Quadratic Part1

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Contents

Notes

This is the one of the killer subtopics in the first half of the syllabus, second only to inequality. Just looking at it though, it doesn't seem like it, especially since we almost learn nothing new compared to add maths SPM. Two reasons :

  • Anything that have already been covered in SPM, will have much more difficult questions in actual STPM, since it is pointless to issue SPM-level questions
  • MOST students never actually did fully understand this topic in SPM. No, not even those with A1. Only a minority really understood the complexity of it. The rest just memorize formulas/workings.

Learning Objectives (Syllabus)

  • use the process of completing the square for a quadratic polynomial
  • derive the quadratic formula
  • solve linear, quadratic, and cubic equations and equations that can be transformed into quadratic or cubic equations
  • use the discriminant of a quadratic equation to determine the properties of its roots
  • prove and use the relationship between the roots and coefficients of a quadratic equation

Prior Knowledge

  • basic algebraic skills
  • surds, complex numbers
  • polynomials, factors, roots


Completing the Square

  • This is one of the main skills that needs to be learnt well here. We will also need to master the various ways it can be used.
  • I prefer to use a direct method of completing rather than using any formulas. It might seem difficult at first but you will get a hang of it.

Preliminary

Complete the following

    • \left(x+1\right)^{2}=x^{2}+\underline{\quad}x+\underline{\quad} \left(x+1\right)^{2}=x^{2}+{\color{Red}2}x+{\color{Blue}1}
    • \left(x+3\right)^{2}=x^{2}+\underline{\quad}x+\underline{\quad} \left(x+3\right)^{2}=x^{2}+{\color{Red}6}x+{\color{Blue}9}
    • \left(x+7\right)^{2}=x^{2}+\underline{\quad}x+\underline{\quad} \left(x+7\right)^{2}=x^{2}+{\color{Red}14}x+{\color{Blue}49}
    • \left(x+\frac{1}{2}\right)^{2}=x^{2}+\underline{\quad}x+\underline{\quad} \left(x+\frac{1}{2}\right)^{2}=x^{2}+{\color{Red}1}x+{\color{Blue}\frac{1}{4}}
    • \left(x+\underline{\quad}\right)^{2}=x^{2}+8x+\underline{\quad} \left(x+{\color{Red}4}\right)^{2}=x^{2}+8x+ {\color{Blue}16}
    • \left(x+\underline{\quad}\right)^{2}=x^{2}+2x+\underline{\quad} \left(x+{\color{Red}1}\right)^{2}=x^{2}+2x+ {\color{Blue}1}
    • \left(x-1\right)^{2}=x^{2}\underline{\qquad} \left(x-1\right)^{2}=x^{2}{\color{Red}-2}x +{\color{Blue}1}
    • \left(x-5\right)^{2}=x^{2}\underline{\qquad} \left(x-5\right)^{2}=x^{2}{\color{Red}-10}x +{\color{Blue}25}
    • \left(x-\frac{3}{2}\right)^{2}=x^{2}\underline{\qquad} \left(x-\frac{3}{2}\right)^{2}=x^{2}{\color{Red}-3}x +{\color{Blue}\frac{9}{4}}
    • \left(x\underline{\qquad}\right)^{2}=x^{2}+10x+\underline{\quad} \left(x{\color{Red}+5}\right)^{2}=x^{2}+10x {+\color{Blue}25}
    • \left(x\underline{\qquad}\right)^{2}=x^{2}-6x\underline{\quad} \left(x{\color{Red}-3}\right)^{2}=x^{2}-6x {\color{Blue}+9}
    • \left(x\underline{\qquad}\right)^{2}=x^{2}+5x\underline{\quad} \left(x{\color{Red}+\frac{5}{2}}\right)^{2}=x^{2}+5x {\color{Blue}+\frac{25}{4}}
    • \left(x\underline{\qquad}\right)^{2}=x^{2}-x\underline{\quad} \left(x{\color{Red}-\frac{1}{2}}\right)^{2}=x^{2}-x {\color{Blue}+\frac{1}{4}}
    • x^{2}+4x=\left(x\underline{\qquad}\right)^{2}\underline{\quad} x^{2}+4x=\left(x{\color{Red}+2}\right)^{2} {\color{Blue}-4} Since we can't write the \,+4 on the left, we bring it over to the right and it becomes \,-4
    • x^{2}-3x=\left(x\underline{\qquad}\right)^{2}\underline{\quad} x^{2}-3x=\left(x{\color{Red}-\frac{3}{2}}\right)^{2} {\color{Blue}-\frac{9}{4}}

