WEBVTT
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Find the solution set of the equation nine 𝑥 squared minus four 𝑥 is equal to 20, giving values to three decimal places.
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In order to solve any quadratic equation of this type, we need to rearrange it so it is in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is equal to zero, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎 is not equal to zero.
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We will do this by subtracting 20 from both sides.
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This gives us nine 𝑥 squared minus four 𝑥 minus 20 is equal to zero.
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Our values of 𝑎, 𝑏, and 𝑐 are nine, negative four, and negative 20, respectively.
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We can solve any quadratic equation of this type 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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The positive and negative signs give us two solutions.
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Substituting in our values of 𝑎, 𝑏, and 𝑐, we have 𝑥 is equal to negative negative four plus or minus the square root of negative four squared minus four multiplied by nine multiplied by negative 20 all divided by two multiplied by nine.
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Using our scientific calculator, the right-hand side simplifies to two plus or minus two root 46 all divided by nine.
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This gives us two possible solutions: either 𝑥 is equal to two plus two root 46 over nine or 𝑥 is equal to two minus two root 46 over nine.
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These are equal to 1.7294 and so on and negative 1.2849 and so on.
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We are asked to give our answers to three decimal places.
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Therefore, 𝑥 is equal to 1.729 or negative 1.285.
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These can be written using set notation such that the solution set of the equation nine 𝑥 squared minus four 𝑥 equals 20 is 1.729 and negative 1.285.