Solve the Quadratic Equation: Finding X in 7x + 1 + (2x + 3)² = (4x + 2)²

Quadratic Equations with Binomial Expansion

Find X

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

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:08 Let's start by finding X.
00:11 First, let's open the parentheses using simple multiplication formulas.
00:30 Now, calculate the squares and products of each term.
00:49 Next, rearrange the equation so that the right side is zero.
00:58 Then, group similar terms together.
01:10 Continue by simplifying the expression as much as possible.
01:21 Identify the coefficients in the equation.
01:30 Use the formula for roots to find possible solutions.
01:44 Substitute the values, then solve to find the answers.
01:49 Finally, calculate the squares and products if needed.
02:00 And that's how we find the solution to the problem!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Find X

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

2

Step-by-step solution

To solve the equation 7x+1+(2x+3)2=(4x+2)2 7x + 1 + (2x + 3)^2 = (4x + 2)^2 , we follow these steps:

  • Step 1: Expand both sides using the square of a binomial formula.
  • Step 2: Simplify the equation to form a standard quadratic equation.
  • Step 3: Use the quadratic formula to find the roots of the equation.

Step 1: Expand the squares.

The left side: (2x+3)2=4x2+12x+9 (2x + 3)^2 = 4x^2 + 12x + 9 .

The right side: (4x+2)2=16x2+16x+4 (4x + 2)^2 = 16x^2 + 16x + 4 .

Step 2: Substitute back into the original equation and simplify:

7x+1+4x2+12x+9=16x2+16x+4 7x + 1 + 4x^2 + 12x + 9 = 16x^2 + 16x + 4 .

Combine like terms:

4x2+19x+10=16x2+16x+4 4x^2 + 19x + 10 = 16x^2 + 16x + 4 .

Step 3: Move all terms to one side:

4x2+19x+1016x216x4=0 4x^2 + 19x + 10 - 16x^2 - 16x - 4 = 0 .

Which simplifies to:

12x2+3x+6=0-12x^2 + 3x + 6 = 0 .

Step 4: Divide by -3 to simplify:

4x2x2=0 4x^2 - x - 2 = 0 .

Step 5: Use the quadratic formula:

x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=4 a = 4 , b=1 b = -1 , c=2 c = -2 .

Calculate the discriminant:

b24ac=(1)244(2)=1+32=33 b^2 - 4ac = (-1)^2 - 4 \cdot 4 \cdot (-2) = 1 + 32 = 33 .

Calculate the roots:

x=1±338 x = \frac{1 \pm \sqrt{33}}{8} .

Therefore, the solution to the problem is x=1±338 x = \frac{1 \pm \sqrt{33}}{8} .

3

Final Answer

1±338 \frac{1\pm\sqrt{33}}{8}

Key Points to Remember

Essential concepts to master this topic
  • Expansion: Use (a + b)² = a² + 2ab + b² for both squared terms
  • Technique: Collect like terms: 4x² + 19x + 10 becomes -12x² + 3x + 6
  • Check: Substitute x=1+338 x = \frac{1 + \sqrt{33}}{8} back into original equation ✓

Common Mistakes

Avoid these frequent errors
  • Incorrectly expanding squared binomials
    Don't forget the middle term when expanding (2x + 3)² = 4x² + 9! This misses the 2ab term (12x) and gives completely wrong coefficients. Always use the formula (a + b)² = a² + 2ab + b² for every squared binomial.

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 do I need to expand both squared terms?

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You must expand both (2x+3)2 (2x + 3)^2 and (4x+2)2 (4x + 2)^2 to see all the terms clearly. Without expanding, you can't combine like terms or form the standard quadratic equation.

What's the formula for squaring a binomial?

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Use (a+b)2=a2+2ab+b2 (a + b)^2 = a^2 + 2ab + b^2 . For example: (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

How do I know when to use the quadratic formula?

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Use the quadratic formula when you have an equation in the form ax2+bx+c=0 ax^2 + bx + c = 0 and it doesn't factor easily. After expanding and simplifying, we got 4x2x2=0 4x^2 - x - 2 = 0 .

Why does the answer have ± and a square root?

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The ± symbol means there are two solutions to this quadratic equation. The 33 \sqrt{33} comes from the discriminant b24ac=1+32=33 b^2 - 4ac = 1 + 32 = 33 in the quadratic formula.

Can I check my answer without substituting back?

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The most reliable way is substitution, but you can also check that your discriminant calculation is correct. Since we got 33 \sqrt{33} , verify that (1)24(4)(2)=33 (-1)^2 - 4(4)(-2) = 33 .

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