Solve the Quadratic Equation: (x+2)² = (2x+3)²

Quadratic Equations with Perfect Square Forms

Solve the following equation:

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

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:07 Let's find the value of X.
00:10 We'll use short multiplication formulas.
00:39 Substitute the given values into the equation, and expand the brackets.
01:03 Now, plug them into our equation.
01:06 Expand the second set of brackets the same way.
01:35 Let's calculate the squares and products.
01:51 Insert these into our equation.
02:10 Rearrange so one side equals zero.
02:28 Group similar terms together.
02:51 Take a look at the coefficients.
03:04 Apply the quadratic formula.
03:30 Substitute the values and find the solution.
03:46 Calculate the squares and products again.
04:11 Find the square root of 4.
04:30 There are two solutions: one for addition, and one for subtraction.
04:49 And that's the solution to our problem.

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve the following equation:

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

2

Step-by-step solution

We will solve the equation (x+2)2=(2x+3)2 (x+2)^2 = (2x+3)^2 by expanding and simplifying both sides:

Step 1: Expand both sides of the equation:
Left side: (x+2)2=x2+4x+4 (x+2)^2 = x^2 + 4x + 4
Right side: (2x+3)2=4x2+12x+9 (2x+3)^2 = 4x^2 + 12x + 9

Step 2: Set the expanded forms equal to each other:
x2+4x+4=4x2+12x+9 x^2 + 4x + 4 = 4x^2 + 12x + 9

Step 3: Rearrange to form a standard quadratic equation:
Subtract x2+4x+4 x^2 + 4x + 4 from both sides:
0=3x2+8x+5 0 = 3x^2 + 8x + 5

Step 4: Rearrange to get:
3x2+8x+5=0 3x^2 + 8x + 5 = 0

Step 5: Solve using the quadratic formula:
Using a=3 a = 3 , b=8 b = 8 , c=5 c = 5 :
x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
x=8±8243523 x = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 3 \cdot 5}}{2 \cdot 3}
x=8±64606 x = \frac{-8 \pm \sqrt{64 - 60}}{6}
x=8±46 x = \frac{-8 \pm \sqrt{4}}{6}
x=8±26 x = \frac{-8 \pm 2}{6}

Step 6: Calculate the solutions:
x1=8+26=66=1 x_1 = \frac{-8 + 2}{6} = \frac{-6}{6} = -1
x2=826=106=53 x_2 = \frac{-8 - 2}{6} = \frac{-10}{6} = -\frac{5}{3}

Verify in the original equation to assure correctness. Hence, both solutions are valid.

Therefore, the solutions are x1=1 x_1 = -1 and x2=53 x_2 = -\frac{5}{3} , which matches choice 3.

3

Final Answer

x1=1,x2=53 x_1=-1,x_2=-\frac{5}{3}

Key Points to Remember

Essential concepts to master this topic
  • Perfect Squares: When both sides are squared expressions, expand carefully
  • Technique: Use quadratic formula with a=3,b=8,c=5 a=3, b=8, c=5
  • Check: Verify both solutions in original equation: (1+2)2=(2(1)+3)2 (-1+2)^2 = (2(-1)+3)^2

Common Mistakes

Avoid these frequent errors
  • Taking square root of both sides without considering signs
    Don't just take (x+2)2=(2x+3)2 \sqrt{(x+2)^2} = \sqrt{(2x+3)^2} and write x+2 = 2x+3! This ignores negative possibilities and gives only one solution instead of two. Always expand both sides completely to find all solutions.

Practice Quiz

Test your knowledge with interactive questions

a = coefficient of x²

b = coefficient of x

c = coefficient of the constant term


What is the value of \( c \) in the function \( y=-x^2+25x \)?

FAQ

Everything you need to know about this question

Why can't I just take the square root of both sides?

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Taking square roots seems easier, but you miss solutions! When a2=b2 a^2 = b^2 , it means a=b a = b OR a=b a = -b . Expanding gives you the complete picture.

How do I expand (2x+3)2 (2x+3)^2 correctly?

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Use the pattern (a+b)2=a2+2ab+b2 (a+b)^2 = a^2 + 2ab + b^2 . So (2x+3)2=(2x)2+2(2x)(3)+32=4x2+12x+9 (2x+3)^2 = (2x)^2 + 2(2x)(3) + 3^2 = 4x^2 + 12x + 9 .

What if I get a negative discriminant?

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A negative discriminant means no real solutions exist. In this problem, we got 4=2 \sqrt{4} = 2 , which is positive, so we have two real solutions.

Do I always get two solutions for quadratic equations?

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Not always! You get two solutions when the discriminant is positive, one solution when it equals zero, and no real solutions when it's negative.

How do I verify my answers are correct?

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Substitute each solution back into the original equation. For x=1 x = -1 : (1+2)2=1 (-1+2)^2 = 1 and (2(1)+3)2=12=1 (2(-1)+3)^2 = 1^2 = 1

Why is my final quadratic 3x2+8x+5=0 3x^2 + 8x + 5 = 0 ?

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After expanding both sides, you move everything to one side. The x2 x^2 terms: x24x2=3x2 x^2 - 4x^2 = -3x^2 , so you get 3x28x5=0 -3x^2 - 8x - 5 = 0 , which equals 3x2+8x+5=0 3x^2 + 8x + 5 = 0 .

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