WEBVTT
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Find the solution set of the equation negative three 𝑥 squared minus 𝑥 plus 12 is equal to zero in the set of real numbers, giving values to one decimal place.
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We recall that any quadratic equation in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, where 𝑎, 𝑏, and 𝑐, are constants and 𝑎 is nonzero, can be solved using the quadratic formula.
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This states that 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all divided by two 𝑎.
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In this question, the values of 𝑎, 𝑏, and 𝑐 are negative three, negative one, and 12, respectively.
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Substituting these three values, we have 𝑥 is equal to negative negative one plus or minus the square root of negative one squared minus four multiplied by negative three multiplied by 12 all divided by two multiplied by negative three.
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Squaring negative one gives us positive one.
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Multiplying four, negative three, and 12 gives us negative 144.
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On the denominator, two multiplied by negative three is negative six.
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The equation simplifies to 𝑥 is equal to one plus or minus the square root of one minus negative 144 all divided by negative six.
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Subtracting negative 144 from one gives us 145.
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𝑥 is equal to one plus or minus the square root of 145 all divided by negative six.
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This gives us two possible solutions.
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Either 𝑥 is equal to one plus the square root of 145 divided by negative six.
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Or 𝑥 is equal to one minus the square root of 145 all divided by negative six.
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Typing the first calculation into our calculator gives us negative 2.173 and so on.
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We are asked to round this to one decimal place.
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This is equal to negative 2.2.
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One minus the square root of 145 divided by negative six is equal to 1.840 and so on.
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To one decimal place, this is equal to 1.8.
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We can therefore conclude that the solution set of the equation negative three 𝑥 squared minus 𝑥 plus 12 equals zero is negative 2.2 and 1.8.