2018 H2 Mathematics Paper 1 Question 4

Equations and Inequalities

Answers

(i)

x=−1−3,  {x=- 1 - \sqrt{3},\;}−1+3,  {- 1 + \sqrt{3},\;}−1,  {- 1,\;}0{0}

(ii)

−1−3<x<−1{- 1 - \sqrt{3} < x < - 1} or 0<x<−1+3{0 < x < - 1 + \sqrt{3}}

Full solutions

(i)

∣2x2+3x−2∣=2−x|2 x^2 + 3 x - 2|=2 - x
2x2+3x−2=2−x or2x2+3x−2=−(2−x)\begin{align} && \quad 2 x^2 + 3 x - 2 &= 2 - x \quad \textrm{ or} \\ && \quad 2 x^2 + 3 x - 2 &= -(2 - x) \\ \end{align}
From (1),{(1),}
2x2+4x−4=0x2+2x−2=0\begin{align*} 2 x^2 + 4 x - 4 &= 0 \\ x^2 + 2 x - 2 &= 0 \\ \end{align*}
x=−2±22−4(−2)2(1)=−2±122=−1±3\begin{align*} x &= \frac{-2\pm\sqrt{2^2-4(-2)}}{2(1)} \\ &= \frac{-2\pm\sqrt{12}}{2} \\ &= -1 \pm \sqrt{3} \end{align*}
From (2),{(2),}
2x2+2x=02x(x+1)=0\begin{align*} 2 x^2 + 2 x &= 0 \\ 2x(x+1) &= 0 \end{align*}
x=0   or   x=−1x=0 \; \textrm{ or } \; x = -1
Hence the exact roots of the equation are:
x=−1−3,  ■x=−1+3,  ■x=−1,  ■x=0  ■\begin{align*} &x = - 1 - \sqrt{3}, \; \blacksquare \\ &x = - 1 + \sqrt{3}, \; \blacksquare \\ &x = - 1, \; \blacksquare \\ &x = 0 \; \blacksquare \end{align*}

(ii)

From the graphs, solution to ∣2x2+3x−2∣<2−x:{|2 x^2 + 3 x - 2| < 2 - x: }
−1−3<x<−1  ■or   0<x<−1+3  ■\begin{gather*} - 1 - \sqrt{3} < x < - 1 \; \blacksquare \\ \textrm{or } \; 0 < x < - 1 + \sqrt{3} \; \blacksquare \end{gather*}