Solve the Equation: Balancing (5-3a)² + a and (a+1)² - 31a

Quadratic Equations with No Real Solutions

(53a)2+a=(a+1)231a (5-3a)^2+a=(a+1)^2-31a

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Solve
00:04 Let's use the shortened multiplication formulas to open the parentheses
00:20 Calculate the multiplications and squares
00:59 Collect like terms
01:09 Simplify what we can
01:17 Isolate A
01:31 Any squared expression is always greater than 0
01:36 Therefore it cannot be equal to a negative, no solution
01:39 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

(53a)2+a=(a+1)231a (5-3a)^2+a=(a+1)^2-31a

2

Step-by-step solution

To solve this problem, we'll follow these steps:

  • Step 1: Expand (53a)2 (5-3a)^2 .
  • Step 2: Expand (a+1)2 (a+1)^2 .
  • Step 3: Combine like terms and set the equation to zero.
  • Step 4: Solve for a a .

Now, let's work through each step:

Step 1: Expand (53a)2 (5-3a)^2 :

(53a)2=2530a+9a2 (5-3a)^2 = 25 - 30a + 9a^2 .

Step 2: Expand (a+1)2 (a+1)^2 :

(a+1)2=a2+2a+1 (a+1)^2 = a^2 + 2a + 1 .

Step 3: Substitute the expressions into the equation:

2530a+9a2+a=a2+2a+131a 25 - 30a + 9a^2 + a = a^2 + 2a + 1 - 31a .

Step 4: Simplify both sides:

Left-hand side: 9a229a+25 9a^2 - 29a + 25 .

Right-hand side: a229a+1 a^2 - 29a + 1 .

Set the equation 9a229a+25=a229a+1 9a^2 - 29a + 25 = a^2 - 29a + 1 .

Simplify the equation:

Subtract a229a+1 a^2 - 29a + 1 from both sides:

9a2a229a+29a+251=0 9a^2 - a^2 - 29a + 29a + 25 - 1 = 0 .

8a2+24=0 8a^2 + 24 = 0 .

8a2=24 8a^2 = -24 .

Divide through by 8:

a2=3 a^2 = -3 .

Since a2=3 a^2 = -3 , there are no real solutions for a a because no real number squared equals a negative number. Thus, there are no solutions in the real number set.

Therefore, the correct answer is No solution.

3

Final Answer

No solution

Key Points to Remember

Essential concepts to master this topic
  • Expansion Rule: Use FOIL to expand binomial squares completely
  • Technique: Simplify both sides: 9a229a+25=a229a+1 9a^2 - 29a + 25 = a^2 - 29a + 1
  • Check: When a2=3 a^2 = -3 , no real solutions exist ✓

Common Mistakes

Avoid these frequent errors
  • Forgetting to expand both squared terms completely
    Don't just expand (53a)2 (5-3a)^2 and skip (a+1)2 (a+1)^2 = missing terms and wrong equation! This leads to incorrect simplification and wrong conclusions. Always expand both binomial squares using FOIL: (a+b)2=a2+2ab+b2 (a+b)^2 = a^2 + 2ab + b^2 .

Practice Quiz

Test your knowledge with interactive questions

Declares the given expression as a sum

\( (7b-3x)^2 \)

FAQ

Everything you need to know about this question

What does it mean when I get a squared variable equals a negative number?

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When you get something like a2=3 a^2 = -3 , it means no real solution exists! Since any real number squared is always positive or zero, there's no real number that when squared gives a negative result.

Did I make a mistake if my equation has no solution?

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Not necessarily! Some equations genuinely have no real solutions. Always double-check your algebra, but if you correctly get x2=negative number x^2 = \text{negative number} , then 'no solution' is the right answer.

How do I expand a binomial square like (53a)2 (5-3a)^2 ?

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Use the formula (AB)2=A22AB+B2 (A-B)^2 = A^2 - 2AB + B^2 . So (53a)2=522(5)(3a)+(3a)2=2530a+9a2 (5-3a)^2 = 5^2 - 2(5)(3a) + (3a)^2 = 25 - 30a + 9a^2 .

Why do the -29a terms cancel out in this problem?

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After expanding and combining like terms, both sides have 29a -29a . When you subtract one side from the other, these identical terms cancel: 29a(29a)=0 -29a - (-29a) = 0 .

Should I try to solve using complex numbers instead?

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For most algebra problems, we work with real numbers only. When a2=3 a^2 = -3 , the answer is simply 'no real solution' unless your teacher specifically asks about complex solutions.

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