Solve x^2 + x - 12 = 0

Mathematics

1 answer:

bogdanovich [222]1 year ago

6 0

Answer:

x₁ = -4

x₂ = 3

Step-by-step explanation:

x²+ x + 12 = 0

x = {-1±√((1²)-(4*1*-12))} / (2*1)

x = {-1±√(1+48)} / 2

x = {-1±√49} / 2

x = {-1±7} / 2

x₁ = {-1-7} / 2 = -8/2 = -4

x₂ = {-1+7} / 2 = 6/2 = 3

Check:

x₁

-4² + (-4) - 12 = 0

16 - 4 - 12 = 0

x₂

3² + 3 - 12 = 0

9 + 3 - 12 = 0

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Firstly, solve the outer exponent:

$(\frac{4mn}{m^{-2}n^{6}})^{-2}=\frac{4^{-2}m^{-2}n^{-2}}{m^{-2*-2}n^{6*-2}}=\frac{4^{-2}m^{-2}n^{-2}}{m^4n^{-12}}$

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Next, convert the negative exponents:

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