QUESTION 1 Solve for $$ x: $$ $$ \begin{array}{l}1.1 .1 \quad(2 x-3)(x+7)=0 \ 1.1 .2 \quad 7 x^{2}+3 x-2=0 ext { (leave your answer correct to TWO decimal places) } \ 1.1 .3 \quad \sqrt{x-1}+3=x \ \ 1.1 .4 \quad x^{2}3(x+6) \ \ ext { Solve for } x ext { and } y ext { simultaneously: } \ 2 y+x=1 \ x^{2}+y^{2}+3 x y+y=0 \ 1.3 \ ext { If } f(x)=0 ext { has roots } \quad x=\frac{-5 \pm \sqrt{3-12 k^{2}}}{4} ext {, for which values of } k ext { will the roots be } \ ext { equal? }\end{array} $$

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QUESTION 1 Solve for $$ x: $$ $$ \begin{array}{l}1.1 .1 \quad(2 x-3)(x+7)=0 \ 1.1 .2 \quad 7 x^{2}+3 x-2=0 	ext { (leave your answer correct to TWO decimal places) } \ 1.1 .3 \quad \sqrt{x-1}+3=x \ \ 1.1 .4 \quad x^{2}>3(x+6) \ \ 	ext { Solve for } x 	ext { and } y 	ext { simultaneously: } \ 2 y+x=1 \ x^{2}+y^{2}+3 x y+y=0 \ 1.3 \ 	ext { If } f(x)=0 	ext { has roots } \quad x=\frac{-5 \pm \sqrt{3-12 k^{2}}}{4} 	ext {, for which values of } k 	ext { will the roots be } \ 	ext { equal? }\end{array} $$

