- Trinomials Real-life Problems
- Trinomials Real-life Problems
- Trinomials Real-life Problems
- Trinomials Real-life Problems
- Trinomials Real-life Problems
- Trinomials Real-life Problems
- Trinomials Real-life Problems
- Trinomials Real-life Problems
Word Problem 1
Earth’s Layers / Real-life Problems / Math
A miner is digging down to study the Earth’s layers. The depth he can reach (in kilometers) is modeled by the trinomial equation:
x² – 6x + 5 = 0
where x represents the possible depth.
Solution
Factor the trinomial:
(x – 5) (x – 1) = 0
Solve for x:
– x – 5 = 0 –> x = 5
– x – 1 = 0 –> x = 1
Verification
Substitute x = 1: 1² – 6(1) + 5 = 1 – 6 + 5 = 0 ✅
Substitute x = 5: 5² – 6(5) + 5 = 25 – 30 + 5 = 0 ✅
Both roots satisfy the equation.
Answer
The miner could dig either 1 km (shallow study near the crust) or 5 km (a deeper layer).
Explanation
This shows how trinomial equations can be used to model real-life situations like measuring underground distances. By solving the equation, we find possible depths where important studies of Earth’s crust and mantle might happen.
Word Problem 2
Science / Tunnel Construction / Math
A tunnel engineer models the cross-sectional area of a tunnel (in m²) using the trinomial equation:
x² + 7x – 60 = 0
where x represents a possible dimension of the tunnel in meters.
Solution
Factor the trinomial:
(x + 12) (x – 5) = 0
Solve each factor:
x + 12 = 0 –> x = –12
x – 5 = 0 –> x = 5
Interpret the solutions:
-12 is not valid (a negative length doesn’t make sense).
5 is valid.
Verification:
Substitute x = 5 back into the original equation:
(5)² + 7(5) – 60 = 25 + 35 – 60 = 0 ✅
Answer
The tunnel’s dimension is 5 meters.
Explanation
This shows how trinomial equations help determine feasible measurements in engineering, rejecting negative values as non-realistic.
Word Problem 3
Ang Kwintas / Real-life Problems / Math
A jeweler designs a triangular pendant. The trinomial equation for its dimensions is:
x² + 4x – 45 = 0
where x is the pendant’s base in cm.
Solution
Factor the trinomial:
(x + 9)(x – 5) = 0
Solve each factor:
x + 9 = 0 –> x = –9
x – 5 = 0 –> x = 5
Interpret the solutions:
–9 is not valid (a negative length doesn’t make sense).
5 is valid.
Verification:
Substitute x = 5 back into the original equation:
(5)² + 4(5) – 45 = 25 + 20 – 45 = 0 ✅
Answer
The pendant’s base is 5 cm.
Explanation
By solving the trinomial, we find the realistic base length, which matches design requirements.
Word Problem 4
Earth’s Crust / Science / Math
Scientists drill into Earth’s crust. The trinomial depth equation is:
x² – 9x + 14 = 0
where x is depth in km.
Solution
Factor the trinomial:
(x – 7)(x – 2) = 0
Solve each factor:
x – 7 = 0 –> x = 7
x – 2 = 0 –> x = 2
Interpret the solutions:
Both 2 and 7 are valid since depth can be positive in both cases.
Verification:
For x = 2: (2)² – 9(2) + 14 = 4 – 18 + 14 = 0 ✅
For x = 7: (7)² – 9(7) + 14 = 49 – 63 + 14 = 0 ✅
Answer
The drill could reach either 2 km (near the crust boundary) or 7 km (deeper, closer to mantle studies).
Explanation
This illustrates how math can model scientific research, giving possible solutions that represent real drilling depths.
Word Problem 5
Greek Mythology / King Midas / Math
King Midas commissions a gate. Its rectangular area (m²) is modeled by:
x² + 11x – 60 = 0
where x is one possible side length in meters.
Solution
Factor the trinomial:
(x + 15)(x – 4) = 0
Solve each factor:
x + 15 = 0 –> x = –15
x – 4 = 0 –> x = 4
Interpret the solutions:
–15 is not valid (a side length cannot be negative).
4 is valid.
Verification:
Substitute x = 4 back into the original equation:
(4)² + 11(4) – 60 = 16 + 44 – 60 = 0 ✅
Answer
One side of the gate is 4 meters.
Explanation
By solving the trinomial equation, we find that a realistic gate side length is 4 meters. The negative solution is rejected since lengths cannot be negative.