Title: 15. Quadratic equations
Course: Demonstrating STACK (v4.4.7)

Solve the following quadratic equation by factorising: \[ [value] = 0. \]

Write this quadratic in factored form: \(=0\).

Solve this equation, giving your answer as a set \(\left(\mbox{e.g. }\left\{x_1 , x_2\right\}\right)\):

\([value] \in \)
Feedback: There are multiple ways of solving quadratic equations but if you can spot how to factorise the quadratic then this method is by far the fastest and simplest. In this example we can factorise the quadratic to give: \[ [value] = 0. \] Now, notice, that for the above equation to be equal to zero, the factors must be made to equal zero. So we find that: \[ [value] = 0, \] or: \[ [value] = 0. \] By rearranging these expressions we find that: \([value] = [value]\) or \([value] = [value]\). Therefore our solution set is: \[ [value] \in \left\{ [value] , [value] \right\}. \]

Solve the following quadratic equation by completing the square: \[ [value] = 0. \]
Give the completed square form: \(=0.\)

Hence solve the quadratic, giving your answer as a set \(\left(\mbox e.g.\ \left\{x_1 , x_2\right\} \right)\):

\([value] \in \)
Feedback: When solving a more complicated quadratic a better option, than factorising, is completing the square (The method by which we remove the [value] term). The general rule (when the \(x^2\) coeffecient is one) is: \[ x^2+ax+b = \left(x+\frac{a}{2} \right)^2-\frac{a^2}{4}+c. \] Once you have removed the \(x\) term you can rearrange the formula to find \((x+\frac{a}{2})^2\). Then you simply need to square root and rearrange to find \(x\). In this example: \[ [value] = 0, \] we complete the square: \[ [value] = 0. \] Rearranging gives: \[ [value] = [value]. \] Square root (remember to take plus and minus): \[ [value] = \pm [value]. \] Finally: \[ [value] = [value] \pm [value]. \] Therefeore, our solution set is: \[ [value] \in \left\{ [value] , [value] \right\}. \]

Give the following quadratic in completed square form i.e. in the form \(\left( x+a \right)^2+b.\)
\[ [value] \]

Solve the quadratic equation \([value]=0\). Give your answer as a set \(\left(\mbox e.g.\ \left\{x_1 , x_2\right\} \right)\):
Feedback: When solving a more complicated quadratic a better option, than factorising, is completing the square (The method by which we remove the [value] term). The general rule is: \[ ax^2+bx+c = a\left(x^2+\frac{b}{a}x+\frac{c}{a}x \right) = a\left(\left(x+\frac{b}{2a} \right)^2-\frac{b^2}{4a^2}+\frac{c}{a} = a\left(x+\frac{b}{2a} \right)^2-\frac{b^2}{4a}+c \right). \] Once you have removed the \(x\) term you can rearrange the formula to find \((x+\frac{b}{2a})^2\). Then you simply need to square root and rearrange to find \(x\). In this example: \[ [value] = 0, \] we extract the [value] coefficient, \([value]\): \[ [value] = 0. \] Then we complete the square: \[ [value] = 0. \] Rearranging gives: \[ [value] = [value]. \] Square root (remember to take plus and minus): \[ [value] = \pm [value]. \] Finally: \[ [value] = [value] \pm [value]. \] Therefeore, our solution set is: \[ [value] \]

Solve the following quadratic equation by using the quadratic formula: \[ [value] = 0. \]
Give your answer as a set \(\left(e.g. \left\{x_1 , x_2\right\} \right)\):

\([value] \in \)
Feedback: A good method for solving any quadratic is the 'quadratic formula'. Solving a general quadratic by completing the square gives a general formula for \(x\) in terms of the coefficients of the quadratic. It is best to simply learn the quadratic formula: For \(ax^2+bx+c = 0 \): \[ x = \frac{ -b\pm\sqrt{b^2-4ac} }{2a}. \] However it can be derived, as mentioned above, by completing the square: \[ ax^2+bx+c = 0 \] Factor out the \(x^2\) coefficient \(a\): \[ a\left(x^2+\frac{b}{a}x+\frac{c}{a}x \right) = 0. \] Complete the square: \[ a\left(\left(x+\frac{b}{2a} \right)^2-\frac{b^2}{4a^2}+\frac{c}{a}\right) = 0. \] Multiply back through by \(a\): \[ a\left(x+\frac{b}{2a} \right)^2-\frac{b^2}{4a}+c = 0. \] Rearranging gives: \[ \left(x+\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}-\frac{c}{a}. \] Simplify RHS into a single fraction: \[ \left(x+\frac{b}{2a}\right)^2 = \frac{b^2-4ac}{4a^2}. \] Square root (remember to take plus and minus): \[ \left(x+\frac{b}{2a}\right) = \pm \frac{\sqrt{b^2-4ac}}{2a}. \] Finally: \[ x = \frac {-b \pm \sqrt{b^2-4ac} }{2a}. \] In this exmaple: \(a = [value]\), \(b = [value]\), \(c = [value]\), so subbing these into the quadratic formula gives: \[ [value] = \frac{[value] \pm \sqrt{[value]-([value])}}[value] = [value] \pm [value]. \] Therefore \( [value] = [value]\) or \( [value] = [value]\) and so the solution set is: \[ [value] \in [value]. \]

