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In this video, we will learn how to solve quadratic equations whose roots are complex numbers.
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We will begin by learning how to solve simple equations which have complex solutions and then look at what the introduction of the complex number set does to our understanding of the quadratic formula and the discriminant.
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Finally, weβll learn how to reconstruct a quadratic equation given a complex root.
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During our exploration of the concept of numbers, we will have come across equations that have no solutions or at least those that we presume to have no real solutions.
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However, the introduction of the set of complex numbers opens up a whole new world when it comes to these kinds of equations.
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Letβs consider the equation two π₯ squared equals negative eight.
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To solve this equation, we begin by dividing through by two.
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That tells us that π₯ squared is equal to negative four.
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We will then find the square root of both sides of the equation.
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In the past, we might have said that this equation, whose solution is found by finding the square root of a negative number, doesnβt actually have any real solutions.
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And of course, this statement would be entirely correct.
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However, weβre no longer just dealing with real numbers.
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We now have the imaginary number set.
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And we recall that these revolve around the single letter π, where π is defined as the solution to the equation π₯ squared equals negative one.
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But we often just say that π is equal to the square root of negative one.
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And this means we can now solve our equation by finding the square root of negative four.
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To find the square root of a negative number, we split it up.
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So, for example, the square root of negative π is the same as the square root of π multiplied by negative one, which in turn is equal to the square root of π multiplied by the square root of negative one.
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And since the square root of negative one is π, we see that the square root of negative π is the same as π root π.
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In this example, π₯ is equal to plus or minus π root four.
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And of course, the square root of four is two.
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And we see that the equation two π₯ squared equals negative eight has two solutions: two π and negative two π.
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Letβs consider another equation that we would have previously been unable to solve.
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Solve the equation five π₯ squared plus one equals negative 319.
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We can begin by solving this equation just as we would any other, by performing a series of inverse operations.
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Weβll start by subtracting one from both sides of the equation.
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Five π₯ squared plus one minus one is simply five π₯ squared.
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And negative 319 minus one is negative 320.
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Next, we divide through by five.
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And we see that π₯ squared is equal to negative 64.
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Our final step is to find the square root of both sides of the equation.
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The square root of π₯ squared is π₯.
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And remember, we can take both the positive and negative roots of negative 64.
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And we see that π₯ is equal to plus or minus the square root of negative 64.
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At this point, we choose to rewrite negative 64 as 64 multiplied by negative one.
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And we then see that the square root of negative 64 is the same as the square root of 64 multiplied by the square root of negative one.
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The square root of negative one though is π, and the square root of 64 is eight.
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And we finished solving our equation.
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π₯ has two solutions.
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Itβs eight π and negative eight π.
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Now in fact we can apply the usual methods for solving equations to help us solve any equation with nonreal roots.
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In the case of a quadratic equation, we might struggle to factorise a quadratic expression.
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But we can apply the other two methods weβre confident in.
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These are the quadratic formula and completing the square.
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And there are advantages and disadvantages to both.
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The quadratic formula can be a little nicer to work with when the coefficient of π₯ squared is not equal to one.
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And weβll see in a moment that we can use part of the formula to help us find the nature and number of roots of the equation.
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The completing the square method though can be fairly efficient when the coefficient of π₯ squared is equal to one.
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Now itβs very much personal preference.
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Weβll use the quadratic formula mainly throughout this video.
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Then we will look at both methods during our next example.
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Letβs remind ourselves of the quadratic formula and the discriminant.
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For a quadratic equation given as ππ₯ squared plus ππ₯ plus π equals zero, where π is not equal to zero, the roots are given by π₯ equals negative π plus or minus the square root of π squared minus four ππ all over two π.
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And we use the discriminant to find the nature of the roots of the equation.
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Itβs the part of the formula that sits inside the square root: π squared minus four ππ.
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It makes sense that if the discriminant is greater than zero, the square root of the discriminant will be a real number.
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And this means that there will be two real roots for our equation.
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If the value of the discriminant is equal to zero, there will be exactly one solution.
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This is known as a repeated root.
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It occurs when the turning point of the curve touches the π₯-axis exactly once.
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And what about if the value is less than zero?
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Well, weβve seen that the square root of a negative number is not a real number.
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So there are no real roots.
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This means that the curve doesnβt actually intersect the π₯-axis.
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Letβs look at an example of a quadratic equation which has nonreal roots.
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Solve the quadratic equation π₯ squared minus four π₯ plus eight equals zero.
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Before we solve this equation, we can, if we want, double-check the nature of the roots of the equation by finding the value of the discriminant.