Guideline

Similarly

  • \,x^{2}+px=
    • {\color{Red}\left(x\qquad \right)^{2}}
    • \left(x{\color{Red}+\frac{p}{2}}\right)^{2}
    • \left(x+\frac{p}{2}\right)^{2}{\color{Red}-\frac{p^{2}}{4}}

Thus

  • \,x^{2}+px+q= \left(x+\frac{p}{2}\right)^{2}-\frac{p^{2}}{4}+q

Important

  • Rearrange and factorize (if needed) so that coefficient of \,x^{2} is \,1
  • Final form is a\left(x+m\right)^{2}+n (need to multiply back if factored at the beginning)
  • CHECK! The fastest way is check that it will give back the same constant term. This will check for most careless mistakes.

Examples

Complete the square

  • \,x^{2}+4x+1
  • Already nicely arranged and the coefficient of \,x^{2} is \,1, so we can do directly
    • ={\color{Red}\left(x\qquad\right)^{2}\qquad+1} (Write out the form first, including the \,+1 (so that we won't forget), and then only we fill in the blanks
    • =\left(x{\color{Red}+2}\right)^{2}\quad\qquad+1
    • =\left(x+2\right)^{2}{\color{Red}-4}+1 =\left(x+2\right)^{2}-3
    • Check that \left(x{\color{Blue}+2}\right)^{{\color{Blue}2}}{\color{Blue}-3} gives us back \,+1

  • \,2x^{2}-3x-7
  • First, we will need to factor the \,2
    • Be sure to factorize every term
      • =2\left(x^{2}-\frac{3}{2}x-\frac{7}{2}\right)
    • Once again, we will do it in one line, but fill in step by step
      • ={\color{Red}2\left[\left(x\qquad\right)^{2}\qquad\qquad-\frac{7}{2}\right]}
      • =2\left[\left(x{\color{Red}-\frac{3}{4}}\right)^{2}\qquad\qquad-\frac{7}{2}\right]
      • =2\left[\left(x-\frac{3}{4}\right)^{2}-{\color{Red}\frac{9}{16}}-\frac{7}{2}\right]
    • Calculate the final constant (please use calculator!), remember to multiply back the \,2
      • =2\left(x-\frac{3}{4}\right)^{2}-\frac{65}{8}
    • CHECK!
      • {\color{Blue}2}\left(x{\color{Blue}-\frac{3}{4}}\right)^{{\color{Blue}2}}{\color{Blue}-\frac{65}{8}}. Use calculator to check it will give back \,-7
  • \,-x^{2}+6x-5
  • Needs to be factored too
    • =-\left(x^{2}-6x+5\right)
    • ={\color{Red}-\left[\left(x\qquad\right)^{2}\qquad+5\right]}
      • =-\left[\left(x{\color{Red}-3}\right)^{2}\qquad+5\right]
      • =-\left[\left(x-3\right)^{2}{\color{Red}-9}+5\right]
    • =-\left(x-3\right)^{2}+4

  • \,1-2x-3x^{2}
  • Try it first! Answer is -3\left(x+\frac{1}{3}\right)^{2}+\frac{4}{3}
    • \begin{align}
1-2x-3x^{2} & = -3x^{2}-2x+1\\
& = -3\left(x^{2}+\frac{2}{3}x-\frac{1}{3}\right)\\
& = -3\left[\left(x+\frac{1}{3}\right)^{2}-\frac{1}{9}-\frac{1}{3}\right]\\
& = -3\left(x+\frac{1}{3}\right)^{2}-\frac{4}{3}\\
\end{align}

Exercise 1

Complete the square for the following

a) \,x^{2}+6x-1 \left(x+3\right)^{2}-10

  • \begin{align}
x^{2}+6x-1  
& = \left(x+3\right)^{2}-9-1\\
&= \left(x+3\right)^{2}-10\\
\end{align}

b) \,2x^{2}-3x+11 2\left(x-\frac{3}{4}\right)^{2}+\frac{79}{8}

  • \begin{align}
2x^{2}-3x+11 & = 2\left(x^{2}-\frac{3}{2}x+\frac{11}{2}\right) \\
& = 2\left[\left(x-\frac{3}{4}\right)^{2}-\frac{9}{16}+\frac{11}{2}\right] \\
& = 2\left(x-\frac{3}{4}\right)^{2}+\frac{79}{8} \\
\end{align}

