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Find the solution set of four to the power of π₯ squared plus eight minus 1,026 times two to the power of π₯ squared plus eight plus 2,048 equals zero in the set of real numbers.
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Believe it or not, this is actually a special type of quadratic equation.
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We are going to need to perform some manipulation to make it look like that though.
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And the key here is spotting that four is the same as two squared.
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And so, this means we can write four to the power of π₯ squared plus eight as two squared to the power of π₯ squared plus eight.
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But since we multiply the exponents in this case, we can reverse this and say that this is the same as two to the power of π₯ squared plus eight all squared.
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And so, we can rewrite our equation somewhat.
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We get two to the power of π₯ squared plus eight all squared minus 1,026 times two to the power of π₯ squared plus eight plus 2,048 equals zero.
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And then we perform a substitution.
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Weβre going to let π¦ be equal to two to the power of π₯ squared plus eight.
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And so our equation simplifies quite a lot.
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We get π¦ squared minus 1,026π¦ plus 2,048 equals zero.
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And then we have a number of ways to solve this.
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Letβs solve by factoring the expression on the left-hand side.
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We know that the first term in each binomial must be π¦, since π¦ times π¦ is π¦ squared.
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And then weβre looking for two numbers whose product is 2,048 and whose sum is the coefficient of π¦.
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So thatβs negative 1,026.
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Well, these numbers are negative 1,024 and negative two.
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And then, since the product of our two binomials is zero, we know that this means at least one of the binomials must themselves be zero.
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So, either π¦ minus 1,024 is zero or π¦ minus two is zero.
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Letβs solve this first equation for π¦ by adding 1,024 to both sides.
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And we get π¦ equals 1,024.
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Similarly, weβll add two to both sides of our second equation.
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So, π¦ is equal to two.
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But remember, we were looking to find the values of π₯ which satisfy our original equation.
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So, weβre going to go back to our substitution where we said π¦ is equal to two to the power of π₯ squared plus eight.
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By replacing π¦ with two to the power of π₯ squared plus eight, we form two new equations in π₯.
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They are two to the power of π₯ squared plus eight equals 1,024 and two to the power of π₯ squared plus eight equals two.
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Well, we actually know that two to the power of 10 is 1,024 and two to the power of one is two.
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So, this means that in our first equation, π₯ squared plus eight must be equal to 10.
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And in our second, π₯ squared plus eight must be equal to two.
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Weβll solve these equations for π₯ by subtracting eight from both sides.
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So, we get π₯ squared equals two in our first equation, and π₯ squared equals negative six in our second.
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Our second step is to take the positive and negative square root of our equations.
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In our first equation, we get π₯ is equal to the positive and negative square root of two.
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And in our second, we get π₯ is equal to the positive and negative square root of negative six.
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But we were looking for the solution set in the set of real numbers, and the square root of negative six does not yield a real result.
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So, there are no solutions that weβre interested in for the equation π₯ squared equals negative six.
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And so, there are two values in our solution set.
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They are the square root of two and the negative square root of two.