Unit 2: Quadratic Equations – Exercise 2.1

(i) \(3x - 1 = 2x^{2}\)
\[ \begin{aligned} 3x - 1 &= 2x^{2} \\ 0 &= 2x^{2} - 3x + 1 \\ \mathbf{2x^{2} - 3x + 1} &= \mathbf{0} \end{aligned} \]

(ii) \(2x(x + 1) = 4(2x + 3)\)
\[ \begin{aligned} 2x(x+1) &= 4(2x+3) \\ 2x^{2} + 2x &= 8x + 12 \\ 2x^{2} + 2x - 8x -12 &= 0 \\ 2x^{2} -6x -12 &= 0 \\ 2(x^{2} - 3x -6) &= 0 \\ \Rightarrow x^{2} - 3x - 6 &= 0 \end{aligned} \]

(iii) \(2x^{2} - 4x = 4x + 7\)
\[ \begin{aligned} 2x^{2} - 4x &= 4x + 7 \\ 2x^{2} - 4x - 4x - 7 &= 0 \\ \mathbf{2x^{2} - 8x - 7} &= \mathbf{0} \end{aligned} \]

(iv) \(4(3x - 2) = 9x^{2}\)
\[ \begin{aligned} 4(3x - 2) &= 9x^{2} \\ 12x - 8 &= 9x^{2} \\ 0 &= 9x^{2} - 12x + 8 \\ \mathbf{9x^{2} - 12x + 8} &= \mathbf{0} \end{aligned} \]

(v) \(2x + \frac{1}{x} = 5 - \frac{1}{x},\; x \neq 0\)
\[ \begin{aligned} 2x + \frac{1}{x} &= 5 - \frac{1}{x} \\ \text{Multiplying by } x:\quad 2x^{2} + 1 &= 5x - 1 \\ 2x^{2} + 1 - 5x + 1 &= 0 \\ \mathbf{2x^{2} - 5x + 2} &= \mathbf{0} \end{aligned} \]

(vi) \(\frac{6x + 6}{20 - x} = \frac{1}{x},\; x \neq 0,20\)
\[ \begin{aligned} \frac{6x+6}{20-x} &= \frac{1}{x} \\ x(6x+6) &= 20 - x \\ 6x^{2}+6x &= 20 - x \\ 6x^{2} + 6x -20 + x &= 0 \\ \mathbf{6x^{2} + 7x - 20} &= \mathbf{0} \end{aligned} \]

(i) \(x^{2} - x - 6 = 0\)
\[ \begin{aligned} x^{2} - x - 6 &= 0 \\ x^{2} - 3x + 2x - 6 &= 0 \\ x(x-3) + 2(x-3) &= 0 \\ (x-3)(x+2) &= 0 \\ \Rightarrow x-3=0 \;\; &\text{or}\;\; x+2=0 \\ \mathbf{x=3} \;&\text{or}\; \mathbf{x=-2} \end{aligned} \] \( \text{S.S} = \{3,\ -2\} \)

(ii) \(x^{2} + 3x - 28 = 0\)
\[ \begin{aligned} x^{2}+3x-28 &= 0 \\ x^{2}+7x-4x-28 &= 0 \\ x(x+7)-4(x+7) &= 0 \\ (x+7)(x-4) &= 0 \\ \Rightarrow x=-7 \;&\text{or}\; x=4 \end{aligned} \] \( \text{S.S} = \{-7,\ 4\} \)

(iii) \(6x^{2}+13x-5=0\)
\[ \begin{aligned} 6x^{2}+13x-5 &= 0 \\ 6x^{2}+15x-2x-5 &= 0 \\ 3x(2x+5)-1(2x+5) &= 0 \\ (2x+5)(3x-1) &= 0 \\ \Rightarrow x=-\frac{5}{2} \;&\text{or}\; x=\frac{1}{3} \end{aligned} \] \( \text{S.S} = \left\{-\frac{5}{2},\ \frac{1}{3}\right\} \)

(iv) \(x^{2} - \frac{3}{2}x = \frac{9}{2}\)
\[ \begin{aligned} x^{2} - \frac{3}{2}x &= \frac{9}{2} \\ \text{Multiply by }2:\; 2x^{2} - 3x &= 9 \\ 2x^{2} - 3x - 9 &= 0 \\ 2x^{2} -6x +3x -9 &= 0 \\ 2x(x-3)+3(x-3) &= 0 \\ (x-3)(2x+3) &= 0 \\ \Rightarrow x=3 \;&\text{or}\; x=-\frac{3}{2} \end{aligned} \] \( \text{S.S} = \left\{3,\ -\frac{3}{2}\right\} \)