c) \,-3x^{2}+6x-3 -3\left(x-1\right)^{2}

  • \begin{align}
-3x^{2}+6x-3 & = -3\left(x^{2}-2x+1\right) \\
& = -3\left[\left(x-1\right)^{2}-1+1\right] \\
& = -3\left(x-1\right)^{2} \\
\end{align}

d) \,1+6x-4x^{2} -4\left(x-\frac{3}{4}\right)^{2}+\frac{13}{4}

  • \begin{align}
1+6x-4x^{2} & = -4x^{2}+6x+1\\
& = -4\left(x^{2}-\frac{3}{2}x-\frac{1}{4}\right) \\
& = -4\left[\left(x-\frac{3}{4}\right)^{2}-\frac{9}{16}-\frac{1}{4}\right] \\
& = -4\left(x-\frac{3}{4}\right)^{2}+\frac{13}{4} \\
\end{align}

e) \,2x+7-3x^{2} -3\left(x-\frac{1}{3}\right)^{2}+\frac{22}{3}

  • \begin{align}
2x+7-3x^{2} & = -3x^{2}+2x+7 \\
& = -3\left(x^{2}-\frac{2}{3}x-\frac{7}{3}\right) \\
& = -3\left[\left(x-\frac{1}{3}\right)^{2}-\frac{1}{9}-\frac{7}{3}\right] \\
& = -3\left(x-\frac{1}{3}\right)^{2}+\frac{22}{3} \\
\end{align}

f) \left(x-3\right)\left(x+2\right) \left(x-\frac{1}{2}\right)^{2}-\frac{25}{4}

  • \begin{align}
\left(x-3\right)\left(x+2\right) & = x^{2}-x-6 \\
& = \left(x-\frac{1}{2}\right)^{2}-\frac{1}{4}-6 \\
& =\left(x-\frac{1}{2}\right)^{2}-\frac{25}{4}
\end{align}

To Find Max/Min

Once we have completed the square, we can obtain the max/min value of the polynomial just by looking at the final form. This, like the direct method of completing the square, requires logical thinking rather than memorizing of steps.

Value of x2

For all real values of \,x, we know that \,x^{2} \geq 0 This is because\begin{align}
\left(\mbox{Positive value}\right)^{2} & = \mbox{Positive value} \\
\left(\mbox{Negative value}\right)^{2} & = \mbox{Positive value} \\
0^{2}& = 0
\end{align}

That is,

  • \,x^{2} \geq 0 \mbox{for all}\; x \in \mathbb{R}

Similarly

  • \,y^{2} \geq 0 \mbox{for all}\; y \in \mathbb{R}

Meanwhile

  • \left(x+2\right)^{2} Is it\geq 2? Or \geq 4? It is \geq 0
    • (This is because if \left(x+2\right)^{2} CAN be zero, which is when x=-2\,, so it still will be \geq 0 )
  • \left(1-x\right)^{2} \geq 0
    • (Remember that both negative and positive values, when squared, will give us positive values, so even though \,1-x is negative (when \,x is positive), it does not matter)

That is,

  • \,\left(x+2\right)^{2} \geq 0 \mbox{for all}\; x \in \mathbb{R}
  • \,\left(1-x\right)^{2} \geq 0 \mbox{for all}\; x \in \mathbb{R}
  • \,\left(\frac{1}{3}x-3\right)^{2} \geq 0 \mbox{for all}\; x \in \mathbb{R}
  • \,\left(3x\right)^{2} \geq 0 \mbox{for all}\; x \in \mathbb{R}
  • \,\left(1-7y\right)^{2} \geq 0 \mbox{for all}\; y \in \mathbb{R}
  • \,\left(x-y-7\right)^{2} \geq 0 \mbox{for all}\; x,y \in \mathbb{R}

Note

Since x^{2} \geq 0 for all real values of \,x,

  • \therefore -x^{2} \leq 0 for all real values of \,x
  • Don't mix up \left(-x\right)^{2} with \,-x^{2}

Max/Min value

If y \geq 1, it means \,y can take the value of \,1, or any value greater than \,1. Thus, we say that the minimum value of \,y is \,1

  • y \geq 1 \,\mbox{minimum value of } y =1
  • y \geq 3 \,\mbox{minimum value of } y \,=3
  • y \leq 4 \,\mbox{maximum value of } y \,=4
  • y \leq -1 \,\mbox{maximum value of } y \,=-1
  • y \geq -7 \,\mbox{mimimum value of } y \,=-7
  • y \geq 0 \,\mbox{mimimum value of } y \,=0

