QUESTION 1 1.1 Solve for $$ \boldsymbol{x} $$ : 1.1.1 $$ \quad 5 x^{2}-\mathbf{4 x}=0 $$ 1.1.2 $$ \quad 3 x^{2}=2(x+2) $$ (Leave your answer correct to TWO decimal places) 1.1.3 $$ 4+5 x6 x^{2} $$ 1.1.4 $$ \quad \mathbf{5}^{-x} \cdot \mathbf{5}^{x-2}=\frac{25^{2 x}}{5} $$ 1.2 If $$ \mathbf{2} $$ and $$ \mathbf{- 4} $$ are the roots of the equation $$ x^{2}+b x+c=\mathbf{0} $$ Determine the values of $$ \boldsymbol{b} $$ and $$ \boldsymbol{c} $$. 1.3 Solve simultaneously for x and y : $$ \begin{array}{l} x-3 y=1 \ x^{2}-2 x y+9 y^{2}=17 \end{array} $$ 1.4 For which values of $$ \boldsymbol{p} $$ will the equation $$ \boldsymbol{x}^{2}+\boldsymbol{x}=\boldsymbol{p} $$ have no real roots?

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QUESTION 1 1.1 Solve for $$ \boldsymbol{x} $$ : 1.1.1 $$ \quad 5 x^{2}-\mathbf{4 x}=0 $$ 1.1.2 $$ \quad 3 x^{2}=2(x+2) $$ (Leave your answer correct to TWO decimal places) 1.1.3 $$ 4+5 x>6 x^{2} $$ 1.1.4 $$ \quad \mathbf{5}^{-x} \cdot \mathbf{5}^{x-2}=\frac{25^{2 x}}{5} $$ 1.2 If $$ \mathbf{2} $$ and $$ \mathbf{- 4} $$ are the roots of the equation $$ x^{2}+b x+c=\mathbf{0} $$ Determine the values of $$ \boldsymbol{b} $$ and $$ \boldsymbol{c} $$. 1.3 Solve simultaneously for x and y : $$ \begin{array}{l} x-3 y=1 \ x^{2}-2 x y+9 y^{2}=17 \end{array} $$ 1.4 For which values of $$ \boldsymbol{p} $$ will the equation $$ \boldsymbol{x}^{2}+\boldsymbol{x}=\boldsymbol{p} $$ have no real roots?

