WEBVTT
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Part a) Write four ๐ฅ squared minus eight ๐ฅ plus five in the form ๐ multiplied by ๐ฅ minus ๐ all squared plus ๐.
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Part b) Use your answer to part a to solve four ๐ฅ squared minus eight ๐ฅ plus five equals 11.
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In part a, weโve been asked to write this quadratic expression in a particular form.
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And the form that weโve been asked for is completed square form.
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We need to determine the values of ๐, ๐, and ๐.
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Thereโs a procedure that we can follow for doing this.
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And first, we notice that the coefficient of ๐ฅ squared is not one.
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Itโs four.
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So our first step is going to be to factorise this four out, but only from the first two terms.
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So we bring this factor four out.
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And inside the bracket, we then have ๐ฅ squared minus two ๐ฅ, because four multiplied by ๐ฅ squared gives four ๐ฅ squared and four multiplied by negative two ๐ฅ gives negative eight ๐ฅ.
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We then have the plus five as before.
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Our next step is going to be to complete the square inside this bracket, so one ๐ฅ squared minus two ๐ฅ.
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Again, this is standard process that we can follow.
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We always have ๐ฅ plus or ๐ฅ minus some number all squared.
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And to find this number, we can halve the coefficient of ๐ฅ.
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So in the general form, if the coefficient was negative ๐, then in the bracket, weโd have negative ๐ over two.
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In our quadratic, the coefficient of ๐ฅ is negative two.
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So inside the bracket, we have negative one.
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We then have to subtract this value squared.
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So in our case, weโre subtracting negative one squared.
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Negative one squared is positive one, so weโre subtracting one.
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Letโs just confirm by expansion that ๐ฅ minus one all squared minus one is indeed equal to ๐ฅ squared minus two ๐ฅ.
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๐ฅ multiplied by ๐ฅ is ๐ฅ squared.
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๐ฅ multiplied by negative one is negative ๐ฅ.
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Negative one multiplied by ๐ฅ is also negative ๐ฅ.
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Negative one multiplied by negative one is positive one.
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And then we still have the negative one on the outside of the brackets.
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The positive one and negative one directly cancel each other out, which is what we wanted to happen.
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This is why we subtracted that value of negative one squared.
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The negative ๐ฅ and then another lot of negative ๐ฅ makes negative two ๐ฅ.
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And again, this is why we halve the coefficient of ๐ฅ, because we knew that when we expanded the brackets, we were gonna multiply this by ๐ฅ twice, giving two lots of it.
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So by halving and then multiplying by two, weโre back to the original coefficient of ๐ฅ.
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We can see that our expansion does indeed give ๐ฅ squared minus two ๐ฅ.
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So weโve completed the square within the bracket correctly.
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Now remember, all of this bracket was multiplied by four.
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So Iโm going to put a big bracket around it and then the four on the outside.
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We then bring down the plus five as before.
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Weโre nearly there, but the final step is just to expand that square bracket, the large bracket, so that we have this exactly in the form we were asked for.
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We have to multiply both ๐ฅ minus one squared by four and negative one by four.
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And then we have the plus five as before.
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Itโs really important that you remember to multiply the constant term in that large bracket by four.
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A common mistake would just be to multiply four by the factor of ๐ฅ minus one squared.
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This is why I drew in that large bracket so weโd remember that everything inside it needed to be multiplied by four.
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Lastly, we just need to simplify.
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We have negative four plus five, which is equal to positive one.
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Weโve now written this quadratic in the requested form.
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The value of ๐ is four, the value of ๐ is one, and the value of ๐ is also one.
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In completed square form then, the quadratic four ๐ฅ squared minus eight ๐ฅ plus five is equal to four multiplied by ๐ฅ minus one all squared plus one.
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Now in part b, weโre told to use our answer to part a to solve this quadratic equation.
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And what youโll notice is that the quadratic expression on the left of the equation is exactly the quadratic expression that we had in part a.
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So this suggests that weโre going to use our completed square form of this quadratic and substitute it into the given equation.
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Replacing the left-hand side then with our completed square form gives four multiplied by ๐ฅ minus one all squared plus one equals 11.
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And we now need to solve for ๐ฅ.
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Your first thought may be that we need to expand the bracket on the left-hand side.
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But actually we donโt need to.
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We can solve for ๐ฅ by rearranging.
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First, we subtract one from each side of this equation, giving four lots of ๐ฅ minus one all squared is equal to 10.
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Next, we divide both sides of this equation by four.
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On the left, this will cancel with the factor four in the numerator, just leaving ๐ฅ minus one all squared.
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And on the right, we have the fraction 10 over four, which can be simplified to five over two.
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Next, we need to take the square root of each side of this equation.
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And in doing so, we must remember that whenever we solve an equation by square-rooting, there are two possible answers.
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So we need to take a plus or minus the square root.
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We have that ๐ฅ minus one equals plus or minus the square root of five over two.
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The final step is to add one to each side of this equation, giving ๐ฅ equals one plus or minus the square root of five over two.
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And we have solved for ๐ฅ.
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Now you may get the marks if you just leave your answer in this form.
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But what youโll notice is that we actually have an irrational denominator, because we have the square root of a fraction.
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One of our rules of surds tells us that if we have the square root of a fraction ๐ over ๐, this is actually equal to the square root of the numerator over the square root of the denominator.
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So the square root of five over two is equal to the square root of five over the square root of two.
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Root two is an irrational number.
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And whenever we have an irrational number in the denominator of the fraction, itโs good practice to rationalise this denominator.
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To do so, we multiply both the numerator and denominator of this fraction by root two.
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This is equivalent to multiplying by one, as both the numerator and denominator of what weโre multiplying by are the same.
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So we arenโt changing the size of the original fraction.
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In the numerator, we have root five multiplied by root two, which is equal to root 10, because if we have the product of two square roots, this is actually equal to the square root of the product.
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So 10 is five times two.
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And in the denominator, root two multiplied by root two is equal to two.
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Two is a rational number.
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So weโve now rationalised the denominator of this fraction.
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We found that the solution to the equation four ๐ฅ squared minus eight ๐ฅ plus five equals 11 is ๐ฅ equals one plus or minus root 10 over two.
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Now we could actually have got to this answer with a rational denominator a little bit more quickly if we hadnโt cancelled down at this stage here.
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If weโd left the fraction as 10 over four rather than five over two, then when we square-rooted, weโd have ๐ฅ minus one equals plus or minus the square root of 10 over four.
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Using our laws of surds, this would be equal to the square root of 10 over the square root of four.
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But four is a square number.
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Its square root is just two.
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So we couldโve replaced it here.
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We could then add one to each side, giving the same solution as before.
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๐ฅ equals one plus or minus root 10 over two.
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But we wouldnโt have needed to go through the process of rationalising the denominator.