To Find Max/Min value

Note : MUST complete the square first (if not already so)

  • \,y=x^{2}
  • Let's start with the simplest. There is no need to complete the square here.
    • \mbox{Since }x^{2}\, \geq 0 \mbox{for all }x \in \mathbb{R}
    • \therefore y \geq 0 \mbox{for all }x \in \mathbb{R}
    • \therefore \,\mbox{minimum value of } y \,=0
    • \,\mbox{and it occurs when } \,x= \,0
  • \,y=x^{2}+1
  • This is slightly more complicated. Just keep in mind that we always start with \left({\color{Red}\heartsuit}\right)^{2} \geq 0 and then we will work from there.
    • \mbox{Since }x^{2}\geq 0 \; \mbox{for all }x \in \mathbb{R}
    • \therefore x^{2}+1 \geq \,1 \mbox{for all }x \in \mathbb{R}
    • \therefore y \geq 1 \mbox{ for all }x \in \mathbb{R} (we can probably skip this step and straight to the max/min value step)
    • \therefore \,\mbox{minimum value of } y \,=1 When does \,x^{2}+1 take this minimum value of \,1? When \,x^{2}=0
    • \,\mbox{and it occurs when } \,x= \,0
  • \,y=x^{2}+7
  • Try this on your own first! (Writing the steps is as important as the value)
    • \begin{align}
& \mbox{Since }x^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore x^{2}+7\geq 7 \mbox{ for all }x \in \mathbb{R} \\
& \therefore\mbox{ minimum value of } y =7\mbox{ and it occurs when }x=0
\end{align}
  • \,y=2x^{2}-3
    • \begin{align}
& \mbox{Since}\;x^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore 2x^{2} \geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore 2x^{2}-3 \geq -3 \mbox{ for all }x \in \mathbb{R} \\
\end{align} (Can probably skip a step here)
    • \therefore\mbox{ minimum value of } y =-3\mbox{ and it occurs when }x=0

  • \,y=-x^{2}+5
  • Be careful with the negative here. Regardless, start with \left({\color{Red}\heartsuit}\right)^{2} \geq 0 and work carefully from there
    • \mbox{Since }x^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}
    • \therefore -x^{2} \leq 0 \mbox{ for all }x \in \mathbb{R}
    • \therefore -x^{2}+5 \leq 5 \mbox{ for all }x \in \mathbb{R}
    • \therefore\mbox{ maximum value of } y =5\mbox{ and it occurs when }x=0

  • \,y=-2x^{2}-1
  • Try this on your own first!
    • \begin{align}
& \mbox{Since }x^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore -2x^{2}\leq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore -2x^{2}-1\leq -1 \mbox{ for all }x \in \mathbb{R} \\
& \therefore\mbox{ maximum value of } y =-1 \mbox{ and it occurs when }x=0
\end{align}
  • \,y=\left(x-1\right)^{2}
  • Note that \left({\color{Red}\heartsuit}\right)^{2} \geq 0. DON'T just look at the \,x, look at everything inside the bracket
    • \mbox{Since }\left(x-1\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}
    • \therefore \,\mbox{ minimum value of } y \,=0 When is \left({\color{Red}\heartsuit}\right)^{2}=0?
    • \,\mbox{ and it occurs when } \,x=1

  • \,y=2\left(x+3\right)^{2}
    • \mbox{Since }\left(x+3\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}
    • \therefore 2\left(x+3\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}
    • \therefore \mbox{ minimum value of } y \,=0
    • \,\mbox{ and it occurs when } \,x=-3

  • \,y=-\left(x-7\right)^{2}
  • Try this on your own!
    • \begin{align}
& \mbox{Since }\left(x-7\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore -\left(x-7\right)^{2}\leq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore \mbox{ maximum value of } y=0 \\
& \mbox{ and it occurs when }x=7
\end{align}

  • \,y=\left(x-2\right)^{2}+3
  • Remember, don't memorize the steps, but understand the logic in each step, and you won't go wrong.
    • \mbox{Since }\left(x-2\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}
    • \therefore\left(x-2\right)^{2}+3\geq 3 \mbox{ for all }x \in \mathbb{R}
    • \therefore \mbox{ minimum value of } y \,=3 When does \left({\color{Red}\heartsuit}\right)^{2}+3 take a minimum value of \,3? When \left({\color{Red}\heartsuit}\right)^{2}=0
    • \,\mbox{ and it occurs when } \,x=2

