body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} You must have JavaScript enabled to use this site.

mathleaks.com

mathleaks.com

Start chapters

Start

History

History

{{ item.displayTitle }}

No history yet!

Progress & Statistics

Progress

Student

Teacher

Expand menu

Minimize

{{ courseTrack.displayTitle }}

{{ statistics.percent }}%

Sign in to view progress

{{ printedBook.courseTrack.name }} {{ printedBook.name }}

Solving Quadratic Equations with Square Roots

Quadratic Equations

Solving Quadratic Equations with Square Roots 1.7 - Solution

Return to Solving Quadratic Equations with Square Roots

a

The area of a rectangle is calculated by multiplying the base with the height.

A = b \cdot h

b = 3 x

,

h = 2 x

A = 3 x \cdot 2 x

MultiplyMultiply

A = 6 x^{2}

The area is then

6 x^{2} .

b

We will determine $x$ by substituting $54$ for $A$ and solve the resulting equation.

A = 6 x^{2}

A = 54

54 = 6 x^{2}

RearrangeEqnRearrange equation

6 x^{2} = 54

LHS / 6 = RHS / 6

x^{2} = 9

LHS = RHS

x = \pm 9

CalcRootCalculate root

x = \pm 3

x > 0

x = 3

We can exclude the negative root, since lengths must be positive. This means that $x = 3$ and that $2 x = 2 \cdot 3 = 6 and 3 x = 3 \cdot 3 = 9 .$ The rectangle's side lengths are $6$ and $9$ units