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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:

a = −9.76 x 10−5 b = 1.21 c = 967

a = −9.76 x 10−5 b = 1.21 c = 967
• Substitute a, b and c into the quadratic formula and simplify:

a = −9.76 x 10−5 b = 1.21 c = 967
z =
−b ± b2 − 4ac
2a

a = −9.76 x 10−5 b = 1.21 c = 967
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

a = −9.76 x 10−5 b = 1.21 c = 967
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:

a = −9.76 x 10−5 b = 1.21 c = 967
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
z1 =
−1.21 + 1.357
−1.952 x 10−4 = −753.07m

a = −9.76 x 10−5 b = 1.21 c = 967
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
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

a = −9.76 x 10−5 b = 1.21 c = 967
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
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:

2000 = −9.76 x 10−5 z2 + 1.21z + 967

2000 = −9.76 x 10−5 z2 + 1.21z + 967
• Rearrange the equation into the form 0 = az2 + bz + c:

2000 = −9.76 x 10−5 z2 + 1.21z + 967
0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000

2000 = −9.76 x 10−5 z2 + 1.21z + 967
0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000
0 = −9.76 x 10−5 z2 + 1.21z − 1033

2000 = −9.76 x 10−5 z2 + 1.21z + 967
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:

2000 = −9.76 x 10−5 z2 + 1.21z + 967
0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000
0 = −9.76 x 10−5 z2 + 1.21z − 1033
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

2000 = −9.76 x 10−5 z2 + 1.21z + 967
0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000
0 = −9.76 x 10−5 z2 + 1.21z − 1033
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

2000 = −9.76 x 10−5 z2 + 1.21z + 967
0 = −9.76 x 10−5 z2 + 1.21z + 967 − 2000
0 = −9.76 x 10−5 z2 + 1.21z − 1033
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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