Big Ideas Math A Bridge To Success Algebra 2: Student Edition 2015

1st Edition

ISBN: 9781680331165

Author: HOUGHTON MIFFLIN HARCOURT

Publisher: Houghton Mifflin Harcourt

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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 A Bridge To Success Algebra 2: 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 A Bridge To Success Algebra 2: 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 A Bridge To Success Algebra 2: 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 A Bridge To Success Algebra 2: Student Edition 2015

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