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4a In the imaginary case that T=0, what would the depth be?
4a • State a, b and c for the quadratic:
4a • State a, b and c for the quadratic: 
a = −9.76 x 10−5 b = 1.21 c = 967
4a • State a, b and c for the quadratic: 
a = −9.76 x 10−5 b = 1.21 c = 967 
• Substitute a, b and c into the quadratic formula and simplify:
4a • State a, b and c for the quadratic: 
a = −9.76 x 10−5 b = 1.21 c = 967 
• Substitute a, b and c into the quadratic formula and simplify: 
z = 
−b ± b2 − 4ac 
2a
4a • State a, b and c for the quadratic: 
a = −9.76 x 10−5 b = 1.21 c = 967 
• Substitute a, b and c into the quadratic formula and simplify: 
z = 
−b ± b2 − 4ac 
2a 
z = 
−1.21 ± 1.212 − (4 x − 9.76 x 10−5 x 967 
2 x − 9.76 x 10−5 = 
−1.21 ± 1.357 
−1.952 x 10−4
4a • State a, b and c for the quadratic: 
a = −9.76 x 10−5 b = 1.21 c = 967 
• Substitute a, b and c into the quadratic formula and simplify: 
z = 
−b ± b2 − 4ac 
2a 
z = 
−1.21 ± 1.212 − (4 x − 9.76 x 10−5 x 967 
2 x − 9.76 x 10−5 = 
−1.21 ± 1.357 
−1.952 x 10−4 
• Solve for the two possible roots:
4a • State a, b and c for the quadratic: 
a = −9.76 x 10−5 b = 1.21 c = 967 
• Substitute a, b and c into the quadratic formula and simplify: 
z = 
−b ± b2 − 4ac 
2a 
z = 
−1.21 ± 1.212 − (4 x − 9.76 x 10−5 x 967 
2 x − 9.76 x 10−5 = 
−1.21 ± 1.357 
−1.952 x 10−4 
• Solve for the two possible roots: 
z1 = 
−1.21 + 1.357 
−1.952 x 10−4 = −753.07m
4a • State a, b and c for the quadratic: 
a = −9.76 x 10−5 b = 1.21 c = 967 
• Substitute a, b and c into the quadratic formula and simplify: 
z = 
−b ± b2 − 4ac 
2a 
z = 
−1.21 ± 1.212 − (4 x − 9.76 x 10−5 x 967 
2 x − 9.76 x 10−5 = 
−1.21 ± 1.357 
−1.952 x 10−4 
• Solve for the two possible roots: 
z1 = 
−1.21 + 1.357 
−1.952 x 10−4 = −753.07m 
z2 = 
−1.21 − 1.357 
−1.952 x 10−4 = ퟏퟑ, ퟏퟓퟎ. ퟔퟏ퐦
4a • State a, b and c for the quadratic: 
a = −9.76 x 10−5 b = 1.21 c = 967 
• Substitute a, b and c into the quadratic formula and simplify: 
z = 
−b ± b2 − 4ac 
2a 
z = 
−1.21 ± 1.212 − (4 x − 9.76 x 10−5 x 967 
2 x − 9.76 x 10−5 = 
−1.21 ± 1.357 
−1.952 x 10−4 
• Solve for the two possible roots: 
z1 = 
−1.21 + 1.357 
−1.952 x 10−4 = −753.07m 
z2 = 
−1.21 − 1.357 
−1.952 x 10−4 = ퟏퟑ, ퟏퟓퟎ. ퟔퟏ퐦 
• Depth cannot be negative so z1 is not a valid answer.
4b More realistically, at what depths does the equation 
predict that the temperature will reach 2000°C?
