Solve the Quadratic: Expanding (x+3)² to Find x in 2x+5

Quadratic Equations with Perfect Square Trinomials

Solve the following equation:

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

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Find X
00:03 Use the abbreviated multiplication formulas
00:13 Substitute appropriate values according to the data and open the parentheses
00:30 Substitute in our equation
00:50 Arrange the equation so that one side equals 0
00:58 Collect terms
01:09 Identify coefficients
01:24 Use the roots formula
01:48 Substitute appropriate values and solve
02:07 Calculate the square and products
02:14 The square root of 0 is always equal to 0
02:21 When the root equals 0, the equation will have only one solution
02:48 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve the following equation:

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

2

Step-by-step solution

To solve the equation (x+3)2=2x+5 (x+3)^2 = 2x + 5 , we proceed as follows:

  • Step 1: Expand the left side. Using the identity (a+b)2=a2+2ab+b2 (a+b)^2 = a^2 + 2ab + b^2 , we find:
    (x+3)2=x2+6x+9 (x+3)^2 = x^2 + 6x + 9 .

  • Step 2: Set the equation to zero by moving all terms to one side:
    x2+6x+9=2x+5 x^2 + 6x + 9 = 2x + 5
    Subtract 2x+5 2x + 5 from both sides:
    x2+6x+92x5=0 x^2 + 6x + 9 - 2x - 5 = 0
    This simplifies to:
    x2+4x+4=0 x^2 + 4x + 4 = 0 .

  • Step 3: Solve the quadratic equation x2+4x+4=0 x^2 + 4x + 4 = 0 . Notice this can be factored as:
    (x+2)2=0 (x+2)^2 = 0 .

  • Step 4: Solve for x x by setting the factor equal to zero:
    x+2=0 x+2 = 0 .
    Thus, x=2 x = -2 .

Therefore, the solution to the equation (x+3)2=2x+5 (x+3)^2 = 2x + 5 is x=2 x = -2 .

3

Final Answer

x=2 x=-2

Key Points to Remember

Essential concepts to master this topic
  • Expansion Rule: Use (a+b)2=a2+2ab+b2 (a+b)^2 = a^2 + 2ab + b^2 to expand squares
  • Standard Form: Move all terms to one side: x2+4x+4=0 x^2 + 4x + 4 = 0
  • Verification: Substitute x=2 x = -2 back: (2+3)2=2(2)+5=1 (-2+3)^2 = 2(-2)+5 = 1

Common Mistakes

Avoid these frequent errors
  • Forgetting to expand the squared binomial completely
    Don't just square the first and last terms like (x+3)2=x2+9 (x+3)^2 = x^2 + 9 = missing the middle term! This gives x22x4=0 x^2 - 2x - 4 = 0 instead of the correct x2+4x+4=0 x^2 + 4x + 4 = 0 . Always remember the middle term 2ab 2ab when expanding (a+b)2 (a+b)^2 .

Practice Quiz

Test your knowledge with interactive questions

a = Coefficient of x²

b = Coefficient of x

c = Coefficient of the independent number


what is the value of \( a \) in the equation

\( y=3x-10+5x^2 \)

FAQ

Everything you need to know about this question

Why do I need to expand (x+3)2 (x+3)^2 instead of taking the square root?

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Taking the square root of both sides only works when the right side is a perfect square or constant. Since we have 2x+5 2x + 5 on the right, we must expand and solve as a quadratic equation.

How do I remember the expansion formula (a+b)2 (a+b)^2 ?

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Think "First, Outer, Inner, Last" or use the pattern: square the first term, add twice the product of both terms, then square the last term. For (x+3)2 (x+3)^2 : x2+2(x)(3)+32=x2+6x+9 x^2 + 2(x)(3) + 3^2 = x^2 + 6x + 9 .

What does it mean that x2+4x+4=(x+2)2 x^2 + 4x + 4 = (x+2)^2 ?

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This is called a perfect square trinomial because it factors into a perfect square! When (x+2)2=0 (x+2)^2 = 0 , the only solution is x=2 x = -2 (called a repeated root).

Could I solve this using the quadratic formula instead?

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Yes! Using x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} with a=1,b=4,c=4 a=1, b=4, c=4 gives x=4±02=2 x = \frac{-4 \pm 0}{2} = -2 . But recognizing the perfect square trinomial is faster!

How can I check if my factoring is correct?

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Expand your factored form! If (x+2)2 (x+2)^2 is correct, then (x+2)2=x2+4x+4 (x+2)^2 = x^2 + 4x + 4 , which matches our simplified equation.

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