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Remember, the discriminant of an equation of the form ππ₯ squared plus ππ₯ plus π equals zero is given by the formula π squared minus four ππ.
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And itβs sometimes denoted by this little triangle.
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In our equation, π is the coefficient of π₯ squared.
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Itβs one.
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π is the coefficient of π₯.
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Itβs negative four.
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And π is the constant.
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Itβs eight.
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The discriminant of our equation is therefore negative four squared minus four multiplied by one multiplied by eight, which is negative 16.
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We know that if the discriminant is greater than zero, the equation has two real roots.
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If itβs equal to zero, it has exactly one real root.
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And if itβs less than zero, it has no real roots.
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Our discriminant is less than zero.
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So the equation π₯ squared minus four π₯ plus eight equals zero has no real roots.
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Knowing that weβre going to get two complex roots, letβs solve this equation by first looking at the quadratic formula.
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The solutions to the equation are found by negative π plus or minus the square root of π squared minus four ππ all over two π.
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We already saw that π squared minus four ππ in our example is equal to negative 16.
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So the solutions to our quadratic equation are given by negative negative four plus or minus the square root of negative 16 all over two multiplied by one.
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That simplifies to four plus or minus the square root of negative 16 all over two.
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And at this stage, weβre going to rewrite the square root of negative 16 as the square root of 16 multiplied by the square root of negative one.
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And this is useful because we know that the square root of 16 is four and we know that the square root of negative one is π.
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So π₯ is equal to four plus or minus four π all over two.
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And we can simplify.
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And we see that the solutions to the quadratic equation are π₯ equals two plus two π and π₯ equals two minus two π.
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Now in fact this isnβt the only method for solving this equation.
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We could have completed the square.
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And this is very much personal preference in an example like this.
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Letβs see what that wouldβve looked like.
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The first thing we do is halve the coefficient of π₯.
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Half of negative four is negative two.
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So we write π₯ minus two all squared.
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Now negative two squared is four.
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So we subtract this four and then add on that value of eight.
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And of course, all this is equal to zero.
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We can simplify our equation somewhat, and we get π₯ minus two all squared plus four equals zero.
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Weβre going to solve this by subtracting four from both sides.
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And that gives us π₯ minus two all squared equals negative four.
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Weβre then going to find the square root of both sides of this equation.
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The square root of π₯ minus two all squared is π₯ minus two.
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And remember, we can take both the positive and negative roots of negative four.
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So we see that π₯ minus two is equal to plus or minus the square root of negative four.
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Now using the same method as earlier, we can see that the square root of negative four is actually the same as two π.
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And we can complete this solution by adding two to both sides of the equation.
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And once again, we see the solutions to our equation to be two plus two π and two minus two π.
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In fact, itβs no accident that the roots of the equation are complex conjugates of one another.
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It really makes a lot of sense, especially given our second method of solving, that this would be true for any quadratic equation with complex roots.
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We can say that the nonreal roots of a quadratic equation with real coefficients occur in complex conjugate pairs.
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And there is actually a lovely little proof of this.
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Letβs say we have a quadratic equation of the form ππ₯ squared plus ππ₯ plus π.
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Weβre going to let πΌ be a solution to this equation.
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And we say that πΌ star is the complex conjugate to πΌ.
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Weβre going to substitute this complex conjugate into our equation.
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And when we do, we get π multiplied by πΌ star all squared plus π multiplied by πΌ star plus π.
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And here we recall the fact that, for any two complex numbers, the conjugate of their product is equal to the product of their conjugates.
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This means that the square of the conjugate of our solution is equal to the conjugate of the square.
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And we see the first part becomes π multiplied by πΌ squared star.
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Since π, π, and π are real numbers β remember, our quadratic equation has real coefficients and we also know that the conjugate of a real number is just that number β this can be further rewritten as shown.
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And finally, we recall the fact that, for two complex numbers π§ one and π§ two, the conjugate of their sum is equal to the sum of their conjugates.
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And we see that π of πΌ star is equal to π multiplied by πΌ squared plus π multiplied by πΌ plus π star.
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Now we already said that πΌ is a solution to the equation.
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This means that ππΌ star plus ππΌ plus π must be equal to zero.
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And we, of course, know that the conjugate of zero is simply zero.
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And weβve seen that since π of πΌ star is equal to zero, πΌ star must also be a solution to this equation.
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And in fact, this is called the complex conjugate root theorem, and it can be extended into solving polynomials.
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Letβs have a look at a number of examples of where this theorem can be used to solve problems involving quadratics.
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The complex numbers π plus ππ and π plus ππ, where π, π, π, and π are real numbers, are the roots of a quadratic equation with real coefficients.
