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If πΏ and π are the roots of the equation two π₯ squared minus three π₯ plus one equals zero, find, in its simplest form, the quadratic equation whose roots are two πΏ squared and two π squared.
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We might recall a relationship between a quadratic equation whose coefficient of π₯ squared is one and the roots of that equation.
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The negative coefficient of π₯ tells us the sum of the roots of that equation, and then the constant term tells us the product.
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Now, of course, weβre told that πΏ and π are the roots of our equation, but itβs not currently in this form.
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So weβre going to divide each term in this equation by two to achieve this.
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Two π₯ divided by two is π₯ squared.
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We then divide the coefficient of π₯ by two and we get negative three over two π₯.
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Finally, we divide the constant term by two.
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And of course, we know that zero divided by two is zero.
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So we now have the equation in the correct form.
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And so weβre able to work out the value of the sum of our roots, thatβs πΏ plus π, and their product, thatβs πΏ times π.
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πΏ plus π, the sum of the roots, is the negative coefficient of π₯.
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Now, since the coefficient of π₯ here is negative three over two, we can say that the sum of our roots must simply be three over two.
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Then the product of the roots is equal to the constant term.
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Well, thatβs quite simply one-half.
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So we have that πΏ plus π is equal to three over two and then πΏπ is equal to one-half.
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Letβs now think about the roots of our new equation.
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They are still in terms of πΏ and π.
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Theyβre two πΏ squared and two π squared.
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This time, if we find an expression for the sum of the roots and the product of the roots, we get two πΏ squared plus two π squared and two πΏ squared times two π squared.
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So we need to use our earlier expressions for πΏ plus π and πΏπ to find values for these expressions.
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Letβs look at the equation we formed for the sum of our roots.
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And weβre going to square it.
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On the left-hand side, we get πΏ plus π squared, and then we can just simply square the numerator and the denominator of three over two, and we get nine over four.
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But of course, πΏ plus π squared is πΏ plus π multiplied by πΏ plus π.
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So letβs distribute these parentheses.
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When we do, we get πΏ squared plus two πΏπ plus π squared equals nine over four.
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And we now notice that we have πΏ squared plus π squared.
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So weβre on our way to find an expression for the sum of our new roots.
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What weβll do next is weβll replace πΏπ with the value of one-half.
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And so we get πΏ squared plus two times a half plus π squared equals nine over four.
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Now, of course, two times a half is one.
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So we get πΏ squared plus one plus π squared equals nine over four.
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And then if we subtract one from both sides, we find that expression for πΏ squared plus π squared.
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Now, in fact, we can write one as four over four.
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And the right-hand side becomes nine over four minus four over four, and thatβs simply five over four.
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Now we notice that we have an expression for πΏ squared plus π squared.
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But we want an expression for two πΏ squared plus two π squared.
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And so to achieve this, weβll multiply the left-hand side by two, which means we also have to do the same to the right-hand side.
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The distributive property of multiplication tells us that this is equal to two πΏ squared plus two π squared.
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Then five over four times two is five over two.
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And we can therefore say that the sum of the roots of our new quadratic equation must be five over two.
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But what about the product of these roots?
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Letβs begin by multiplying two πΏ squared by two π squared.
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And we get four πΏ squared π squared.
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Weβre going to clear some space and go back to one of our earlier equations.
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Letβs look at the equation πΏπ equals one-half.
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It should be quite clear to us that if we square this, we can get the πΏ squared π squared part.
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πΏπ squared is πΏ squared π squared.
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And then one-half squared is a quarter.
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Of course, we want four πΏπ squared.
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So weβre going to multiply both sides of this equation by four.
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A quarter times four is simply equal to one.
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And so we can say that four πΏ squared π squared, which is the product of our roots, is simply one.
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We replace these values in the general form.
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We know that the sum of our roots is five over two and the product is one.
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And so we see that our new quadratic equation is π₯ squared minus five over two π₯ plus one equals zero.
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Weβll perform one more step, and weβll make each term integer by multiplying through by two.
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And when we do, we see that the quadratic equation whose roots are two πΏ squared and two π squared is two π₯ squared minus five π₯ plus two equals zero.