Square of sum: Using variables

Equations with Variables on One SideUsing short multiplication formulasEquations with denominatorsComplete the missing numbersIdentify the greater valueUsing fractionsUsing quadrilateralsData with powers and rootsMore than one factorizationSystem of equations with no solutionUsing multiple rulesEquations with variables on both sidesNumber of termsSolving the problemTransition between expressionsResulting in a quadratic equation
Question 1

Given the rectangle ABCD

AB=X

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

We mark the length of the diagonal A the rectangle in m

Check the correct argument:

XXXmmmAAABBBCCCDDD

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

Given the rectangle ABCD

AB=X

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

We mark the length of the diagonal A the rectangle in m

Check the correct argument:

XXXmmmAAABBBCCCDDD

Video Solution

Step-by-Step Solution

Given that:

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

Given that AB equals X

We will substitute accordingly in the formula:

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 focus on triangle ABC and use the Pythagorean theorem:

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

Let's substitute the 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

We'll 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