Square of sum: Using short multiplication formulas

Equations with Variables on One SideUsing variablesEquations 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

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

Question 2

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

Question 3

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

Question 4

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 5

Solve the following equation:

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

Examples with solutions for Square of sum: Using short multiplication formulas

Exercise #1

(x+1)2=x2 (x+1)^2=x^2

Video Solution

Step-by-Step Solution

Let's examine the given equation:

(x+1)2=x2 (x+1)^2=x^2 First, let's simplify the equation, using the perfect square binomial formula:

(a±b)2=a2±2ab+b2 (a\pm b)^2=a^2\pm2ab+b^2 ,

We'll start by opening the parentheses on the left side using the perfect square formula and then move terms and combine like terms, in the final step we'll solve the simplified equation we get:

(x+1)2=x2↓x2+2⋅x⋅1+12=x2x2+2x+1=x22x=−1/:2x=−12 (x+1)^2=x^2 \\ \downarrow\\ x^2+2\cdot x\cdot1+1^2=x^2\\ x^2+2x+1= x^2\\ 2x=-1\hspace{6pt}\text{/}:2\\ \boxed{x=-\frac{1}{2}} Therefore, the correct answer is answer A.

Answer

x=−12 x=-\frac{1}{2}

Exercise #2

(x+2)2−12=x2 (x+2)^2-12=x^2

Video Solution

Step-by-Step Solution

Let's examine the given equation:

(x+2)2−12=x2 (x+2)^2-12=x^2 First, let's simplify the equation, for this we'll use the perfect square binomial formula:

(a±b)2=a2±2ab+b2 (a\pm b)^2=a^2\pm2ab+b^2 ,

We'll start by opening the parentheses on the left side using the perfect square formula and then move terms and combine like terms, in the final step we'll solve the simplified equation we get:

(x+2)2−12=x2↓x2+2⋅x⋅2+22−12=x2x2+4x+4−12=x24x=8/:4x=2 (x+2)^2-12=x^2 \\ \downarrow\\ x^2+2\cdot x\cdot2+2^2-12=x^2\\ x^2+4x+4-12= x^2\\ 4x=8\hspace{6pt}\text{/}:4\\ \boxed{x=2} Therefore, the correct answer is answer C.

Answer

x=2 x=2

Exercise #3

(x+3)2=(x−3)2 (x+3)^2=(x-3)^2

Video Solution

Step-by-Step Solution

Let's examine the given equation:

(x+3)2=(x−3)2 (x+3)^2=(x-3)^2 First, let's simplify the equation, for this we'll use the perfect square formula for a binomial squared:

(a±b)2=a2±2ab+b2 (a\pm b)^2=a^2\pm2ab+b^2 ,

We'll start by opening the parentheses on both sides simultaneously using the perfect square formula mentioned, then we'll move terms and combine like terms, and in the final step we'll solve the simplified equation we get:

(x+3)2=(x−3)2↓x2+2⋅x⋅3+32=x2−2⋅x⋅3+32x2+6x+9=x2−6x+9x2+6x+9−x2+6x−9=012x=0/:12x=0 (x+3)^2=(x-3)^2 \\ \downarrow\\ x^2+2\cdot x\cdot3+3^2= x^2-2\cdot x\cdot3+3^2 \\ x^2+6x+9= x^2-6x+9 \\ x^2+6x+9- x^2+6x-9 =0\\ 12x=0\hspace{6pt}\text{/}:12\\ \boxed{x=0} Therefore, the correct answer is answer A.

Answer

x=0 x=0

Exercise #4

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 #5

Solve the following equation:

(x+3)2+2x2=18 (x+3)^2+2x^2=18

Video Solution

Answer

x1=1,x2=−3 x_1=1,x_2=-3

Question 1

Find X

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

Question 2

Find X

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

Question 3

Solve the following equation:

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

Question 4

Calculate x according to the figure shown below below.

\( x>0 \)

x+1x+1x+1xxxx+2x+2x+2

Question 5

Solve the following equation:

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

Exercise #6

Find X

7=5x2+8x+(x+4)2 7=5x^2+8x+(x+4)^2

Video Solution

Answer

−43±106 -\frac{4}{3}\pm\frac{\sqrt{10}}{6}

Exercise #7

Find X

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

Video Solution

Answer

x1=13,x2=−1 x_1=\frac{1}{3},x_2=-1

Exercise #8

Solve the following equation:

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

Video Solution

Answer

−12[1±5] -\frac{1}{2}[1\pm\sqrt{5}\rbrack

Exercise #9

Calculate x according to the figure shown below below.

x>0

x+1x+1x+1xxxx+2x+2x+2

Video Solution

Answer

x=3 x=3

Exercise #10

Solve the following equation:

(−x+1)2=(2x+1)2 (-x+1)^2=(2x+1)^2

Video Solution

Answer

x1=0,x2=−2 x_1=0,x_2=-2

Question 1

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

Find X

Question 2

Solve the following equation:

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

Question 3

Solve the equation

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

Question 4

Find X

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

Question 5

Solve the following equation:

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

Exercise #11

(1x+12)2(1x+13)2=8164 \frac{(\frac{1}{x}+\frac{1}{2})^2}{(\frac{1}{x}+\frac{1}{3})^2}=\frac{81}{64}

Find X

Video Solution

Answer

x=1,−177 x=1,-\frac{17}{7}

Exercise #12

Solve the following equation:

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

Video Solution

Answer

x=−2 x=-2

Exercise #13

Solve the equation

2x2−2x=(x+1)2 2x^2-2x=(x+1)^2

Video Solution

Answer

Answers a + b

Exercise #14

Find X

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

Video Solution

Answer

1±338 \frac{1\pm\sqrt{33}}{8}

Exercise #15

Solve the following equation:

x3+1(x+1)2=x \frac{x^3+1}{(x+1)^2}=x

Video Solution

Answer

x=12 x=\frac{1}{2}

Question 1

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

x+1x+1x+1

Question 2

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.

and+2and+2and+2AAABBBOOO

Question 3

The square below has an area of 36.

\( x>0 \)

Calculate x.

363636x+1x+1x+1

Question 4

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.

YYYXXXAAABBBCCCDDDEEE

Exercise #16

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

x+1x+1x+1

Video Solution

Answer

x2+2x+1 x^2+2x+1

Exercise #17

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.

and+2and+2and+2AAABBBOOO

Video Solution

Answer

π2[y2+4y+4] \frac{\pi}{2}[y^2+4y+4]

Exercise #18

The square below has an area of 36.

x>0

Calculate x.

363636x+1x+1x+1

Video Solution

Answer

x=5 x=5

Exercise #19

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.

YYYXXXAAABBBCCCDDDEEE

Video Solution

Answer

(x+y)2=4s[sy2+sx2+1] (x+y)^2=4s\lbrack\frac{s}{y^2}+\frac{s}{x^2}+1\rbrack