4b • Substitute the temperature into the equation:
4b • Substitute the temperature into the equation: 
2000 = −9.76 x 10−5 z2 + 1.21z + 967
4b • Substitute the temperature into the equation: 
2000 = −9.76 x 10−5 z2 + 1.21z + 967 
• Rearrange the equation into the form 0 = az2 + bz + c:
4b • Substitute the temperature into the equation: 
2000 = −9.76 x 10−5 z2 + 1.21z + 967 
• Rearrange the equation into the form 0 = az2 + bz + c: 
0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000
4b • Substitute the temperature into the equation: 
2000 = −9.76 x 10−5 z2 + 1.21z + 967 
• Rearrange the equation into the form 0 = az2 + bz + c: 
0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000 
0 = −9.76 x 10−5 z2 + 1.21z − 1033
4b • Substitute the temperature into the equation: 
2000 = −9.76 x 10−5 z2 + 1.21z + 967 
• Rearrange the equation into the form 0 = az2 + bz + c: 
0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000 
0 = −9.76 x 10−5 z2 + 1.21z − 1033 
• Substitute the values of a, b and c into the quadratic formula and solve for the two roots:
4b • Substitute the temperature into the equation: 
2000 = −9.76 x 10−5 z2 + 1.21z + 967 
• Rearrange the equation into the form 0 = az2 + bz + c: 
0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000 
0 = −9.76 x 10−5 z2 + 1.21z − 1033 
• Substitute the values of a, b and c into the quadratic formula and solve for the two roots: 
z = 
−1.21 ± 1.212 − (4 x − 9.76 x 10−5 x − 1033) 
2 x − 9.76 x 10−5 = 
−1.21 ± 1.03 
−1.952 x 10−4
4b • Substitute the temperature into the equation: 
2000 = −9.76 x 10−5 z2 + 1.21z + 967 
• Rearrange the equation into the form 0 = az2 + bz + c: 
0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000 
0 = −9.76 x 10−5 z2 + 1.21z − 1033 
• Substitute the values of a, b and c into the quadratic formula and solve for the two roots: 
z = 
−1.21 ± 1.212 − (4 x − 9.76 x 10−5 x − 1033) 
2 x − 9.76 x 10−5 = 
−1.21 ± 1.03 
−1.952 x 10−4 
z1 = 
−1.21 + 1.03 
−1.952 x 10−4 = ퟗퟐퟐ. ퟏퟑ퐦
4b • Substitute the temperature into the equation: 
2000 = −9.76 x 10−5 z2 + 1.21z + 967 
• Rearrange the equation into the form 0 = az2 + bz + c: 
0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000 
0 = −9.76 x 10−5 z2 + 1.21z − 1033 
• Substitute the values of a, b and c into the quadratic formula and solve for the two roots: 
z = 
−1.21 ± 1.212 − (4 x − 9.76 x 10−5 x − 1033) 
2 x − 9.76 x 10−5 = 
−1.21 ± 1.03 
−1.952 x 10−4 
z1 = 
−1.21 + 1.03 
−1.952 x 10−4 = ퟗퟐퟐ. ퟏퟑ퐦 
z2 = 
−1.21 − 1.03 
−1.952 x 10−4 = ퟏퟏ, ퟒퟕퟓ. ퟒퟏ퐦

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Quadratic Equation Depths for 2000°C Temperature

  • 1. 4a In the imaginary case that T=0, what would the depth be?
  • 2. 4a • State a, b and c for the quadratic:
  • 3. 4a • State a, b and c for the quadratic: a = −9.76 x 10−5 b = 1.21 c = 967
  • 4. 4a • State a, b and c for the quadratic: a = −9.76 x 10−5 b = 1.21 c = 967 • Substitute a, b and c into the quadratic formula and simplify:
  • 5. 4a • State a, b and c for the quadratic: a = −9.76 x 10−5 b = 1.21 c = 967 • Substitute a, b and c into the quadratic formula and simplify: z = −b ± b2 − 4ac 2a
  • 6. 4a • State a, b and c for the quadratic: a = −9.76 x 10−5 b = 1.21 c = 967 • Substitute a, b and c into the quadratic formula and simplify: z = −b ± b2 − 4ac 2a z = −1.21 ± 1.212 − (4 x − 9.76 x 10−5 x 967 2 x − 9.76 x 10−5 = −1.21 ± 1.357 −1.952 x 10−4
  • 7. 4a • State a, b and c for the quadratic: a = −9.76 x 10−5 b = 1.21 c = 967 • Substitute a, b and c into the quadratic formula and simplify: z = −b ± b2 − 4ac 2a z = −1.21 ± 1.212 − (4 x − 9.76 x 10−5 x 967 2 x − 9.76 x 10−5 = −1.21 ± 1.357 −1.952 x 10−4 • Solve for the two possible roots:
  • 8. 4a • State a, b and c for the quadratic: a = −9.76 x 10−5 b = 1.21 c = 967 • Substitute a, b and c into the quadratic formula and simplify: z = −b ± b2 − 4ac 2a z = −1.21 ± 1.212 − (4 x − 9.76 x 10−5 x 967 2 x − 9.76 x 10−5 = −1.21 ± 1.357 −1.952 x 10−4 • Solve for the two possible roots: z1 = −1.21 + 1.357 −1.952 x 10−4 = −753.07m
  • 9. 4a • State a, b and c for the quadratic: a = −9.76 x 10−5 b = 1.21 c = 967 • Substitute a, b and c into the quadratic formula and simplify: z = −b ± b2 − 4ac 2a z = −1.21 ± 1.212 − (4 x − 9.76 x 10−5 x 967 2 x − 9.76 x 10−5 = −1.21 ± 1.357 −1.952 x 10−4 • Solve for the two possible roots: z1 = −1.21 + 1.357 −1.952 x 10−4 = −753.07m z2 = −1.21 − 1.357 −1.952 x 10−4 = ퟏퟑ, ퟏퟓퟎ. ퟔퟏ퐦
  • 10. 4a • State a, b and c for the quadratic: a = −9.76 x 10−5 b = 1.21 c = 967 • Substitute a, b and c into the quadratic formula and simplify: z = −b ± b2 − 4ac 2a z = −1.21 ± 1.212 − (4 x − 9.76 x 10−5 x 967 2 x − 9.76 x 10−5 = −1.21 ± 1.357 −1.952 x 10−4 • Solve for the two possible roots: z1 = −1.21 + 1.357 −1.952 x 10−4 = −753.07m z2 = −1.21 − 1.357 −1.952 x 10−4 = ퟏퟑ, ퟏퟓퟎ. ퟔퟏ퐦 • Depth cannot be negative so z1 is not a valid answer.