  • \,y=\left(x+4\right)^{2}-11
    • \mbox{Since }\left(x+4\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}
    • \therefore\left(x+4\right)^{2}-11\geq -11 \mbox{ for all }x \in \mathbb{R}
    • \therefore \mbox{ minimum value of } y=-11
    • \,\mbox{ and it occurs when }x=-4

  • \,y=3\left(x-2\right)^{2}-2
  • \begin{align}
& \mbox{Since }\left(x-2\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore 3\left(x-2\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore 3\left(x-2\right)^{2}-2\geq -2 \mbox{ for all }x \in \mathbb{R} \\
& \therefore \mbox{ minimum value of } y=-2 \\
& \mbox{ and it occurs when }x=2
\end{align}
  • \,y=-\left(x+1\right)^{2}+7
  • \begin{align}
& \mbox{Since }\left(x+1\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore -\left(x+1\right)^{2}\leq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore -\left(x+1\right)^{2}+7\leq 7 \mbox{ for all }x \in \mathbb{R} \\
& \therefore \mbox{ maximum value of } y=7 \\
& \mbox{ and it occurs when }x=-1
\end{align}
  • \,y=-\frac{7}{2}\left(x-\frac{3}{2}\right)^{2}-\frac{5}{4}
  • \begin{align}
& \mbox{Since }\left(x-\frac{3}{2}\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore -\frac{7}{2}\left(x-\frac{3}{2}\right)^{2}\leq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore -\frac{7}{2}\left(x-\frac{3}{2}\right)^{2}-\frac{5}{4}\leq -\frac{5}{4} \mbox{ for all }x \in \mathbb{R} \\
& \therefore \mbox{ maximum value of } y=-\frac{5}{4} \\
& \mbox{ and it occurs when }x=\frac{3}{2}
\end{align}

Max/Min Point

If maximum/minimum value of \,y=b when \,x=a

  • \therefore maximum/minimum point is \left(a,b\right)

Exercise 2

For every question in Exercise 1 (let the each polynomial to be equals \,y)

  • a)State the maximum/minimum value of \,y and the value of \,x when it occurs
  • b) the maximum/minimum point of its graph

a) \,x^{2}+6x-1 =\left(x+3\right)^{2}-10

  • \begin{align}
& \mbox{Since }\left(x+3\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore \left(x+3\right)^{2}-10\geq -10 \mbox{ for all }x \in \mathbb{R} \\
& \therefore \mbox{ minimum value of } y=-10 \\
& \mbox{ and it occurs when }x=-3 
\end{align}
  • \,\mbox{ Minimum point is }\left(-3,-10\right)

b) \,2x^{2}-3x+11 =2\left(x-\frac{3}{4}\right)^{2}+\frac{79}{8}

  • \begin{align}
& \mbox{Since }\left(x-\frac{3}{4}\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore 2\left(x-\frac{3}{4}\right)^{2}+\frac{79}{8}\geq \frac{79}{8}\mbox{ for all }x \in \mathbb{R} \\
& \therefore \mbox{ minimum value of } y=\frac{79}{8} \\
& \mbox{ and it occurs when }x=\frac{3}{4} 
\end{align}
  • \,\mbox{ Minimum point is }\left(\frac{3}{4},\frac{79}{8}\right)

c) \,-3x^{2}+6x-3 =-3\left(x-1\right)^{2}

  • \begin{align}
& \mbox{Since }\left(x-1\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore -3\left(x-1\right)^{2}\leq 0\mbox{ for all }x \in \mathbb{R} \\
& \therefore \mbox{ maximum value of } y=0 \\
& \mbox{ and it occurs when }x=1 
\end{align}
  • \,\mbox{ Minimum point is }\left(1,0\right)

d) \,1+6x-4x^{2} =-4\left(x-\frac{3}{4}\right)^{2}+\frac{13}{4}

  • \begin{align}
& \mbox{Since }\left(x-\frac{3}{4}\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore -4\left(x-\frac{3}{4}\right)^{2}\leq 0\mbox{ for all }x \in \mathbb{R} \\
& \therefore -4\left(x-\frac{3}{4}\right)^{2}+\frac{13}{4}\leq \frac{13}{4}\mbox{ for all }x \in \mathbb{R} \\
& \therefore \mbox{ maximum value of } y=\frac{13}{4} \\
& \mbox{ and it occurs when }x=\frac{3}{4}
\end{align}
  • \,\mbox{ Maximum point is }\left(\frac{3}{4},\frac{13}{4}\right)