(v) \(\frac{3x-8}{x-2} = \frac{5x-2}{x+5}\)
\[ \begin{aligned} \frac{3x-8}{x-2} &= \frac{5x-2}{x+5} \\ (3x-8)(x+5) &= (5x-2)(x-2) \\ 3x^{2}+15x-8x-40 &= 5x^{2}-10x-2x+4 \\ 3x^{2}+7x-40 &= 5x^{2}-12x+4 \\ 0 &= 2x^{2} -19x +44 \\ 2x^{2}-11x-8x+44 &= 0 \\ x(2x-11)-4(2x-11)&=0 \\ (2x-11)(x-4)&=0 \\ \Rightarrow x=\frac{11}{2} \;&\text{or}\; x=4 \end{aligned} \] \( \text{S.S} = \left\{\frac{11}{2},\ 4\right\} \)

(vi) \(\frac{1}{x-1} - \frac{1}{x+3} = \frac{1}{35}\)
\[ \begin{aligned} \frac{1}{x-1} - \frac{1}{x+3} &= \frac{1}{35} \\ \frac{(x+3)-(x-1)}{(x-1)(x+3)} &= \frac{1}{35} \\ \frac{4}{x^{2}+2x-3} &= \frac{1}{35} \\ 140 &= x^{2}+2x-3 \\ 0 &= x^{2}+2x-143 \\ x^{2}+13x-11x-143 &=0 \\ x(x+13)-11(x+13)&=0 \\ (x+13)(x-11)&=0 \\ \Rightarrow x=-13 \;&\text{or}\; x=11 \end{aligned} \] \( \text{S.S} = \{-13,\ 11\} \)

(i) \(2x^{2}+5x+2=0\)
\[ \begin{aligned} 2x^{2}+5x+2 &=0 \\ x^{2}+\frac{5}{2}x+1&=0 \\ x^{2}+\frac{5x}{2} &= -1 \\ \text{Add } \left(\frac{5}{4}\right)^{2}:&\quad x^{2}+\frac{5x}{2}+\frac{25}{16} = -1+\frac{25}{16} \\ \left(x+\frac{5}{4}\right)^{2} &= \frac{9}{16} \\ x+\frac{5}{4} &= \pm \frac{3}{4} \\ x &= -\frac{5}{4}\pm\frac{3}{4} \\ \Rightarrow x=-\frac{1}{2} \;&\text{or}\; x=-2 \end{aligned} \]

(ii) \(x^{2}+x=42\)
\[ \begin{aligned} x^{2}+x &= 42 \\ \text{Add } \left(\frac{1}{2}\right)^{2}:&\quad x^{2}+x+\frac{1}{4}=42+\frac{1}{4} \\ \left(x+\frac{1}{2}\right)^{2}&=\frac{169}{4} \\ x+\frac{1}{2}&=\pm\frac{13}{2} \\ x &= -\frac{1}{2}\pm\frac{13}{2} \\ \Rightarrow x=6 \;&\text{or}\; x=-7 \end{aligned} \]

(iii) \(12x^{2}+7x=12\)
\[ \begin{aligned} 12x^{2}+7x &=12 \\ x^{2}+\frac{7x}{12}&=1 \\ \text{Add } \left(\frac{7}{24}\right)^{2}:&\quad x^{2}+\frac{7x}{12}+\frac{49}{576}=1+\frac{49}{576} \\ \left(x+\frac{7}{24}\right)^{2}&=\frac{625}{576} \\ x+\frac{7}{24}&=\pm\frac{25}{24} \\ x&= -\frac{7}{24}\pm\frac{25}{24} \\ \Rightarrow x=\frac{3}{4} \;&\text{or}\; x=-\frac{4}{3} \end{aligned} \]

(iv) \(\frac{x+3}{2x-7}=\frac{2x-1}{x-3}\)
\[ \begin{aligned} (x+3)(x-3)&=(2x-1)(2x-7) \\ x^{2}-9 &= 4x^{2}-16x+7 \\ 0&=3x^{2}-16x+16 \\ x^{2}-\frac{16}{3}x &= -\frac{16}{3} \\ \text{Add } \left(\frac{8}{3}\right)^{2}:&\quad \left(x-\frac{8}{3}\right)^{2} = -\frac{16}{3}+\frac{64}{9} = \frac{16}{9} \\ x-\frac{8}{3}&=\pm\frac{4}{3} \\ x=\frac{8}{3}\pm\frac{4}{3} &\Rightarrow x=4 \;\text{or}\; x=\frac{4}{3} \end{aligned} \]

(v) \(\frac{1}{1+x}-\frac{1}{3-x}=\frac{6}{35}\)
\[ \begin{aligned} \frac{(3-x)-(1+x)}{(1+x)(3-x)}&=\frac{6}{35} \\ \frac{2-2x}{3+2x-x^{2}}&=\frac{6}{35} \\ 70-70x &= 18+12x-6x^{2} \\ 6x^{2}-82x+52&=0 \\ x^{2}-\frac{41}{3}x &= -\frac{26}{3} \\ \text{Add } \left(\frac{41}{6}\right)^{2}:&\quad \left(x-\frac{41}{6}\right)^{2}= \frac{1369}{36} \\ x-\frac{41}{6}&=\pm\frac{37}{6} \;\Rightarrow\; x= \frac{41\pm37}{6} \\ x=13 \;&\text{or}\; x=\frac{2}{3} \end{aligned} \]