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Solve the quadratic equation by following steps: - step0: Solve by factoring: $$5x^{2}-4x=0$$ - step1: Factor the expression: $$x\left(5x-4\right)=0$$ - step2: Separate into possible cases: $$\begin{align}&5x-4=0\&x=0\end{align}$$ - step3: Solve the equation: $$\begin{align}&x=\frac{4}{5}\&x=0\end{align}$$ - step4: Rewrite: $$x_{1}=0,x_{2}=\frac{4}{5}$$ Solve the equation $$ 4+5x>6x^{2} $$. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: $$4+5x>6x^{2}$$ - step1: Move the expression to the left side: $$4+5x-6x^{2}>0$$ - step2: Rewrite the expression: $$4+5x-6x^{2}=0$$ - step3: Factor the expression: $$\left(1+2x\right)\left(4-3x\right)=0$$ - step4: Separate into possible cases: $$\begin{align}&1+2x=0\&4-3x=0\end{align}$$ - step5: Solve the equation: $$\begin{align}&x=-\frac{1}{2}\&x=\frac{4}{3}\end{align}$$ - step6: Determine the test intervals: $$\begin{align}&x<-\frac{1}{2}\&-\frac{1}{2}\frac{4}{3}\end{align}$$ - step7: Choose a value: $$\begin{align}&x_{1}=-2\&x_{2}=0\&x_{3}=3\end{align}$$ - step8: Test the chosen value: $$\begin{align}&x<-\frac{1}{2}\textrm{ }\textrm{is not a solution}\&-\frac{1}{2}\frac{4}{3}\textrm{ }\textrm{is not a solution}\end{align}$$ - step9: The final solution is $$-\frac{1}{2} 6x^{2} $$ Rearranging gives: $$ 6x^{2} - 5x - 4 < 0 $$ Now we can find the roots of the equation $$ 6x^{2} - 5x - 4 = 0 $$ using the quadratic formula: Calculating the discriminant: $$ D = (-5)^{2} - 4 \cdot 6 \cdot (-4) = 25 + 96 = 121 $$ Now applying the quadratic formula: $$ x = \frac{5 \pm \sqrt{121}}{12} = \frac{5 \pm 11}{12} $$ Calculating the two roots: 1. $$ x_{1} = \frac{16}{12} = \frac{4}{3} $$ 2. $$ x_{2} = \frac{-6}{12} = -\frac{1}{2} $$ The inequality $$ 6x^{2} - 5x - 4 < 0 $$ holds between the roots: $$ -\frac{1}{2} < x < \frac{4}{3} $$ #### 1.1.4 $$ 5^{-x} \cdot 5^{x-2} = \frac{25^{2x}}{5} $$ Rewriting $$ 25^{2x} $$ as $$ (5^{2})^{2x} = 5^{4x} $$: $$ 5^{-x} \cdot 5^{x-2} = \frac{5^{4x}}{5} $$ This simplifies to: $$ 5^{-x + x - 2} = 5^{4x - 1} $$ Thus: $$ 5^{-2} = 5^{4x - 1} $$ Equating the exponents gives: $$ -2 = 4x - 1 \implies 4x = -1 \implies x = -\frac{1}{4} $$ ### 1.2 Determine $$ b $$ and $$ c $$ for roots $$ 2 $$ and $$ -4 $$ Using Vieta's formulas, the sum of the roots $$ r_1 + r_2 = -b $$ and the product $$ r_1 \cdot r_2 = c $$: 1. $$ 2 + (-4) = -b \implies -2 = -b \implies b = 2 $$ 2. $$ 2 \cdot (-4) = c \implies c = -8 $$ Thus, the values are: $$ b = 2, \quad c = -8 $$ ### 1.3 Solve simultaneously for $$ x $$ and $$ y $$: The system of equations is: 1. $$ x - 3y = 1 $$ 2. $$ x^{2} - 2xy + 9y^{2} = 17 $$ Substituting $$ x = 3y + 1 $$ into the second equation: $$ (3y + 1)^{2} - 2(3y + 1)y + 9y^{2} = 17 $$ Expanding and simplifying leads to: $$ 9y^{2} + 6y + 1 - 6y - 2y + 9y^{2} = 17 $$ Combining like terms gives: $$ 18y^{2} - 16 = 0 \implies 18y^{2} = 16 \implies y^{2} = \frac{8}{9} \implies y = \pm \frac{2\sqrt{2}}{3} $$ Substituting back to find $$ x $$: 1. For $$ y = \frac{2\sqrt{2}}{3} $$: $$ x = 3\left(\frac{2\sqrt{2}}{3}\right) + 1 = 2\sqrt{2} + 1 $$ 2. For $$ y = -\frac{2\sqrt{2}}{3} $$: $$ x = 3\left(-\frac{2\sqrt{2}}{3}\right) + 1 = -2\sqrt{2} + 1 $$ Thus, the solutions are: $$ (x,y) = (2\sqrt{2} + 1, \frac{2\sqrt{2}}{3}) \quad \text{and} \quad (x,y) = (-2\sqrt{2} + 1, -\frac{2\sqrt{2}}{3}) $$ ### 1.4 For which values of $$ p $$ will $$ x^{2} + x = p $$ have no real roots? The discriminant must be less than zero: \[ D = 1^{2} - 4 ### 1.1 Solutions for $$ x $$: 1.1.1 $$ x = 0 $$ or $$ x = \frac{4}{5} $$ 1.1.2 $$ x \approx -0.87 $$ or $$ x \approx 1.54 $$ 1.1.3 $$ -\frac{1}{2} < x < \frac{4}{3} $$ 1.1.4 $$ x = -\frac{1}{4} $$ ### 1.2 Values of $$ b $$ and $$ c $$: $$ b = 2 $$, $$ c = -8 $$ ### 1.3 Simultaneous Solutions for $$ x $$ and $$ y $$: $$ (x, y) = (2\sqrt{2} + 1, \frac{2\sqrt{2}}{3}) $$ and $$ (x, y) = (-2\sqrt{2} + 1, -\frac{2\sqrt{2}}{3}) $$ ### 1.4 Values of $$ p $$ for No Real Roots: $$ p < -\frac{1}{4} $$

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