  • 11. 4b More realistically, at what depths does the equation predict that the temperature will reach 2000°C?
  • 12. 4b • Substitute the temperature into the equation:
  • 13. 4b • Substitute the temperature into the equation: 2000 = −9.76 x 10−5 z2 + 1.21z + 967
  • 14. 4b • Substitute the temperature into the equation: 2000 = −9.76 x 10−5 z2 + 1.21z + 967 • Rearrange the equation into the form 0 = az2 + bz + c:
  • 15. 4b • Substitute the temperature into the equation: 2000 = −9.76 x 10−5 z2 + 1.21z + 967 • Rearrange the equation into the form 0 = az2 + bz + c: 0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000
  • 16. 4b • Substitute the temperature into the equation: 2000 = −9.76 x 10−5 z2 + 1.21z + 967 • Rearrange the equation into the form 0 = az2 + bz + c: 0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000 0 = −9.76 x 10−5 z2 + 1.21z − 1033
  • 17. 4b • Substitute the temperature into the equation: 2000 = −9.76 x 10−5 z2 + 1.21z + 967 • Rearrange the equation into the form 0 = az2 + bz + c: 0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000 0 = −9.76 x 10−5 z2 + 1.21z − 1033 • Substitute the values of a, b and c into the quadratic formula and solve for the two roots:
  • 18. 4b • Substitute the temperature into the equation: 2000 = −9.76 x 10−5 z2 + 1.21z + 967 • Rearrange the equation into the form 0 = az2 + bz + c: 0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000 0 = −9.76 x 10−5 z2 + 1.21z − 1033 • Substitute the values of a, b and c into the quadratic formula and solve for the two roots: z = −1.21 ± 1.212 − (4 x − 9.76 x 10−5 x − 1033) 2 x − 9.76 x 10−5 = −1.21 ± 1.03 −1.952 x 10−4
  • 19. 4b • Substitute the temperature into the equation: 2000 = −9.76 x 10−5 z2 + 1.21z + 967 • Rearrange the equation into the form 0 = az2 + bz + c: 0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000 0 = −9.76 x 10−5 z2 + 1.21z − 1033 • Substitute the values of a, b and c into the quadratic formula and solve for the two roots: z = −1.21 ± 1.212 − (4 x − 9.76 x 10−5 x − 1033) 2 x − 9.76 x 10−5 = −1.21 ± 1.03 −1.952 x 10−4 z1 = −1.21 + 1.03 −1.952 x 10−4 = ퟗퟐퟐ. ퟏퟑ퐦
  • 20. 4b • Substitute the temperature into the equation: 2000 = −9.76 x 10−5 z2 + 1.21z + 967 • Rearrange the equation into the form 0 = az2 + bz + c: 0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000 0 = −9.76 x 10−5 z2 + 1.21z − 1033 • Substitute the values of a, b and c into the quadratic formula and solve for the two roots: z = −1.21 ± 1.212 − (4 x − 9.76 x 10−5 x − 1033) 2 x − 9.76 x 10−5 = −1.21 ± 1.03 −1.952 x 10−4 z1 = −1.21 + 1.03 −1.952 x 10−4 = ퟗퟐퟐ. ퟏퟑ퐦 z2 = −1.21 − 1.03 −1.952 x 10−4 = ퟏퟏ, ퟒퟕퟓ. ퟒퟏ퐦