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Solve the equation by following steps: - step0: Solve for $$x$$: $$3=x$$ - step1: Swap the sides: $$x=3$$ Solve the equation $$ 3-12k^{2}=0 $$. Solve the quadratic equation by following steps: - step0: Solve using square roots: $$3-12k^{2}=0$$ - step1: Add or subtract both sides: $$-12k^{2}=-3$$ - step2: Divide both sides: $$\frac{-12k^{2}}{-12}=\frac{-3}{-12}$$ - step3: Divide the numbers: $$k^{2}=\frac{1}{4}$$ - step4: Simplify the expression: $$k=\pm \sqrt{\frac{1}{4}}$$ - step5: Simplify the expression: $$k=\pm \frac{1}{2}$$ - step6: Separate into possible cases: $$\begin{align}&k=\frac{1}{2}\&k=-\frac{1}{2}\end{align}$$ - step7: Rewrite: $$k_{1}=-\frac{1}{2},k_{2}=\frac{1}{2}$$ Solve the equation $$ 7x^{2}+3x-2=0 $$. Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: $$7x^{2}+3x-2=0$$ - step1: Solve using the quadratic formula: $$x=\frac{-3\pm \sqrt{3^{2}-4\times 7\left(-2\right)}}{2\times 7}$$ - step2: Simplify the expression: $$x=\frac{-3\pm \sqrt{3^{2}-4\times 7\left(-2\right)}}{14}$$ - step3: Simplify the expression: $$x=\frac{-3\pm \sqrt{65}}{14}$$ - step4: Separate into possible cases: $$\begin{align}&x=\frac{-3+\sqrt{65}}{14}\&x=\frac{-3-\sqrt{65}}{14}\end{align}$$ - step5: Rewrite the fraction: $$\begin{align}&x=\frac{-3+\sqrt{65}}{14}\&x=-\frac{3+\sqrt{65}}{14}\end{align}$$ - step6: Rewrite: $$x_{1}=-\frac{3+\sqrt{65}}{14},x_{2}=\frac{-3+\sqrt{65}}{14}$$ Solve the equation $$ x^{2}-3(x+6)>0 $$. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: $$x^{2}-3\left(x+6\right)>0$$ - step1: Rearrange the terms: $$x^{2}-3x-18>0$$ - step2: Rewrite the expression: $$x^{2}-3x-18=0$$ - step3: Factor the expression: $$\left(x-6\right)\left(x+3\right)=0$$ - step4: Separate into possible cases: $$\begin{align}&x-6=0\&x+3=0\end{align}$$ - step5: Solve the equation: $$\begin{align}&x=6\&x=-3\end{align}$$ - step6: Determine the test intervals: $$\begin{align}&x<-3\&-36\end{align}$$ - step7: Choose a value: $$\begin{align}&x_{1}=-4\&x_{2}=2\&x_{3}=7\end{align}$$ - step8: Test the chosen value: $$\begin{align}&x<-3\textrm{ }\textrm{is the solution}\&-36\textrm{ }\textrm{is the solution}\end{align}$$ - step9: The final solution is $$x \in \left(-\infty,-3\right)\cup \left(6,+\infty\right):$$ $$x \in \left(-\infty,-3\right)\cup \left(6,+\infty\right)$$ Solve the system of equations $$ 2y+x=1;x^{2}+y^{2}+3xy+y=0 $$. Solve the system of equations by following steps: - step0: Solve using the substitution method: $$\left\{ \begin{array}{l}2y+x=1\x^{2}+y^{2}+3xy+y=0\end{array}\right.$$ - step1: Solve the equation: $$\left\{ \begin{array}{l}x=1-2y\x^{2}+y^{2}+3xy+y=0\end{array}\right.$$ - step2: Substitute the value of $$x:$$ $$\left(1-2y\right)^{2}+y^{2}+3\left(1-2y\right)y+y=0$$ - step3: Simplify: $$1-y^{2}=0$$ - step4: Move the constant to the right side: $$-y^{2}=0-1$$ - step5: Remove 0: $$-y^{2}=-1$$ - step6: Change the signs: $$y^{2}=1$$ - step7: Simplify the expression: $$y=\pm \sqrt{1}$$ - step8: Simplify: $$y=\pm 1$$ - step9: Separate into possible cases: $$y=1\cup y=-1$$ - step10: Rearrange the terms: $$\left\{ \begin{array}{l}x=1-2y\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=1-2y\y=-1\end{array}\right.$$ - step11: Calculate: $$\left\{ \begin{array}{l}x=-1\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=3\y=-1\end{array}\right.$$ - step12: Check the solution: $$\left\{ \begin{array}{l}x=-1\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=3\y=-1\end{array}\right.$$ - step13: Rewrite: $$\left(x,y\right) = \left(-1,1\right)\cup \left(x,y\right) = \left(3,-1\right)$$ Let's solve each part of the problem step by step. ### 1.1.1 Solve for $$ x $$: $$ (2x-3)(x+7)=0 $$ To find the values of $$ x $$, we set each factor to zero: 1. $$ 2x - 3 = 0 $$ leads to $$ x = \frac{3}{2} $$ 2. $$ x + 7 = 0 $$ leads to $$ x = -7 $$ Thus, the solutions are: $$ x = \frac{3}{2}, \quad x = -7 $$ ### 1.1.2 Solve for $$ x $$: $$ 7x^{2}+3x-2=0 $$ Using the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$: - Here, $$ a = 7 $$, $$ b = 3 $$, $$ c = -2 $$. Calculating the discriminant: $$ b^2 - 4ac = 3^2 - 4 \cdot 7 \cdot (-2) = 9 + 56 = 65 $$ Now applying the quadratic formula: $$ x = \frac{-3 \pm \sqrt{65}}{14} $$ Calculating the approximate values: $$ x_1 \approx -0.79, \quad x_2 \approx 0.36 $$ ### 1.1.3 Solve for $$ x $$: $$ \sqrt{x-1}+3=x $$ Rearranging gives: $$ \sqrt{x-1} = x - 3 $$ Squaring both sides: $$ x - 1 = (x - 3)^2 $$ Expanding: $$ x - 1 = x^2 - 6x + 9 $$ Rearranging: $$ 0 = x^2 - 7x + 10 $$ Factoring: $$ 0 = (x - 5)(x - 2) $$ Thus, the solutions are: $$ x = 5, \quad x = 2 $$ ### 1.1.4 Solve for $$ x $$: $$ x^{2} > 3(x+6) $$ Rearranging gives: $$ x^{2} - 3x - 18 > 0 $$ Factoring: $$ (x - 6)(x + 3) > 0 $$ The critical points are $$ x = 6 $$ and $$ x = -3 $$. Testing intervals: - For $$ x < -3 $$: Positive - For $$ -3 < x < 6 $$: Negative - For $$ x > 6 $$: Positive Thus, the solution is: $$ x \in (-\infty, -3) \cup (6, +\infty) $$ ### 2. Solve for $$ x $$ and $$ y $$ simultaneously: 1. $$ 2y + x = 1 $$ (Equation 1) 2. $$ x^{2} + y^{2} + 3xy + y = 0 $$ (Equation 2) From Equation 1, we can express $$ x $$ in terms of $$ y $$: $$ x = 1 - 2y $$ Substituting into Equation 2: $$ (1 - 2y)^{2} + y^{2} + 3(1 - 2y)y + y = 0 $$ Expanding and simplifying leads to: $$ y^2 - 3y + 1 = 0 $$ Using the quadratic formula: $$ y = \frac{3 \pm \sqrt{5}}{2} $$ Substituting back to find $$ x $$: 1. For $$ y = \frac{3 + \sqrt{5}}{2} $$, $$ x = 1 - 2\left(\frac{3 + \sqrt{5}}{2}\right) = -2 - \sqrt{5} $$ 2. For $$ y = \frac{3 - \sqrt{5}}{2} $$, $$ x = 1 - 2\left(\frac{3 - \sqrt{5}}{2}\right) = -2 + \sqrt{5} $$ Thus, the solutions are: $$ (x, y) = (-2 - \sqrt{5}, \frac{3 + \sqrt{5}}{2}) \quad \text{and} \quad (x, y) = (-2 + \sqrt{5}, \frac{3 - \sqrt{5}}{2}) $$ ### 3. Roots of $$ f(x) = 0 $$: The roots are given by: $$ x = \frac{-5 \pm \sqrt{3 - 12k^{2}}}{4} $$ For the roots to be equal, the discriminant must be zero: $$ 3 - 12k^{2} = 0 $$ Solving for $$ k $$: $$ 12k^{2} = 3 \implies k^{2} = \frac{1}{4} \implies k = \pm \frac{1}{2} $$ ### Summary of Solutions: 1. $$ x = \frac{3}{2}, -7 $$ 2. $$ x \approx -0.79, 0.36 $$ 3. $$ x = 5, 2 $$ 4. $$ x \in (-\infty, -3) \cup (6, +\infty) $$ 5. $$ (x, y) = (-1, 1) $$ or $$ (3, -1) $$ 6. $$ k = \pm \frac{1}{2} $$ for equal roots. Solutions: 1. $$ x = 1.5, -7 $$ 2. $$ x \approx -0.79, 0.36 $$ 3. $$ x = 5, 2 $$ 4. $$ x < -3 $$ or $$ x > 6 $$ 5. $$ (x, y) = (-1, 1) $$ or $$ (3, -1) $$ 6. $$ k = 0.5 $$ or $$ k = -0.5 $$

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