e) \,2x+7-3x^{2} =-3\left(x-\frac{1}{3}\right)^{2}+\frac{22}{3}

  • \begin{align}
& \mbox{Since }\left(x-\frac{1}{3}\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore -3\left(x-\frac{1}{3}\right)^{2}\leq 0\mbox{ for all }x \in \mathbb{R} \\
& \therefore -3\left(x-\frac{1}{3}\right)^{2}+\frac{22}{3}\leq \frac{22}{3}\mbox{ for all }x \in \mathbb{R} \\
& \therefore \mbox{ maximum value of } y=\frac{22}{3} \\
& \mbox{ and it occurs when }x=\frac{1}{3}
\end{align}
  • \,\mbox{ Minimum point is }\left(\frac{1}{3},\frac{22}{3}\right)

f) \left(x-3\right)\left(x+2\right) =\left(x-\frac{1}{2}\right)^{2}-\frac{25}{4}

  • \begin{align}
& \mbox{Since }\left(x-\frac{1}{2}\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R} \\
& \therefore \left(x-\frac{1}{2}\right)^{2}-\frac{25}{4}\geq -\frac{25}{4} \mbox{ for all }x \in \mathbb{R} \\
& \therefore \mbox{ minimum value of } y=-\frac{25}{4} \\
& \mbox{ and it occurs when }x=\frac{1}{2} 
\end{align}
  • \,\mbox{ Minimum point is }\left(\frac{1}{2},-\frac{25}{4}\right)

Sign of Quadratic Polynomial

Preliminary

  • If y \geq 1,
    • \therefore y \,>\,0 \mbox{ for all }x \in \mathbb{R}
    • \,y\mbox{ is always } \,\mbox{positive}
  • If y \leq -3,
    • \therefore y <\,\,0 \mbox{ for all }x \in \mathbb{R}
    • \,y\mbox{ is always } \,\mbox{negative}
  • If y \geq -2 y\, could be positive or negative
  • If y \leq 4 y\, could be positive or negative

Thus

  • If y\, has a positive minimum value
    • \therefore y>0\mbox{ for all }x \in \mathbb{R}
    • or we say \,y\mbox{ is always positive}
  • If y\, has a negative maximum value
    • \therefore y<0\mbox{ for all }x \in \mathbb{R}
    • or we say \,y\mbox{ is always negative}
  • If it is neither, but we are asked to prove always positive/negative, it means we made a mistake in the completing the square

To Show Sign of Quadratic Polynomial

From above, we can see that we would need to find max/min value. Thus, we will first need to complete the square.

Important Note: Whenever question ask us to show something, never ever start with that something. That something must be the last line.


Example

  • Show that x^{2}-4x+8>0\, for all x \in \mathbb{R}
  • First, remember, never start with what we need to show.
    • Thus, the working should look like this
      • \begin{align}
& \ldots \\
& \ldots \\
& \ldots \\
& \therefore x^{2}-4x+8>0 \mbox{ for all }x \in \mathbb{R}\frac{\qquad}{}\mbox{Shown}\\
\end{align}
    • And NOT
      • \begin{align}
& x^{2}-4x+8>0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore \ldots \\
& \ldots \\
& \ldots \\
& \ldots \\
\end{align}
    • We do know that we need to complete the square of the LHS, so we will start from there
    • \,x^{2}-4x+8= \left(x-2\right)^{2}+4
      • Note again that it is WRONG to write
        • \begin{align}
& x^{2}-4x+8>0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore \left(x-2\right)^{2}+4>0 \mbox{ for all }x \in \mathbb{R}\\ \ldots \\
\end{align}
    • So what we have till now is
      • \begin{align}
& x^{2}-4x+8=\left(x-2\right)^{2}+4 \\
& \ldots\\
& \ldots\\
& \ldots\\
& \therefore x^{2}-4x+8>0 \mbox{ for all }x \in \mathbb{R}\\
\end{align}
      • x^{2}-4x+8=\left(x-2\right)^{2}+4 Let's just start again from \left({\color{Red}\heartsuit}\right)^{2}\geq 0 and see where we can go...
      • \mbox{Since }\left(x-2\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}
      • \therefore\left(x-2\right)^{2}+4\geq 4 \mbox{ for all }x \in \mathbb{R} Hmmm... but we don't need to prove that it is minimum \,4, we need to prove that it is >0\,, so instead of this line, we will write
      • \therefore \left(x-2\right)^{2}+4> 0 \mbox{ for all }x \in \mathbb{R}
      • Hurray... finished.... nope
      • \therefore x^{2}-4x+8>0 \mbox{ for all }x \in \mathbb{R}

