Questions and Answers #7 Quadratics

Questions and Answers Sheet 7 Quadratics Question #1 How to ''Re-write completing the square'': 𝑥2 + 𝑥 + 1 Answer: Remember the formula for the square of a binomial: (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏 2 Now, when you see 𝑥 2 + 𝑥 + 1, you want to think of 𝑥 2 + 𝑥 as the first two terms you get in expanding the binomial (x+c)2 for some c; that is, 𝑥 2 + 𝑥 + ⋯ = (𝑥 + 𝑐)2 1 Since the middle term should be 2cx, and you have x, that means that you want 2c=1, or 𝑐 = . 2 1 2 2 1 2 But if you have (𝑥 + ) , you get 𝑥 + 𝑥 + . Since all you have is 𝑥 + 𝑥, you complete the square by 2 4 1 adding the missing . 4 1 Since you are not allowed to just add constants willy-nilly, you must also cancel it out by subtracting . 4 So: 𝑥 2 + 𝑥 + 2 = (𝑥 2 + 𝑥 + ⋯ ) + 1 1 = (𝑥 2 + 2 ( ) 𝑥 + ⋯ ) + 1 figuring out what c is 2 2 1 1 2 1 2 2 4 = (𝑥 + 2 ( ) 𝑥 + ( ) ) − + 1 completing the square 1 2 3 2 4 = (𝑥 + ) + Question #2 Solving Radical Equations 𝑥 − 7 = √𝑥 − 5 Answer: We can avoid squaring both sides. Let 𝑥 − 5 = 𝑢2 , where 𝑢 ≥ 0 Then √𝑥 − 5 = 𝑢 Also, 𝑥 = 𝑢2 + 5 so 𝑥 − 7 = 𝑢2 − 2 Thus our equation can be rewritten as 𝑢2 − 2 = 𝑢 or equivalently 𝑢2 − 𝑢 − 2 = 0 But 𝑢2 − 𝑢 − 2 = (𝑢 − 2)(𝑢 + 1) Thus the solutions of 𝑢2 − 𝑢 − 2 = 0 are u=2 and u=-1 Since 𝑢 ≥ 0, we reject the solution u=-1 We conclude that u=2 , and therefore 𝑥 = 𝑢2 + 5 = 9. Question #3 A quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 has equal roots at a=2c How could we find the sum of reciprocals of the roots of this equation? Answer: Since the roots are equal, you must have 𝑏 2 = 4𝑎𝑐 = 8𝑐 2 Also, the roots satisfy 𝑏 𝑐 𝑎 𝑎 0 = 𝑥2 + 𝑥 + 𝑏 1 𝑎 2 = 𝑥2 + + so the product of the roots is 𝑏 1 2 1 ; that is, 4𝑐 2 = . Therefore, 𝑏 2 = 2(4𝑐 2 ) = 1 2 The sum of the roots is − ; but it also is equal to 2a. So 2𝑎 = − 𝑎 This gives 2𝑎2 = −𝑏, and we know |b|=1, so we must have b=-1 4𝑎 So the sum of the reciprocals is − = 4𝑎 Now, 𝑎2 = 𝑏 1 2 √2 so either 𝑎 = or 𝑎 = − 𝑏 𝑎 2 √2 2 Thus, the equation is either √2 2 𝑥2 − 𝑥 + √2 4 or − √2 2 𝑥2 − 𝑥 − √2 4 Both satisfy the desired conditions: they have a double root at a=2c. In one case, the sum of the reciprocals is 2√2, in the other it is −2√2. Question #4 Complex squares, what are the steps: 2𝑥 2 − 𝑥(5 + 𝑚) + (3 + 𝑚) = 0 Answer: To complete the square on the left hand side of your equation, first factor out the 2: 1) 2𝑥 2 − 𝑥(5 + 𝑚) + (3 + 𝑚) 5+𝑚 = 2 (𝑥 2 − 2 + 3+𝑚 2 ) Write down what you want: 5+𝑚 3+𝑚 2) 𝑥 2 − 𝑥+ 2 2 = (𝑥 − ℎ)2 + 𝑘 Write the right hand side of (2) as 3) 𝑥 2 − 2ℎ𝑥 + ℎ2 + 𝑘 Looking at the x term of the left hand side of (2) with the x term of (3), write down an equation and solve for h: 5+𝑚 5+𝑚 − 𝑥 = −2ℎ𝑥 ⇒ ℎ = 2 4 Now use equations (2) and (3) to