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Given that π is not equal to zero, what conditions, if any, must π, π, π, and π satisfy?
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In this question, we are told that π plus ππ and π plus ππ are roots to our quadratic equation with real coefficients.
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This equation would usually be of the form ππ₯ squared plus ππ₯ plus π equals zero, though π, π, and π are not to be confused with the letters π, π, and π in our complex numbers.
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So weβll rewrite this as ππ₯ squared plus ππ₯ plus π equals zero.
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Now we know that the nonreal roots of a quadratic equation with real coefficients occur in complex conjugate pairs.
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And remember, to find the conjugate, we change the sign of the imaginary part.
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So the conjugate of π plus ππ is π minus ππ.
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And π plus ππ and π plus ππ must be complex conjugates of one another by this theorem.
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This means that the conjugate of π plus ππ must be equal to π plus ππ.
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So we say that π minus ππ equals π plus ππ.
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And for two complex numbers to be equal, their real parts must be equal.
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So here we equate π and π.
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But their imaginary parts must also be equal.
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So we equate the imaginary parts.
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And we see that negative π equals π.
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So the conditions that π, π, π, and π must satisfy here is that π must be equal to π and negative π must be equal to π.
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This time, weβre going to use our knowledge of the nature of complex roots of quadratic equations to reconstruct an equation given one of its roots.
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Find the quadratic equation with real coefficients which has five plus π as one of its roots.
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Weβre told that five plus π is a root of the quadratic equation.
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And remember, we know that the nonreal roots of a quadratic equation which has real coefficients occur in complex conjugate pairs.
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To find the complex conjugate, we change the sign of the imaginary part.
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And we can therefore see that the roots of our equation are five plus π and five minus π.
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And this means that our quadratic equation is of the form π₯ minus five plus π multiplied by π₯ minus five minus π is equal to zero.
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And this comes from the fact that when we solve a quadratic equation by factoring, we equate each expression inside the parentheses to zero.
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So in this case, we would have π₯ minus five plus π equals zero and π₯ minus five minus π equals zero.
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We would solve this first equation by adding five plus π to both sides.
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And we see that π₯ equals five plus π.
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And we solve the second equation by adding five minus π to both sides.
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And we get that second root π₯ equals five minus π.
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Weβre going to need to distribute these parentheses.
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Letβs use the grid method here since thereβs a number of bits and pieces that could trip us up.
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π₯ multiplied by π₯ is π₯ squared.
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π₯ multiplied by negative five plus π is negative π₯ five plus π.
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Similarly, we get negative π₯ multiplied by five minus π.
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And negative five plus π multiplied by negative five minus π gives us a positive five minus π multiplied by five plus π.
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And weβll distribute these brackets using the FOIL method.
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Multiplying the first term in the first bracket by the first term in the second bracket gives us 25.
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We multiply the outer two terms β thatβs five π β and the inner two terms β thatβs negative five π.
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And five π minus five π is zero.
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So these cancel each other out.
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And then we multiply the last terms.
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Negative π multiplied by π is negative π squared.
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And since π squared is equal to negative one, we see that these brackets distribute to be 25 minus negative one, which is equal to 26.
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So our quadratic equation is currently of the form π₯ squared minus π₯ multiplied by five plus π minus π₯ multiplied by five minus π plus 26.
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We collect like terms.
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And we get π₯ squared minus five plus π plus five minus π multiplied by π₯ plus 26.
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π minus π is zero.
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And weβre left with π₯ squared minus 10π₯ plus 26 equals zero.
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Now actually there is a formula that we can use that will save us some time.
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If we have a quadratic equation with real roots and a compact solution π plus ππ, the equation of that quadratic is π₯ squared minus two ππ₯ plus π squared plus π squared equals zero.
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π is the real part of the solution.
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Here thatβs five.
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And π is the imaginary part.
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In our solution, thatβs one.
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We can substitute what we know about our complex number into the formula.
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And we get π₯ squared minus two times five times π₯ plus five squared plus one squared.
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Two times five is 10, and five squared plus one squared is 26.
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And we see once again we have the same quadratic equation.
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And we should be able to see now why this method can be a bit of a time saver.
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In this video, weβve learned that we can use the quadratic formula or completing the square to solve equations with no real roots by giving our answers as complex numbers.
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Weβve also seen that these solutions occur in complex conjugate pairs.
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And weβve also learned that we can reconstruct a quadratic equation given one of its complex solutions.
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If the solution is π plus ππ, the quadratic equation is π₯ squared minus two ππ₯ plus π squared plus π squared equals zero.