  • Show that 2x-x^{2}-3\, is always negative.
  • Try this on your own first! Pay special attention to how you write the proof, as a wrong way will get you zero marks
    • \begin{align}
2x-x^{2}-3 & = -x^{2}+2x-3 \\
& = -\left(x^{2}-2x+3\right) \\
& =-\left[\left(x-1\right)^{2}+2\right]\\
& =-\left(x-1\right)^{2}-2 \\
\end{align}
    • \begin{align}
&\mbox{Since }\left(x-1\right)^{2} \geq 0\mbox{ for all } x \in \mathbb{R}\\
&\therefore -\left(x-1\right)^{2} \leq 0\mbox{ for all } x \in \mathbb{R}\\
&\therefore -\left(x-1\right)^{2}-2 < 0 \mbox{ for all } x \in \mathbb{R}\\
&\therefore 2x-x^{2}-3 < 0 \mbox{ for all } x \in \mathbb{R}\frac{\qquad}{}\mbox{Shown}\\
\end{align}


To find Max/Min vs To Prove Sign

Suppose we have \,x^{2}+3, since x^{2}\geq0 \mbox{ for all }x \in \mathbb{R}

  • If we need to find max/min, \therefore x^{2}+3\, \geq 3 \mbox{ for all }x \in \mathbb{R}
  • If we need to prove it is always positive, \therefore x^{2}+3\, >0 \mbox{ for all }x \in \mathbb{R}

Maximum/Minimum Value & Sign

  • If given \left(x+p\right)^{2} + q > 0 \mbox{ for all } x \in \mathbb{R}
  • \therefore q >0\, (it must have a positive minimum value)
  • If given -\left(x+p\right)^{2} + q< 0 \mbox{ for all } x \in \mathbb{R}
  • \therefore q <0\, (it must have a negative maximum value)

Example

  • Find range of \,p if x^{2}+x+p>0\, for all x \in \mathbb{R}
  • Be very aware that this is a given question, not a prove. Though it uses the same concept, the working is different (less rigid, actually)
    • We must use what is given to get a solution. Even if we can't see yet how it will give the answer, just write down what is given first
    • \mbox{Given }x^{2}+x+p>0 \mbox{ for all } x \in \mathbb{R} Let's complete the square
    • \left(x+\frac{1}{2}\right)^{2}-\frac{1}{4}+p>0 \mbox{ for all } x \in \mathbb{R} Note that we can complete the square directly from the inequality, it is not necessary to purposely write LHS=.... like we have to do if it was prove. Make sure you understand why.
    • We can see that the minimum value is \left(x+\frac{1}{2}\right)^{2}{\color{Red}-\frac{1}{4}+p}, so this value must be positive
    • \therefore -\frac{1}{4}+p>0
    • \therefore p>\frac{1}{4}

Exercise 3

1) Prove that \,12x^{2}+6x+11>0 for all x \in \mathbb{R}.

    • \begin{align}
12x^{2}+6x+11 & = 12\left(x^{2}+\frac{1}{2}x+\frac{11}{12}\right)\\
& = 12\left[\left(x+\frac{1}{4}\right)^{2}-\frac{1}{16}+\frac{11}{12}\right]\\
& = 12\left(x+\frac{1}{4}\right)^{2}+\frac{41}{4}\\
\end{align}
    • \begin{align}
& \mbox{Since }\left(x+\frac{1}{4}\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore 12\left(x+\frac{1}{4}\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore 12\left(x+\frac{1}{4}\right)^{2}+\frac{41}{4}>0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore 12x^{2}+6x+11>0 \mbox{ for all }x \in \mathbb{R}\frac{\qquad}{}\mbox{Proven}\\
\end{align}

2) Prove that \left(x+3\right)\left(1-x\right)-7<0 for all x \in \mathbb{R}.

