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Given that πΏ and π are the roots of the equation two π₯ squared minus 10π₯ plus one equals zero, find, in its simplest form, the quadratic equation whose roots are πΏ over three and π over three.
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We begin by recalling some key facts about a quadratic equation written in the form ππ₯ squared plus ππ₯ plus π equals zero, where π, π, and π are constants and π is nonzero.
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If the two roots of the quadratic are π sub one and π sub two, their sum is equal to negative π over π.
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The product of the two roots, π one multiplied by π two, is equal to π over π.
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In this question, we are given the equation two π₯ squared minus 10π₯ plus one equals zero.
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The values of π, π, and π here are two, negative 10, and one.
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We are also told that πΏ and π are the two roots of the equation.
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Therefore, the sum of these, πΏ plus π, is equal to negative negative 10 over two.
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This is equal to five.
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The product of the roots πΏ multiplied by π is equal to one over two or one-half.
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We can now use these values to help find the quadratic equation whose roots are πΏ over three and π over three.
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Letβs begin by considering the sum of these roots.
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We have πΏ over three plus π over three.
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As the denominators are the same, we can simply add the numerators, giving us πΏ plus π over three.
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We have already worked out that πΏ plus π is equal to five.
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This means that the sum of the roots is equal to five over three or five-thirds.
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Negative π over π is equal to five-thirds.
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We will now consider the product of our two roots.
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We need to multiply πΏ over three and π over three.
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We do this by multiplying the numerators and denominators separately, giving us πΏπ over nine.
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As πΏπ is equal to one-half, we have a half over nine or a half divided by nine.
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This is equal to one eighteenth.
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The product of our roots π over π is equal to one eighteenth.
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We can now solve these two equations to find the values of π, π, and π.
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We note that the denominators on the right-hand side are different.
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However, on the left-hand side of both our equations, the denominator is π.
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We can therefore multiply the numerator and denominator of the right-hand side of our first equation by six.
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Five over three or five-thirds is the same as 30 over 18.
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Our next step is to let π equal 18.
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From the first equation, this means that negative π is equal to 30, which means that π is equal to negative 30.
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From the second equation, if π over π is equal to one over 18 and π equals 18, then π must be equal to one.
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The quadratic equation whose roots are πΏ over three and π over three is 18π₯ squared minus 30π₯ plus one equals zero.