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

Solve the following equation:

\( 2x^2-10x-12=0 \)

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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