find the value of k: 3+𝑚 3+𝑚 = ℎ2 + 𝑘 ⇒ 𝑘 = − ℎ2 ⇒ 𝑘 2 = 3+𝑚 2 −( 2 So 𝑥 − 2 5+𝑚 2 4 5+𝑚 2 2 ) 𝑥+ 3+𝑚 2 = (𝑥 − ℎ) + 𝑘 = (𝑥 − 5+𝑚 2 4 ) + 3+𝑚 2 −( 5+𝑚 2 4 ) 2 and 2𝑥 − 𝑥(5 + 𝑚) + (3 + 𝑚) = 2 (𝑥 2 − 5+𝑚 2 𝑥+ 3+𝑚 2 = 2 ((𝑥 − 2 ) 5+𝑚 3+𝑚 5+𝑚 2 ) + −( ) ) 4 2 4 = 2 (𝑥 − 5+𝑚 2 ) + 3+ 𝑚 − 2( 4 5+𝑚 2 4 ) Question #5 Inequality (𝑎 + 𝑏)(𝑎 + 𝑏 + 𝑐) < 0 Prove (𝑏 − 𝑐)2 > 4𝑎(𝑎 + 𝑏 + 𝑐) Answer: Consider the quadratic 𝑓(𝑥) = 𝑎𝑥 2 − (𝑏 − 𝑐)𝑥 + (𝑎 + 𝑏 + 𝑐) 𝑓(1)𝑓(0) = 2(𝑎 + 𝑐)(𝑎 + 𝑏 + 𝑐) < 0 Thus if 𝑎 ≠ 0 , then this has a real root in (0,1) and so (𝑏 − 𝑐)2 ≥ 4𝑎(𝑎 + 𝑏 + 𝑐) If (𝑏 − 𝑐)2 + 4𝑎(𝑎 + 𝑏 + 𝑐), then we have a double root in (0,1) in which case, f(0) and f(1) will have the same sign. Thus (𝑏 − 𝑐)2 > 4𝑎(𝑎 + 𝑏 + 𝑐) If a=0, then 𝑐(𝑏 + 𝑐) < 0, and so we cannot have b=c and thus (𝑏 − 𝑐)2 > 0 = 4𝑎(𝑎 + 𝑏 + 𝑐) And if you want a more ''direct'' approach, we show that (𝑝 + 𝑞 + 𝑟)𝑟 < 0 ⇒ 𝑞 2 > 4𝑝𝑟 using the following identity: (𝑝 + 𝑞 + 𝑟)𝑟 = (𝑝 (1 + +𝑝 (𝑟 − 𝑞2 4𝑝 ) ((1 + 𝑞 2𝑝 𝑞 ) 𝑞 2𝑝 2𝑝 2 𝑞 2 ) + (𝑟 − 𝑞2 4𝑝 2 ) 2 ) +( ) ) 2𝑝 If (𝑝 + 𝑞 + 𝑟)𝑟 < 0, then we must have 𝑝 (𝑟 − 𝑞2 4𝑝 )<0 as all the other terms on the right side are non-negative. Of course, this was gotten by completing the square in 𝑝𝑥 2 + 𝑞𝑥 + 𝑟 and setting x=0 and x=1 and multiplying. Question #6 Quadratic equation 3𝑥 3 = 24 Answer: May be you know this already but it's not clear to me from your post that you have the progression from the original problem statement to the final two factors exactly correct. 3𝑥 3 = 24 ⇒ 3𝑥 3 − 24 = 0 ⇒ 3(𝑥 3 − 8) = 0 ⇒ 𝑥3 − 8 = 0 ⇒ (𝑥 − 2)(𝑥 2 + 2𝑥 + 4) = 0 ⇒ 𝑥−2= 0, 𝑥 2 + 2𝑥 + 4 = 0 The first factor results in x=2. The second factor must be solved using the quadratic formula or by completing the square since it is not factorable. Question #7 Proving Quadratic Formula How does 𝑏 2𝑎 become 𝑏2 4𝑎2 ? Answer: Remember how to complete the square: 𝐴𝑥 2 + 𝐵𝑥 = 𝐴 (𝑥 + 𝐵 2𝐴 2 ) − 𝐵2 4𝐴2 So now 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 complete square 𝑏 𝑎 (𝑥 + (𝑥 + 2 ) − 2𝑎 𝑏 2 2𝑎 𝑥1, 2 + 𝑥1, 2 = ) = 𝑏 2𝑎 𝑏2 = −𝑐 4𝑎 𝑏 2 −4𝑎𝑐 =± 4𝑎2 √𝑏 2 −4𝑎𝑐 2𝑎 −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 Question #8 How can I convert 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 to a FOIL-style (x+d)(x-e)=0 equation? Answer: Step 1 There's several problems here. First, notice: (𝑥 + 𝑑)(𝑥 − 𝑒) = 𝑥 2 − 𝑒𝑥 + 𝑑𝑥 − 𝑒𝑑 = 𝑥 2 + (𝑑 − 𝑒)𝑥 − 𝑒𝑑 Where in the world are you going to get a leading coefficient? In other words, this method doesn't suffice without some manipulations of the standard form. That manipulation is: 1 (𝑎𝑥 2 + 𝑏𝑥 + 𝑐) = 𝑥 2 + 𝑝𝑥 + 𝑞 2 𝑝= 𝑞= 𝑏 𝑎 𝑐 𝑎 Step 2 The second issue is that it can't be generally figured out what d and e are. d, by definition, is the opposite of the first root of the equation. e, by