(vi) \(\frac{3x-1}{4x+7}=1-\frac{6}{x+7}\)
\[ \begin{aligned} \frac{3x-1}{4x+7}&=\frac{x+1}{x+7} \\ (3x-1)(x+7)&=(x+1)(4x+7) \\ 3x^{2}+20x-7&=4x^{2}+11x+7 \\ 0&=x^{2}-9x+14 \\ x^{2}-9x&=-14 \\ \text{Add } \left(\frac{9}{2}\right)^{2}:&\quad \left(x-\frac{9}{2}\right)^{2}= \frac{25}{4} \\ x-\frac{9}{2}&=\pm\frac{5}{2} \\ x=7 \;&\text{or}\; x=2 \end{aligned} \]

(i) \(2x^{2}-5x+3=0\)
\[ \begin{aligned} a=2,\; b=-5,\; c=3 &: \quad x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} = \frac{5\pm\sqrt{25-24}}{4} = \frac{5\pm1}{4} \\ \Rightarrow x=\frac{3}{2} \;&\text{or}\; x=1 \end{aligned} \]

(ii) \(2x^{2}-7x-15=0\)
\[ \begin{aligned} x&=\frac{7\pm\sqrt{49+120}}{4}=\frac{7\pm\sqrt{169}}{4}=\frac{7\pm13}{4} \\ \Rightarrow x=5 \;&\text{or}\; x=-\frac{3}{2} \end{aligned} \]

(iii) \(2x^{2}+7x=15 \;\Rightarrow\; 2x^{2}+7x-15=0\)
\[ \begin{aligned} x&=\frac{-7\pm\sqrt{49+120}}{4}=\frac{-7\pm13}{4} \\ \Rightarrow x=\frac{3}{2} \;&\text{or}\; x=-5 \end{aligned} \]

(iv) \(x^{2}+11=7x \;\Rightarrow\; x^{2}-7x+11=0\)
\[ \begin{aligned} x&=\frac{7\pm\sqrt{49-44}}{2}=\frac{7\pm\sqrt{5}}{2} \\ \Rightarrow &\; \mathbf{x=\frac{7+\sqrt{5}}{2}} \quad\text{or}\quad \mathbf{x=\frac{7-\sqrt{5}}{2}} \end{aligned} \]

(v) \(\frac{x+4}{x-4}+\frac{x-2}{x-3}=6\frac{1}{3}=\frac{19}{3}\)
\[ \begin{aligned} \frac{(x-3)(x+4)+(x-4)(x-2)}{(x-4)(x-3)}&=\frac{19}{3} \\ \frac{2x^{2}-5x-4}{x^{2}-7x+12}&=\frac{19}{3} \\ 6x^{2}-15x-12&=19x^{2}-133x+228 \\ 0&=13x^{2}-118x+240 \\ x&=\frac{118\pm\sqrt{13924-12480}}{26}=\frac{118\pm\sqrt{1444}}{26}=\frac{118\pm38}{26} \\ \Rightarrow x=6 \;&\text{or}\; x=\frac{40}{13} \end{aligned} \]

(vi) \(\frac{3x-3}{x+1}=\frac{2x-1}{x-1}\)
\[ \begin{aligned} (3x-3)(x-1)&=(2x-1)(x+1) \\ 3x^{2}-6x+3 &= 2x^{2}+x-1 \\ x^{2}-7x+4&=0 \\ x&=\frac{7\pm\sqrt{49-16}}{2}=\frac{7\pm\sqrt{33}}{2} \end{aligned} \]

(i) \(x^{2}-3x-18=0\)
Vertex \(x=1.5\), table points: graph cuts x-axis at \(x=-3\) and \(x=6\).
\[ \text{Solution Set} = \{-3,\ 6\} \]

(ii) \(x^{2}-5x-14=0\)
Vertex \(x=2.5\), graph cuts x-axis at \(x=-2\) and \(x=7\).
\[ \text{Solution Set} = \{-2,\ 7\} \]

(iii) \(2x^{2}+13x+6=0\)
Vertex \(x=-3.25\), x-intercepts: \(x=-6\) and \(x=-0.5\).
\[ \text{Solution Set} = \{-6,\ -0.5\} \]

(iv) \(4x^{2}+12x-27=0\)
Vertex \(x=-1.5\), intercepts at \(x=-4.5\) and \(x=1.5\).
\[ \text{Solution Set} = \{-4.5,\ 1.5\} \]