Solve for x: 1.1 x(5x+2)=0; 1.2 x(2x−3)=4 (correct to TWO decimal places); 1.3 x²-x-6 >= 6; 1.4 2^(x+2) + 2^(x+2) = 42; 1.5 sqrt(4x−11) = 2−x

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a = 3$, $b = -8$, $c = 4$.

$$x = \frac{8 + 4}{6} = \frac{12}{6} = 2,$$ $$x = \frac{8 - 4}{6} = \frac{4}{6} = \frac{2}{3}.$$

x = 2,\ x = \dfrac{2}{3}

Question 1 (b)

$$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

where $a = 4$, $b = -8$, $c = -3$.

$$y = \frac{8 \pm 4\sqrt{7}}{2 \cdot 4} = \frac{8 \pm 4\sqrt{7}}{8} = \frac{2 \pm \sqrt{7}}{2}.$$

y = \dfrac{2 + \sqrt{7}}{2},\ y = \dfrac{2 - \sqrt{7}}{2}

Question 2 (a)

$$x^2 + (x^2 - 2x + 1) = 9$$ $$2x^2 - 2x + 1 = 9$$ $$2x^2 - 2x - 8 = 0.$$

$$y = x - 1 = \frac{1 \pm \sqrt{17}}{2} - 1 = \frac{1 \pm \sqrt{17} - 2}{2} = \frac{-1 \pm \sqrt{17}}{2}.$$

\left( \dfrac{1 + \sqrt{17}}{2},\ \dfrac{-1 + \sqrt{17}}{2} \right),\ \left( \dfrac{1 - \sqrt{17}}{2},\ \dfrac{-1 - \sqrt{17}}{2} \right)

Question 2 (c)

$$x = 1,\ x = 2.$$

The graph is a parabola opening upwards with vertex at $\left( \frac{3}{2}, -\frac{1}{4} \right)$ and two distinct real roots.

x=1,\ x=2