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Core Connections Algebra 2, 2013

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Core Connections Algebra 2, 2013 View details

2. Section 4.2

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Exercise 95 Page 198

Practice makes perfect

a

Since our variable is in a square root, let's start with raising both sides of the equation to the power of 2.

sqrt(x+15)=5+sqrt(x)

LHS^2=RHS^2

(sqrt(x+15))^2=(5+sqrt(x))^2

▼

Simplify

( sqrt(a) )^2 = a

x+15=(5+sqrt(x))^2

ExpandPosPerfectSquare

(a+b)^2=a^2+2ab+b^2

x+15=25+10sqrt(x)+(sqrt(x))^2

( sqrt(a) )^2 = a

x+15=25+10sqrt(x)+x

Now we will gather all variables on one side of the equation and all of the constant terms on the other side of the equation.

x+15=25+10sqrt(x)+x

LHS-x=RHS-x

15=25+10sqrt(x)

LHS-25=RHS-25

-10=10sqrt(x)

.LHS /10.=.RHS /10.

-1=sqrt(x)

Rearrange equation

sqrt(x)=-1

Normally the next step would be to raise both sides of the equation to the power of 2 to eliminate the square root. Let's see what would happen.

sqrt(x)=-1

LHS^2=RHS^2

(sqrt(x))^2=(-1)^2

( sqrt(a) )^2 = a

x=(-1)^2

Calculate power

x=1

It looks like x is the solution to our equation. To make sure it is a correct answer, let's substitute 1 for x in the original equation.

sqrt(x+15)=5+sqrt(x)

x= 1

sqrt(1+15)? =5+sqrt(1)

▼

Simplify

Add terms

sqrt(16)? =5+sqrt(1)

Calculate root

4? =5+1

Add terms

4≠6

We ended with the false statement, which means that x=1 is not a correct answer. Therefore our equation has no solution, because the square root of any number cannot be negative.

b

To solve an equation, we should gather all terms on one side of the equation. But before we do that, let's first expand the expression inside the parentheses.

(y-6)^2+10=3y

ExpandNegPerfectSquare

(a-b)^2=a^2-2ab+b^2

y^2-12y+36+10=3y

LHS-3y=RHS-3y

y^2-15y+36+10=0

Add terms

y^2-15y+46=0

Now we have a quadratic equation in terms of only the y-variable.

y^2-15y+46=0 ⇔ 1y^2+( -15)y+ 46=0 To solve the above equation, let's recall the Quadratic Formula. x=- b±sqrt(b^2-4ac)/2a We can substitute a= 1, b= -15, and c= 46 into this formula to solve the quadratic equation.

y=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

y=-( -15)±sqrt(( -15)^2-4( 1)( 46))/2( 1)

▼

Solve for y

a-(- b)=a+b

y=15±sqrt((-15)^2-4(1)(46))/2(1)

Calculate power

y=15±sqrt(225-4(1)(46))/2(1)

a * 1=a

y=15±sqrt(225-4(46))/2

(- a)b = - ab

y=15±sqrt(225-184)/2

Subtract term

y=15±sqrt(41)/2

This result tells us that we have two solutions for y. One of them will use the positive sign, and the other one will use the negative sign.

Subchapter links

4.2.1 Solving Inequalities with One or Two Variables p.188-190

4.2.2 Using Systems to Solve a Problem p.193-194

4.2.3 Application of Systems of Linear Inequalities p.197-198

4.2.4 Using Graphs to Find Solutions p.200-201