Solve the following quadratic equation by using the quadratic formula: \[ [value] = 0. \]
Give your answer as a set \(\left(\mbox {e.g.} \ \left\{x_1 , x_2\right\} \right)\):

\([value] \in \)
Feedback: A good method for solving any quadratic is the 'quadratic formula'. Solving a general quadratic by completing the square gives a general formula for \(x\) in terms of the coefficients of the quadratic. It is best to simply learn the quadratic formula: For \(ax^2+bx+c = 0 \): \[ x = \frac{ -b\pm\sqrt{b^2-4ac} }{2a}. \] However it can be derived, as mentioned above, by completing the square: \[ ax^2+bx+c = 0 \] Factor out the \(x^2\) coefficient \(a\): \[ a\left(x^2+\frac{b}{a}x+\frac{c}{a}x \right) = 0. \] Complete the square: \[ a\left(\left(x+\frac{b}{2a} \right)^2-\frac{b^2}{4a^2}+\frac{c}{a}\right) = 0. \] Multiply back through by \(a\): \[ a\left(x+\frac{b}{2a} \right)^2-\frac{b^2}{4a}+c = 0. \] Rearranging gives: \[ \left(x+\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2}-\frac{c}{a}. \] Simplify RHS into a single fraction: \[ \left(x+\frac{b}{2a}\right)^2 = \frac{b^2-4ac}{4a^2}. \] Square root (remember to take plus and minus): \[ \left(x+\frac{b}{2a}\right) = \pm \frac{\sqrt{b^2-4ac}}{2a}. \] Finally: \[ x = \frac {-b \pm \sqrt{b^2-4ac} }{2a}. \] In this exmaple: \(a = [value]\), \(b = [value]\), \(c = [value]\), so subbing these into the quadratic formula gives: \[ [value] = \frac{[value] \pm \sqrt{[value]-([value])}}[value] = [value] \pm [value]. \] Therefore \( [value] = [value]\) or \( [value] = [value]\) and so the solution set is: \[ [value] \in [value]. \]

Find the values of \(x\) and \(y\), from the following expressions. Give your answer as a set containing the pairs of points which satisfy the equation, e.g.  \(\left( \left\{[x_1, y_1],[x_2, y_2] \right\}\right)\):
\[ \left. \begin{array}{rcl} [value] &=&[value] \\ [value] &=&[value] \end{array} \right\} \]
\([x,y] \in \)
Feedback: By rearranging the linear equation, \([value] = [value] \) we obtain: \[ y = [value], \] which we can substitute into the quadratic to give: \[ [value] = [value]. \] This can be rearranged into the more familiar quadratic form: \[ [value] = 0. \] Solving this gives \( x = 3\) or \(x = -1 \). By substituting these back into the linear equation you can find the corresponding values of y. ( \(y = -1\) and \(y = 3\) respectively). Therefore, the solution set is: \[ (x,y) \in \left\{ [3,-1] , [-1,3] \right\}. \]

Re-write the following in "completed square" form: \[ [value]. \]

(i.e. the form \(\left([value]+a\right)^2+b\)).
Feedback: Our aim is to 'destroy' the \(x\) term. We do this by considering the general quadratic: \[ x^2+2ax+b. \] Consider, \[\left(x-a \right)^2 = x^2-2ax+a^2. \] Therefore we can see that: \[ x^2+2ax+b = \left( x-a \right)^2-a^2+b. \] So in this example we have [value]. Therefore we complete the square: \[ [value] = [value].\]

Solve the following quadratic equation by completing the square: \[ [value] = 0. \]
Give the completed square form: \(=0\).

Give your answer as a set \(\left(\mbox {e.g.}\ \left\{x_1 , x_2\right\} \right)\):

\([value] \in \)
Feedback: When solving a more complicated quadratic; a better option, than factorising, is completing the square (The method by which we remove the [value] term). The general rule is: \[ ax^2+bx+c = a\left(x^2+\frac{b}{a}x+\frac{c}{a}x \right) = a\left(\left(x+\frac{b}{2a} \right)^2-\frac{b^2}{4a^2}+\frac{c}{a} = a\left(x+\frac{b}{2a} \right)^2-\frac{b^2}{4a}+c \right). \] Once you have removed the \(x\) term you can rearrange the formula to find \(\left( x+\frac{b}{2a} \right)^2\). Then you simply need to square root and rearrange to find \(x\). In this example: \[ [value] = 0, \] we extract the [value] coefficient, \([value]\): \[ [value] = 0. \] Then we complete the square: \[ [value] = 0. \] Multiply through by \([value]\): \[ [value] = 0. \] Rearranging gives: \[ [value] = [value]. \] Square root (remember to take plus and minus): \[ [value] = \pm [value]. \] Finally: \[ [value] = [value] \pm [value]. \] Therefeore, our solution set is: \[ [value] \in \left\{ [value] , [value] \right\}. \]