A = \sqrt21(21 - 13)(21 - 14)(21 - 15) = \sqrt21 \cdot 8 \cdot 7 \cdot 6 - Abu Waleed Tea
Unlock the Power of Square Roots: Solve A = √[21(21 – 13)(21 – 14)(21 – 15)] with Confidence
Unlock the Power of Square Roots: Solve A = √[21(21 – 13)(21 – 14)(21 – 15)] with Confidence
Mathematics often presents challenges in the form of elegant algebraic expressions—like the powerful cancellation technique behind the square root equation:
A = √[21(21 – 13)(21 – 14)(21 – 15)]
At first glance, this may look like a standard expression, but it reveals a deeper mathematical insight involving factoring, simplification, and clever substitution. Let’s unpack this expression step-by-step to uncover how it simplifies effortlessly—and why it matters.
Understanding the Context
Breaking Down the Expression: A = √[21(21 – 13)(21 – 14)(21 – 15)]
Start by calculating each term inside the parentheses:
- 21 – 13 = 8
- 21 – 14 = 7
- 21 – 15 = 6
Key Insights
So the expression becomes:
A = √[21 × 8 × 7 × 6]
Instead of leaving it as a product of four numbers, we simplify by reordering and grouping factors:
A = √(21 × 8 × 7 × 6) = √[21 × 7 × 8 × 6]
Notice the pairing: 21 and 7 are multiples, and 8 and 6 share common structure. This sets the stage for factoring to reveal perfect squares—the key to simplifying square roots.
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Step 1: Factor and Rearrange
Let’s factor each number into primes:
- 21 = 3 × 7
- 8 = 2³
- 7 = 7
- 6 = 2 × 3
Putting it all together:
A = √[(3 × 7) × (2³) × (7) × (2 × 3)]
Now combine like terms:
- 2³ × 2 = 2⁴ (since 2³ × 2 = 2⁴ = 16)
- 3 × 3 = 3²
- 7 × 7 = 7²
So the radicand becomes:
A = √(2⁴ × 3² × 7²)
Step 2: Extract Perfect Squares
Using the property √(a×b) = √a × √b, we separate each squared term:
A = √(2⁴) × √(3²) × √(7²)