definition, is the second root. However, the relationships between p and q and d and e are: d-e=p -ed=q This information doesn't allow us to determine d and e in terms of the coefficients p and q. There's also the common sensical fact, that like I said, you have to know the roots to know e and d. But the biggest issue is what you want. You cannot get rid of the imaginary solutions. They will always be solutions to particular quadratics. It is impossible to ''get rid of them.'' For example, notice: 𝑥2 + 1 = 0 There are no solutions to this equation in 𝑅. The only solutions are in 𝐶. They are i and -i. Now, let me show why they have to be those particular quantities in a very simply way: 𝑥 2 + 1 = (𝑥 − 𝑟1 )(𝑥 − 𝑟2 ) = (𝑥 − 𝑖)(𝑥 + 𝑖) = 𝑥(𝑥 + 𝑖) − 𝑖(𝑥 + 𝑖) 𝑥 2 + 𝑖𝑥 − 𝑖𝑥 − 𝑖 2 = 𝑥 2 − (−1) = 𝑥2 + 1 Do you see how those solutions have to be i and -i? This applies to all quadratics. Their roots are whatever makes the following relation true and the fundamental relation (that is, that the quadratic is equal to 0) true. I hope this makes sense. Addendum: It may help to see precisely where ''that formula'' comes from. It is derived directly from the standard form equation. Step 3 First, make this observation: (𝑥 − 𝑣)2 = 𝑥 2 − 2𝑣𝑥 + 𝑣 2 You can turn the manipulated form into a square: 𝑥 2 + 𝑝𝑥 + 𝑞 = 0 𝑥 2 + 𝑝𝑥 = −𝑞 𝑝 2 𝑝 2 𝑥 2 + 𝑝𝑥 + ( ) = −𝑞 + ( ) 2 𝑝 2 𝑝 2 (𝑥 + ) = −𝑞 + ( ) = 2 2 −4𝑞+𝑝2 2 𝑝 −4𝑞+𝑝2 2 4 𝑥 + = ±√ 4 1 = √−4𝑞 + 𝑝2 2 1 𝑝 𝑥 = ± √−4𝑞 + 𝑝2 − = 2 2 −𝑝±√−4𝑞+𝑝2 2 Substituting back in for p and q gives us: 2 𝑥= 𝑏 𝑐 𝑏 − ±√−4 +( ) 𝑎 𝑎 𝑎 2 𝑐 𝑏2 = −𝑏±𝑎 √−4𝑎+ 2 𝑐 2𝑎 𝑏 2 − 4𝑎𝑐 𝑎2 2𝑎 −𝑏 ± 𝑎√ = = −𝑏±√𝑏 2 −4𝑐 2𝑎 Also, André Nicolas provides an even more simple derivation of the quadratic formula in this question: Why can ALL quadratic equations be solved by the quadratic formula? Question #9 Find the least a, for which two equations have a common root 2𝑥 2 − 3𝑥 + 1 = 0 and 2𝑥 2 − (𝑎 + 3)𝑥 + 3𝑎 = 0 Answer: I think you can just do it in a concrete way. The roots of 2𝑥 2 − 3𝑥 + 1 = 0 are 2 formula, the roots of 2𝑥 − (𝑎 + 3)𝑥 + 3𝑎 = 0 are 𝑎+3−√𝑎2 −18𝑎+9 and 8 𝑎+3+√𝑎2 −18𝑎+9 8 If they have common roots, we have the following possibilities: 𝑎+3−√𝑎2 −18𝑎+9 8 = 1 2 or 𝑎+3−√𝑎2 −18𝑎+9 8 =1 or 𝑎+3+√𝑎2 −18+9 8 = 1 2 or 𝑎+3+√𝑎2 −18𝑎+9 8 =1 Each one of them is a quadratic equation in a which can be solved. 1 2 and 1. By quadratic Question #10 How to find original equations of type 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐. Given 3 coordinate points? (40,30)(60,28)(20,25) Answer: Your first problem is in the title. The function you're looking for is 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐. And you have 3 points on that parabola. So just plug them in. 1600a+40b+c=30 3600a+60b+c=28 400a+20b+c=25