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2.2 Completing the square
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2.4 Substitution
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It is not always possible to solve a quadratic equation by factorisation and it can take a long time to complete the square. The method of completing the square provides a way to derive a formula that can be used to solve any quadratic equation. The quadratic formula provides an easy and fast way to solve quadratic equations.
Consider the standard form of the quadratic equation \(ax^2 + bx + c = 0\). Divide both sides by \(a\) \((a \ne 0)\) to get
\[x^2 + \frac{bx}{a} + \frac{c}{a} = 0\]Now using the method of completing the square, we must halve the coefficient of \(x\) and square it. We then add and subtract \(\left( \dfrac{b}{2a} \right)^2\) so that the equation remains true.
\begin{align*} x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} -\frac{b^2}{4a^2} + \frac{c}{a} &= 0 \\ \left( x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} \right) -\frac{b^2}{4a^2} + \frac{c}{a} &= 0 \\ \left( x + \frac{b}{2a} \right)^2 -\frac{b^2-4ac}{4a^2} &= 0 \end{align*}
We add the constant to both sides and take the square root of both sides of the equation, being careful to include a positive and negative answer.
\begin{align*} \left( x + \frac{b}{2a} \right)^2 &= \frac{b^2-4ac}{4a^2} \\ \sqrt{\left( x + \frac{b}{2a} \right)^2} &= \pm \sqrt{\frac{b^2-4ac}{4a^2}} \\ x + \frac{b}{2a} &= \pm \dfrac{\sqrt{b^2-4ac}}{2a} \\ x &= - \dfrac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a} \\ x &= \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \end{align*}
Therefore, for any quadratic equation \(ax^2+bx+c = 0\) we can determine two roots \[x = \dfrac{-b + \sqrt{b^2-4ac}}{2a} \text{ or } x = \dfrac{-b - \sqrt{b^2-4ac}}{2a}\]
It is important to notice that the expression \({b}^{2}-4ac\) must be greater than or equal to zero for the roots of the quadratic to be real. If the expression under the square root sign is less than zero, then the roots are non-real (imaginary).
Solve for \(x\) and leave your answer in simplest surd form: \(2x^2+3x=7\)
The expression cannot be factorised, so the general quadratic formula must be used.
\[2x^2+3x-7=0\]
\[a = 2; \qquad b = 3; \qquad c = -7\]
Always write down the formula first and then substitute the values of \(a\), \(b\) and \(c\).
\begin{align*} x &= \frac{-b±\sqrt{{b}^{2}-4ac}}{2a} \\ &= \frac{-\left(3\right)±\sqrt{{\left(3\right)}^{2}-4\left(2\right)\left(-7\right)}}{2\left(2\right)} \\ &= \frac{-3±\sqrt{65}}{4} \end{align*}The two roots are \(x=\dfrac{-3+\sqrt{65}}{4}\) or \(x=\dfrac{-3-\sqrt{65}}{4}\).
Find the roots of the function \(f(x)= x^2-5x+8\).
To determine the roots of \(f(x)\), we let \(x^2-5x+8 = 0\).
The expression cannot be factorised, so the general quadratic formula must be used.
\[a = 1; \qquad b = -5; \qquad c = 8\]
There are no real roots for \(f(x)= x^2-5x+8\) since the expression under the square root is negative (\(\sqrt{-7}\) is not a real number). This means that the graph of the quadratic function has no \(x\)-intercepts; the entire graph lies above the \(x\)-axis.
Solve the following using the quadratic formula.
