Square of sum: Using short multiplication formulas

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:

Given that:

$\frac{AB}{BC}=\sqrt{\frac{x}{2}}$

Given that AB equals X

We will substitute accordingly in the formula:

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

$x\sqrt{2}=BC\sqrt{x}$

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

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

$\sqrt{x}\times\sqrt{2}=BC$

Now let's focus on triangle ABC and use the Pythagorean theorem:

$AB^2+BC^2=AC^2$

Let's substitute the known values:

$x^2+(\sqrt{x}\times\sqrt{2})^2=m^2$

$x^2+x\times2=m^2$

We'll add 1 to both sides:

$x^2+2x+1=m^2+1$

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

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

Find X

$(3x+1)^2+8=12$

$x_1=\frac{1}{3},x_2=-1$

Find X

$7=5x^2+8x+(x+4)^2$

$-\frac{4}{3}\pm\frac{\sqrt{10}}{6}$

Solve the following equation:

$(x+3)^2+2x^2=18$

$x_1=1,x_2=-3$

$(x+2)^2-12=x^2$

$x=2$

Calculate x according to the figure shown below below.

x>0

$x=3$

$(x+1)^2=x^2$

$x=-\frac{1}{2}$

Find X

$7x+1+(2x+3)^2=(4x+2)^2$

$\frac{1\pm\sqrt{33}}{8}$

Solve the equation

$2x^2-2x=(x+1)^2$

Answers a + b

Solve the following equation:

$(-x+1)^2=(2x+1)^2$

$x_1=0,x_2=-2$

$\frac{(\frac{1}{x}+\frac{1}{2})^2}{(\frac{1}{x}+\frac{1}{3})^2}=\frac{81}{64}$

Find X

$x=1,-\frac{17}{7}$

Solve the following equation:

$(x+3)^2=2x+5$

$x=-2$

Write an algebraic expression for the area of the square below.

$x^2+2x+1$

Solve the following equation:

$\frac{1}{(x+1)^2}+\frac{1}{x+1}=1$

$-\frac{1}{2}[1\pm\sqrt{5}\rbrack$

Given a circle whose center O. From the center of the circle go out 2 radii that cut the circle at the points A and B.

Given AO⊥OB.

The side AB is equal to and+2.

Express band and the area of the circle.

$\frac{\pi}{2}[y^2+4y+4]$

$(x+3)^2=(x-3)^2$

$x=0$

The square below has an area of 36.

x>0

Calculate x.

$x=5$

Solve the following equation:

$\frac{x^3+1}{(x+1)^2}=x$

$x=\frac{1}{2}$

Shown below is the rectangle ABCD.

AB = y

AD = x

Express the square of the sum of the sides of the rectangle using the area of the triangle DEC.

$(x+y)^2=4s\lbrack\frac{s}{y^2}+\frac{s}{x^2}+1\rbrack$