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Cerrena [4.2K]
1 year ago
12

What are the solutions of x² + 6x - 16 = 0?

Mathematics
1 answer:
AveGali [126]1 year ago
6 0

Step-by-step explanation:

x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\cdot \:1\cdot \left(-16\right)}}{2\cdot \:1}\\\sqrt{6^2-4\cdot \:1\cdot \left(-16\right)}\\\sqrt{6^2+64}\\\sqrt{36+64}\\\sqrt{100}\\\sqrt{10^2}\\=10\\x_1=\frac{-6+10}{2\cdot \:1},\:x_2=\frac{-6-10}{2\cdot \:1}\\\\\bold{x_1:2}\\\frac{-6+10}{2\cdot \:1}\\\frac{4}{2\cdot \:1}\\\frac{4}{2}\\=2\\\\\bold{x_2:-8}\\\frac{-6-10}{2\cdot \:1}\\\frac{-16}{2\cdot \:1}\\\frac{-16}{2}\\-\frac{16}{2}\\=-8

Answer:

x₁ = 2

x₂ = -8

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The sum of two consecutive odd integers is 236. What is smaller integers
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I'll say the first integer is x.  The next consecutive odd number would be x+2.  If the sum of the odd integers is 236, the equation would be
x + (x + 2) = 236
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divide both sides by 2
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