    • \begin{align}
\left(x+3\right)\left(1-x\right)-7 & =3-2x-x^{2}-7 \\
& =-x^{2}-2x-4 \\
& =-\left(x^{2}+2x+4\right)\\
& =-\left[\left(x+1\right)^{2}-1+4\right]\\
& =-\left(x+1\right)^{2}-3
\end{align}
    • \begin{align}
& \mbox{Since }\left(x+1\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore -\left(x+1\right)^{2}\leq 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore -\left(x+1\right)^{2}-3 < 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore \left(x+3\right)\left(1-x\right)-7<0 \mbox{ for all }x \in \mathbb{R}\frac{\qquad}{}\mbox{Proven}\\
\end{align}

3) Prove that \,2x^{2}+7>6x for all x \in \mathbb{R}.

    • \begin{align}
2x^{2}-6x+7 &= 2\left(x^{2}-3x+\frac{7}{2}\right)\\
&= 2\left[\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}+\frac{7}{2}\right]\\
&= 2\left(x-\frac{3}{2}\right)^{2}+\frac{5}{2}\\
\end{align}
    • \begin{align}
& \mbox{Since }\left(x-\frac{3}{2}\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore 2\left(x-\frac{3}{2}\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore 2\left(x-\frac{3}{2}\right)^{2}+\frac{5}{2} > 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore 2x^{2}-6x+7 > 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore 2x^{2}+7>6x \mbox{ for all }x \in \mathbb{R}\frac{\qquad}{}\mbox{Proven}\\
\end{align}

4) Prove that \,x^{2}+x+1 is always positive for all x \in \mathbb{R}.

    • \begin{align}
x^{2}+x+1 & =\left(x+\frac{1}{2}\right)^{2}-\frac{1}{4}+1\\
& =\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}\\
\end{align}
    • \begin{align}
& \mbox{Since }\left(x+\frac{1}{2}\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore \left(x+\frac{1}{2}\right)^{2}+\frac{3}{4} > 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore x^{2}+x+1 > 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore x^{2}+x+1 \mbox{ is always positive for all }x \in \mathbb{R}\frac{\qquad}{}\mbox{Proven}\\
\end{align}

5) Find range of \,p if \,x^{2}-6x+p>0 for all x \in \mathbb{R}. \,p>9

    • \begin{align}
x^{2}-6x+p & =\left(x-3\right)^{2}-9+p
\end{align}
    • \begin{align}
& \mbox{If }x^{2}-6x+p>0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore \left(x-3\right)^{2}-9+p>0 \mbox{ for all }x \in \mathbb{R}\\
& \mbox{Since }\left(x-3\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore -9+p>0\\
& \therefore p>9\\
\end{align}

6) Given \,p-3x-6x^{2}<0 for all x \in \mathbb{R}. Find range of \,p. p<-\frac{3}{8}

    • \begin{align}
p-3x-6x^{2} & =-6\left(x^{2}+\frac{1}{2}x-\frac{p}{6}\right)\\
&= -6\left[\left(x+\frac{1}{4}\right)^{2}-\frac{1}{16}-\frac{p}{16}\right]\\
&= -6\left(x+\frac{1}{4}\right)^{2}+p+\frac{3}{8}\\
\end{align}
    • \begin{align}
&\mbox{Given }p-3x-6x^{2}<0  \mbox{ for all }x \in \mathbb{R}\\
&\therefore -6\left(x+\frac{1}{4}\right)^{2}+p+\frac{3}{8}<0  \mbox{ for all }x \in \mathbb{R}\\
&\mbox{Since }-6\left(x+\frac{1}{4}\right)^{2}\leq 0 \mbox{ for all }x \in \mathbb{R}\\
&\therefore p+\frac{3}{8}<0 \\
&\therefore p<-\frac{3}{8}
\end{align}

7) Show that if \,x^{2}+px+q>0 for all x \in \mathbb{R}, \,p^{2}>4q.

  • \begin{align}
& \mbox{If }x^{2}+px+q>0  \mbox{ for all }x \in \mathbb{R}\\
&\left(x+\frac{p}{2}\right)^{2}-\frac{p^{2}}{4}+q>0\\
&\mbox{Since }\left(x+\frac{p}{2}\right)^{2}\geq 0 \mbox{ for all }x \in \mathbb{R}\\
& \therefore -\frac{p^{2}}{4}+q>0 \\
& q>\frac{p^{2}}{4} \\
& p^{2}>4q\frac{\qquad}{}\mbox{Shown}
\end{align}
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