WEBVTT
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Find the solution set of the equation three over 𝑥 plus three minus four over 𝑥 minus three is equal to three in the set of real values, giving values to one decimal place.
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It might not be immediately obvious how we can solve the equation in this question.
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One strategy for an equation of this type is to try and eliminate the denominators by firstly finding the lowest common denominator.
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When adding one-third and one-quarter, the easiest way to find a common denominator is to multiply the two denominators together.
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Three multiplied by four is equal to 12.
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Therefore, our common denominator is equal to 12.
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Whilst our equation is more complicated, we can use the same method.
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We have denominators of 𝑥 plus three and 𝑥 minus three.
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Multiplying these together, we have 𝑥 plus three multiplied by 𝑥 minus three.
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By multiplying each of the three terms in our equation by this, we will be able to eliminate the denominators.
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Multiplying the first term by 𝑥 plus three and 𝑥 minus three gives us three 𝑥 plus three 𝑥 minus three divided by 𝑥 plus three.
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This simplifies to three multiplied by 𝑥 minus three.
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The second term becomes negative four multiplied by 𝑥 plus three multiplied by 𝑥 minus three divided by 𝑥 minus three.
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This time we can cancel a factor of 𝑥 minus three from the numerator and denominator, leaving us with negative four multiplied by 𝑥 plus three.
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On the right-hand side, we have three multiplied by 𝑥 plus three multiplied by 𝑥 minus three.
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At this stage, we might notice that 𝑥 plus three multiplied by 𝑥 minus three is the factorization of the difference of two squares.
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We know that 𝑥 squared minus 𝑎 squared is equal to 𝑥 plus 𝑎 multiplied by 𝑥 minus 𝑎.
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This means that 𝑥 plus three multiplied by 𝑥 minus three is equal to 𝑥 squared minus nine.
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The right-hand side of our equation becomes three multiplied by 𝑥 squared minus nine.
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Our next step is to distribute the parentheses or expand the brackets.
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This gives us three 𝑥 minus nine.
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On the left-hand side, we also have negative four 𝑥 minus 12.
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And on the right-hand side, we have three 𝑥 squared minus 27.
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The left-hand side simplifies to negative 𝑥 minus 21.
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We can then add 𝑥 and 21 to both sides of the equation such that three 𝑥 squared plus 𝑥 minus six equal zero.
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This is a quadratic equation written in the form 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero.
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We know that one way to solve an equation of this type, where 𝑎, 𝑏, and 𝑐 are constants and 𝑎 is nonzero, is 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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If the value of 𝑏 squared minus four 𝑎𝑐, known as the discriminant, is greater than or equal to naught, the equation will have real solutions.
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In our equation, the values of 𝑎, 𝑏, and 𝑐 are three, one, and negative six, respectively.
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This means that 𝑥 is equal to negative one plus or minus the square root of one squared minus four multiplied by three multiplied by negative six all divided by two multiplied by three.
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This simplifies to negative one plus or minus the square root of 73 all divided by six.
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We have two possible solutions either 𝑥 is equal to negative one plus the square root of 73 divided by six or 𝑥 is equal to negative one minus the square root of 73 divided by six.
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Typing these into the calculator, we have 𝑥 is equal to 1.257 and so on or 𝑥 is equal to negative 1.590 and so on.
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We have been asked to give our solutions to one decimal place.
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The solution set of the equation three over 𝑥 plus three minus four over 𝑥 minus three equals three contains the values 1.3 and negative 1.6 correct to one decimal place.