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For what values of  is \(2x^2+8x\le-6\)? Express your answer in interval notation.

 May 18, 2022
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 Add 6 to both sides:                          \(2x^2+8x+6\le 0\)

Divide by 2:                                         \(x^2+4x+3\le0\)

Factorise:                                            \((x+3)(x+1)\le 0\)

Critical values:                                     \(x_1=-3,\text{and, }x_2=-1\)

Sketch the Quadratic, it is open-upwards and intersects the x-axis at x=-3 and x=-1.

Thus, the desired region is below the x-axis that is, between -3 and -1

Therefore,                                \(-3\le x\le-1\) is the desired solution.

In interval notation:   \(x \in[-3,-1]\)

 May 19, 2022

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