BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

15th Edition

ISBN: 9781608408405

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Cengage Learning

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Question

Chapter 3.5, Problem 53E

To determine

To evaluate: the values for m so that the system has no solution, one solution and two solutions and justify using graph.

Expert Solution & Answer

Answer to Problem 53E

When m=6 , it has no solution.

When m=489 , it has one solution.

When m=6 , it has two solutions.

Explanation of Solution

Given:

3y=−x2+8x−7y=mx+3

Taking the given system of equations,

3y=−x2+8x−7y=mx+3

Substituting ‘ y’ in the first equation,

3(mx+3)=−x2+8x−7x2−8x+3mx+9+7=0x2−8x+3mx+16=0x2−(8−3m)x+16=0

So,

Finding for the no solution,

If the equation has no solution then it means that its discriminant should be less than 0.

b2−4ac<0(8−3m)2−4(1)(16)<064−48m+9m2−64<0−48m+9m2<09m2<48mm<489

It means that if the value of m is less than 489 then, it has no solution.

Justifying it using the graph, taking m = 4,

Equation becomes,

x2−(8−3(4))+16=0x2−(8−12)+16=0x2+4x+16=0

The graph of the above equation is shown below,

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 3.5, Problem 53E , additional homework tip 1

It is justified that it has no solution when m = 4.

Finding for the one solution,

If the equation has one solution then it means that its discriminant should be equal to 0.

b2−4ac=0(8−3m)2−4(1)(16)=064−48m+9m2−64=0−48m+9m2=09m2=48mm=489

It means that if the value of m is equal to 489 then, it has one solution.

Justifying it using the graph, taking m=489 ,

Equation becomes,

x2−(8−3(489))x+16=0x2−(8−16)x+16=0x2+8x+16=0

The graph of the above equation is shown below,

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 3.5, Problem 53E , additional homework tip 2

It is justified that it has one solution when

m=489 .

Finding for the two solutions,

If the equation has two solutions then it means that its discriminant should be greater than to 0.

b2−4ac>0(8−3m)2−4(1)(16)>064−48m+9m2−64>0−48m+9m2>09m2>48mm>489

It means that if the value of m is greater than 489 then, it has two solutions.

Justifying it using the graph, taking m=6 ,

Equation becomes,

x2−(8−3(6))x+16=0x2−(8−18)x+16=0x2+10x+16=0

The graph of the above equation is shown below,

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015, Chapter 3.5, Problem 53E , additional homework tip 3

It is justified that it has two solution when

m>489 .

Chapter 3.5, Problem 52E

Chapter 3.5, Problem 54E

Chapter 3 Solutions

BIG IDEAS MATH Algebra 2: Common Core Student Edition 2015

Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E

Ch. 3.1 - Prob. 11E Ch. 3.1 - Prob. 12E Ch. 3.1 - Prob. 13E Ch. 3.1 - Prob. 14E Ch. 3.1 - Prob. 15E Ch. 3.1 - Prob. 16E Ch. 3.1 - Prob. 17E Ch. 3.1 - Prob. 18E Ch. 3.1 - Prob. 19E Ch. 3.1 - Prob. 20E Ch. 3.1 - Prob. 21E Ch. 3.1 - Prob. 22E Ch. 3.1 - Prob. 23E Ch. 3.1 - Prob. 24E Ch. 3.1 - Prob. 25E Ch. 3.1 - Prob. 26E Ch. 3.1 - Prob. 27E Ch. 3.1 - Prob. 28E Ch. 3.1 - Prob. 29E Ch. 3.1 - Prob. 30E Ch. 3.1 - Prob. 31E Ch. 3.1 - Prob. 32E Ch. 3.1 - Prob. 33E Ch. 3.1 - Prob. 34E Ch. 3.1 - Prob. 35E Ch. 3.1 - Prob. 36E Ch. 3.1 - Prob. 37E Ch. 3.1 - Prob. 38E Ch. 3.1 - Prob. 39E Ch. 3.1 - Prob. 40E Ch. 3.1 - Prob. 41E Ch. 3.1 - Prob. 42E Ch. 3.1 - Prob. 43E Ch. 3.1 - Prob. 44E Ch. 3.1 - Prob. 45E Ch. 3.1 - Prob. 46E Ch. 3.1 - Prob. 47E Ch. 3.1 - Prob. 48E Ch. 3.1 - Prob. 49E Ch. 3.1 - Prob. 50E Ch. 3.1 - Prob. 51E Ch. 3.1 - Prob. 52E Ch. 3.1 - Prob. 53E Ch. 3.1 - Prob. 54E Ch. 3.1 - Prob. 55E Ch. 3.1 - Prob. 56E Ch. 3.1 - Prob. 57E Ch. 3.1 - Prob. 58E Ch. 3.1 - Prob. 59E Ch. 3.1 - Prob. 60E Ch. 3.1 - 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