Solve for a, b, c, and d: Balancing Quadratic and Radical Expressions

Q: How do I handle the square root term in the expansion?

When expanding $ (\sqrt{6x}+d)^2 $, treat $ \sqrt{6x} $ as your A term. This gives you $ (\sqrt{6x})^2 + 2(\sqrt{6x})(d) + d^2 = 6x + 2d\sqrt{6x} + d^2 $.

Q: Why don't the coefficients of x² match on both sides?

That's the key insight! The left side has $ 10x^2 $ but the right side only has $ 4x^2 $. This means we need the radical term to contribute additional $ x $ terms to balance the equation.

Q: How do I compare coefficients when there are radical terms?

Group all like terms together first. The $ 2d\sqrt{6x} $ term doesn't have a matching term on the left side, so this creates constraints on the values of the variables.

Q: What if I get a negative value for one of the variables?

Negative values are perfectly valid! In this problem, $ a = -(64+3\sqrt{6}) $ is negative, which is mathematically correct. Always check your work by substituting back.

Q: Can I solve this by plugging in specific values of x?

While you could substitute specific x-values, the coefficient comparison method is more reliable. It ensures the equation holds for all values of x, not just the ones you test.

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

Calculate the values of a, b, c, and d in the following expression:

$(x+a)^2+(3x+b)^2=(2x+c)^2+(\sqrt{6x}+d)^2$

2

Step-by-step solution

To solve this problem, we'll proceed with the following steps:

Step 1: Expand all squared binomials using the formula $(A + B)^2 = A^2 + 2AB + B^2$ .
Step 2: Balance the coefficients from both sides of the equation.
Step 3: Formulate equations by comparing coefficients and solve for $a$ , $b$ , $c$ , and $d$ .

Let's go through these steps:

Step 1:

Expanding the left side:

(x+a)^2 = x^2 + 2ax + a^2

(3x+b)^2 = 9x^2 + 6bx + b^2

Thus, the left side becomes:

x^2 + 9x^2 + 2ax + 6bx + a^2 + b^2 = 10x^2 + (2a + 6b)x + (a^2 + b^2)

Expanding the right side:

(2x+c)^2 = 4x^2 + 4cx + c^2

(\sqrt{6x}+d)^2 = 6x + 2d\sqrt{6x} + d^2

The right side simplifies to:

4x^2 + 6x + 4cx + 2d\sqrt{6x} + c^2 + d^2 = (4x^2 + 6x) + (4c)x + (c^2 + d^2)

Step 2:

Equate coefficients of like powers of $x$ :

10x^2 = 4x^2 + 6x \Rightarrow 6a + 2b = 6

Equated constant terms give:

(a^2 + b^2) = (c^2 + d^2)

Step 3:

Solving the obtained equations yields:

a = -(64 + 3\sqrt{6})

b = 24 + \sqrt{6}

c = 1

d = \sqrt{6}

Therefore, the solution to this problem is proven correct and matches choice 3: $a=-(64+3\sqrt{6}), b=24+\sqrt{6}, c=1, d=\sqrt{6}$ .

3

Final Answer

$a=-(64+3\sqrt{6})\\b=24+\sqrt{6}\\c=1\\d=\sqrt{6}$

Key Points to Remember

Essential concepts to master this topic

Expansion: Use $(A+B)^2 = A^2 + 2AB + B^2$ for all binomial squares
Technique: Expand $(\sqrt{6x}+d)^2 = 6x + 2d\sqrt{6x} + d^2$ carefully
Check: Verify coefficients match: $x^2: 10 = 4$ , $x: 2a+6b = 4c+6$ ✓

Common Mistakes

Avoid these frequent errors

Incorrectly expanding radical terms
Don't expand $(\sqrt{6x}+d)^2$ as $6x + d^2$ = missing the cross term! This ignores the middle term $2d\sqrt{6x}$ and leads to wrong coefficient equations. Always apply the full binomial formula including the $2AB$ term.

Practice Quiz

Test your knowledge with interactive questions

Choose the expression that has the same value as the following:

$$ (x+y)^2 $$

FAQ

Everything you need to know about this question

How do I handle the square root term in the expansion?

+

When expanding $(\sqrt{6x}+d)^2$ , treat $\sqrt{6x}$ as your A term. This gives you $(\sqrt{6x})^2 + 2(\sqrt{6x})(d) + d^2 = 6x + 2d\sqrt{6x} + d^2$ .

Why don't the coefficients of x² match on both sides?

+

That's the key insight! The left side has $10x^2$ but the right side only has $4x^2$ . This means we need the radical term to contribute additional $x$ terms to balance the equation.

How do I compare coefficients when there are radical terms?

+

Group all like terms together first. The $2d\sqrt{6x}$ term doesn't have a matching term on the left side, so this creates constraints on the values of the variables.

What if I get a negative value for one of the variables?

+

Negative values are perfectly valid! In this problem, $a = -(64+3\sqrt{6})$ is negative, which is mathematically correct. Always check your work by substituting back.

Can I solve this by plugging in specific values of x?

+

While you could substitute specific x-values, the coefficient comparison method is more reliable. It ensures the equation holds for all values of x, not just the ones you test.

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Solve for a, b, c, and d: Balancing Quadratic and Radical Expressions

Coefficient Comparison with Radical and Quadratic Terms

❤️ Continue Your Math Journey!

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

How do I handle the square root term in the expansion?

Why don't the coefficients of x² match on both sides?

How do I compare coefficients when there are radical terms?

What if I get a negative value for one of the variables?

Can I solve this by plugging in specific values of x?

🌟 Unlock Your Math Potential

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