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 70E
Solution

a.

Show that the sum of the two solutions of the equation ax2+bx+c=0, a≠0 and b2−4ac>0 is −ba .

The sum of the two solutions is x1+x2 =−ba

Given:

The equation is, ax2+bx+c=0

Concept Used:

  • The sum of the zeroes is equal to the negative of the coefficient of x by the coefficient of x2. 
  • The product of the zeroes is equal to the constant term by the coefficient of x2.

Calculation:

  ax2+bx+c=0x=−b±b2−4ac2ax1=−b+b2−4ac2a,x2=−b−b2−4ac2a

The sum of two solution of equation is −

  x1+x2=−b+b2−4ac2a+−b−b2−4ac2ax1+x2 =−b+b2−4ac−b−b2−4ac2ax1+x2 =−b−b2ax1+x2 =−2b2ax1+x2 =−ba

So the sum of the two solutions of the equation is −ba

b.

Show that the product of the two solutions of the equation ax2+bx+c=0, a≠0 and b2−4ac>0 is ca .

The product of the two solutions is x1⋅x2=ca

Given:

The equation, ax2+bx+c=0

Concept Used:

  • The sum of the zeroes is equal to the negative of the coefficient of x by the coefficient of x2. 
  • The product of the zeroes is equal to the constant term by the coefficient of x2.

Calculation:

  ax2+bx+c=0x=−b±b2−4ac2ax1=−b+b2−4ac2a,x2=−b−b2−4ac2a

The product of two solution of equation is −

  x1⋅x2=(−b+b2−4ac2a)⋅(−b−b2−4ac2a)x1⋅x2=−(−b+b2−4ac2a)⋅(b+b2−4ac2a)x1⋅x2=−[(b2−4ac)2−b2]4a2x1⋅x2=−(b2−4ac−b2)4a2x1⋅x2=−(−4ac)4a2x1⋅x2=ca

So the product of the two solutions of the equation is ca

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Chapter P.5, Problem 69E
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Chapter P.5, Problem 71E

Chapter P Solutions

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

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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