Problem 1QR

Chapter P Solutions

Precalculus: Graphical, Numerical, Algebraic Common Core 10th Edition

Ch. P.1 - Prob. 1QRCh. P.1 - Prob. 2QRCh. P.1 - Prob. 3QRCh. P.1 - Prob. 4QRCh. P.1 - Prob. 5QRCh. P.1 - Prob. 6QRCh. P.1 - Prob. 7QRCh. P.1 - Prob. 8QRCh. P.1 - Prob. 9QRCh. P.1 - Prob. 10QR
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