\(3{t}^{2}+t-4=0\)
\begin{align*} 3t^2 + t - 4 &= 0\\ t &= \dfrac{-(1) \pm \sqrt{1^2 - 4(3)(-4)}}{2(3)}\\ & = \dfrac{-1 \pm \sqrt{1+48}}{6}\\ &=\dfrac{-1 \pm \sqrt{49}}{6}\\ &=\dfrac{-1 \pm 7}{6}\\ t = \dfrac{-1+7}{6}= \frac{6}{6}=1 &\text{ or } t = \dfrac{-1-7}{6}=\frac{-8}{6} = \frac{-4}{3} \end{align*}
\({x}^{2}-5x-3=0\)
\begin{align*} x^2 - 5x - 3 &= 0\\ x &= \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-3)}}{2(1)}\\ &= \dfrac{5 \pm \sqrt{25 + 12}}{2}\\ &=\dfrac{5 \pm \sqrt{37}}{2}\\ \text{therefore } x =\dfrac{5 + \sqrt{37}}{2} &\text{ or } x =\dfrac{5 - \sqrt{37}}{2} \end{align*}
\(2{t}^{2}+6t+5=0\)
\begin{align*} 2t^2 + 6t + 5 &= 0\\ t&=\dfrac{-6 \pm \sqrt{(6)^2 - 4(2)(5)}}{2(2)}\\ &= \dfrac{-6 \pm \sqrt{36 - 40}}{4}\\ &=\dfrac{-6 \pm \sqrt{-4}}{4}\\ \text{No real solution} \end{align*}
\(2p(2p+1)=2\)
\begin{align*} 4p^2 + 2p - 2 &= 0\\ 2p^2 + p - 1 &= 0\\ p &= \dfrac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)}\\ &= \dfrac{-1 \pm \sqrt{1+8}}{4}\\ &= \dfrac{-1 \pm \sqrt{9}}{4} \\ &= \dfrac{-1 \pm 3}{4} \\ \text{therefore } p =\dfrac{-1 + 3}{4} = \frac{1}{2} &\text{ or } p =\dfrac{-1 - 3}{4} = -1 \end{align*}
\(-3{t}^{2}+5t-8=0\)
\begin{align*} -3t^2 + 5t -8 &= 0\\ t&= \dfrac{-5 \pm \sqrt{5^2 -4(-3)(-8)}}{2(-3)}\\ &= \dfrac{-5 \pm \sqrt{25-96}}{-6}\\ &=\dfrac{-5 \pm \sqrt{-71}}{-6}\\ \text{No real solution} \end{align*}
\(5{t}^{2}+3t-3=0\)
\begin{align*} 5t^2 + 3t - 3 &= 0\\ t &= \dfrac{-3 \pm \sqrt{3^2 - 4(5)(-3)}}{2(5)}\\ &= \dfrac{-3 \pm \sqrt{9+60}}{10}\\ &=\dfrac{-3 \pm \sqrt{69}}{10}\\ \text{therefore } t =\dfrac{-3 + \sqrt{69}}{10} &\text{ or } t =\dfrac{-3 - \sqrt{69}}{10} \end{align*}
\({t}^{2}-4t+2=0\)
\(9(k^2 - 1) = 7k\)
\begin{align*} 9k^2 - 7k - 9 &= 0\\ k &= \dfrac{-(-7) \pm \sqrt{(-7)^2 - 4(9)(-9)}}{2(9)}\\ &= \dfrac{7 \pm \sqrt{49 + 324}}{18}\\ &=\dfrac{7 \pm \sqrt{373}}{18}\\ k = \dfrac{7 + \sqrt{373}}{18} &\text{ or } k =\dfrac{7 - \sqrt{373}}{18} \end{align*}
\(3f -2 = -2f^2\)
\begin{align*} 2f^2 + 3f - 2 &= 0\\ f &= \dfrac{-3 \pm \sqrt{3^2-4(2)(-2)}}{2(2)}\\ &= \dfrac{-3 \pm \sqrt{9+16}}{4}\\ &= \dfrac{-3 \pm \sqrt{25}}{4}\\ &= \dfrac{-3 \pm 5}{4}\\ \text{therefore } f =\dfrac{-3 + 5}{4} = \frac{1}{2}&\text{ or } f =\dfrac{-3 - 5}{4} = -2 \end{align*}
\({t}^{2}+t+1=0\)
\begin{align*} t^2 + t +1 &= 0\\ t&= \dfrac{-1 \pm \sqrt{1-4(1)(1)}}{2(1)}\\ &= \dfrac{-1 \pm \sqrt{-3}}{2}\\ \text{No real solution} \end{align*}
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2.2 Completing the square
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2.4 Substitution
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