Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
1st Edition
ISBN: 9781680331141
Author: HOUGHTON MIFFLIN HARCOURT
Publisher: Houghton Mifflin Harcourt
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Chapter 9.5, Problem 54E

a.

To determine

To solve the equation with graphing method.

a.

Expert Solution
Check Mark

Answer to Problem 54E

Graph:

  Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015, Chapter 9.5, Problem 54E , additional homework tip  1

Explanation of Solution

Given:

Equation:

  3x2+11x+6=0

Calculation:

On comparing the equation with ax2+bx+c

  a=3,b=11,c=6

Step:1 Finding the axis of symmetry.

  x=b2a

So,

  x=112(3)

Therefore,

  x=116

So,

  x=1.83

Step:2 a=1 , therefore a>0 so graph will open up.

Step:3

Find the plotting points:

  y=x2+4x+4

For x=1.83 find y

  y=3(1.83)2+11(1.83)+6

Opening brackets,

  y=3(3.34)20.13+6

Again

  y=10.0214.13

So,

  y=4.083

Again, for x=1 find y

  y=3(1)2+11(1)+6

So,

  y=311+6

So,

  y=2

Again, for x=1 find y

  y=3(1)2+11(1)+6

So,

  y=3+11+6

So,

  y=20

Therefore plotting points are:

    x-1-1.831
    y-2-4.08320

Plot the graph:

Plot all the points and axis of symmetry.

  Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015, Chapter 9.5, Problem 54E , additional homework tip  2

b.

To determine

To solve the equation with factoring method.

b.

Expert Solution
Check Mark

Answer to Problem 54E

  x=3 or x=23

Explanation of Solution

Given:

Equation:

  3x2+11x+6=0

Calculation:

Split the middle term in such a way that the sum comes 11 and product comes 18

  3x2+2x+9x+6=0

Taking common

  x(3x+2)+3(3x+2)=0

So,

  (x+3)(3x+2)=0

Therefore

  (x+3)=0 or (3x+2)=0

So,

  x=3 or 3x=2

So,

  x=3 or x=23

c.

To determine

To solve the equation with quadratic formula.

c.

Expert Solution
Check Mark

Answer to Problem 54E

  x=3 or x=23

Explanation of Solution

Given:

Equation:

  3x2+11x+6=0

Calculation:

On comparing the equation with ax2+bx+c

  a=3,b=11,c=6

Finding the value of x

  x=b±b24ac2a

Substituting the values,

  x=(11)±(11)24(3)(6)2(3)

Opening brackets,

  x=11±121726

So,

  x=11±496

Solving root,

  x=11±76

Therefore

  x=11+76 or x=1176

So,

  x=46 or x=186

So,

  x=3 or x=23

Chapter 9 Solutions

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015

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