# For what values of a and b, and x = — 2 are solutions of the equation ax2 + bx — 6 = 0.

Put x = 3/4

ax2 + bx — 6 = 0   9a + 12b – 96 = 0 divide by 3

3a + 4b – 32 = 0

3a + 4b = 32 (1)

Put x = –2

ax2 + bx — 6 = 0

a(–2)2 + b(–2) – 6 = 0

4a – 2b – 6 = 0

4a – 2b = 6 (2)

Eliminate (1) and (2)

3a + 4b = 32

4a – 2b = 6 ×2 a = 4

Put a = 4 in equation (1).

3a + 4b = 32

3(4) + 4b = 32

12 + 4b = 32

4b = 32 – 12

4b = 20

b = 5

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