Square of sum: Using variables

Equations with Variables on One SideUsing short multiplication formulasEquations with denominatorsComplete the missing numbersIdentify the greater valueUsing fractionsUsing quadrilateralsMore than one factorizationSystem of equations with no solutionData with powers and rootsUsing multiple rulesNumber of termsEquations with variables on both sidesSolving the problemTransition between expressionsResulting in a quadratic equation
Question 1

The rectangle ABCD is shown below.

AB = X

The ratio between AB and BC is \( \sqrt{\frac{x}{2}} \).


The length of diagonal AC is labelled m.

XXXmmmAAABBBCCCDDD

Determine the value of m:

Question 2

Consider the following relationships between the variables x and y:

\( x^2+4=-6y \)

\( y^2+9=-4x \)

Which answer is correct?

Question 3

Solve the following system of equations:

\( \begin{cases} \sqrt{x}+\sqrt{y}=\sqrt{\sqrt{61}+6} \\ xy=9 \end{cases} \)

Examples with solutions for Square of sum: Using variables

Exercise #1

The rectangle ABCD is shown below.

AB = X

The ratio between AB and BC is x2 \sqrt{\frac{x}{2}} .


The length of diagonal AC is labelled m.

XXXmmmAAABBBCCCDDD

Determine the value of m:

Video Solution

Step-by-Step Solution

We know that:

ABBC=x2 \frac{AB}{BC}=\sqrt{\frac{x}{2}}

We also know that AB equals X.

First, we will substitute the given data into the formula accordingly:

xBC=x2 \frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}

x2=BCx x\sqrt{2}=BC\sqrt{x}

x2x=BC \frac{x\sqrt{2}}{\sqrt{x}}=BC

x×x×2x=BC \frac{\sqrt{x}\times\sqrt{x}\times\sqrt{2}}{\sqrt{x}}=BC

x×2=BC \sqrt{x}\times\sqrt{2}=BC

Now let's look at triangle ABC and use the Pythagorean theorem:

AB2+BC2=AC2 AB^2+BC^2=AC^2

We substitute in our known values:

x2+(x×2)2=m2 x^2+(\sqrt{x}\times\sqrt{2})^2=m^2

x2+x×2=m2 x^2+x\times2=m^2

Finally, we will add 1 to both sides:

x2+2x+1=m2+1 x^2+2x+1=m^2+1

(x+1)2=m2+1 (x+1)^2=m^2+1

Answer

m2+1=(x+1)2 m^2+1=(x+1)^2

Exercise #2

Consider the following relationships between the variables x and y:

x2+4=−6y x^2+4=-6y

y2+9=−4x y^2+9=-4x

Which answer is correct?

Video Solution

Answer

(x+2)2+(y+3)2=0 (x+2)^2+(y+3)^2=0

Exercise #3

Solve the following system of equations:

{x+y=61+6xy=9 \begin{cases} \sqrt{x}+\sqrt{y}=\sqrt{\sqrt{61}+6} \\ xy=9 \end{cases}

Video Solution

Answer

x=612−2.5 x=\frac{\sqrt{61}}{2}-2.5

y=612+2.5 y=\frac{\sqrt{61}}{2}+2.5

or

x=612+2.5 x=\frac{\sqrt{61}}{2}+2.5

y=612−2.5 y=\frac{\sqrt{61}}{2}-2.5