QUESTION I DEE KON 170 1.1 Solve for $$ x $$. $$ 1.11 \quad(2 x-3)(x+7)=0 $$ $$ 1.127 x^{2}+3 x-2=0 $$ (leave your axaser correct to TWO decimal places) $$ 1.13 \sqrt{x-1}+3=x $$ $$ 114 \quad x^{2}3(x+6) $$ 12 Solve for $$ x $$ and $$ y $$ simulianoously: $$ \begin{array}{l} 2 y+x=1 \ x^{2}+y^{2}+3 x y+y=0 \end{array} $$ 13 If $$ f(x)=0 $$ has roo's $$ x=\frac{-5 \pm \sqrt{3-12 k^{2}}}{4} $$, for which values of $$ k $$ will the rcots be equal?

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QUESTION I DEE KON 170 1.1 Solve for $$ x $$. $$ 1.11 \quad(2 x-3)(x+7)=0 $$ $$ 1.127 x^{2}+3 x-2=0 $$ (leave your axaser correct to TWO decimal places) $$ 1.13 \sqrt{x-1}+3=x $$ $$ 114 \quad x^{2}>3(x+6) $$ 12 Solve for $$ x $$ and $$ y $$ simulianoously: $$ \begin{array}{l} 2 y+x=1 \ x^{2}+y^{2}+3 x y+y=0 \end{array} $$ 13 If $$ f(x)=0 $$ has roo's $$ x=\frac{-5 \pm \sqrt{3-12 k^{2}}}{4} $$, for which values of $$ k $$ will the rcots be equal?

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Let's solve each part of the problem step by step. ### 1.11 Solve for $$ x $$ in $$ (2x-3)(x+7)=0 $$ To solve this equation, we set each factor equal to zero: 1. $$ 2x - 3 = 0 $$ 2. $$ x + 7 = 0 $$ **Step 1: Solve $$ 2x - 3 = 0 $$** $$ 2x = 3 \implies x = \frac{3}{2} $$ **Step 2: Solve $$ x + 7 = 0 $$** $$ x = -7 $$ Thus, the solutions are $$ x = \frac{3}{2} $$ and $$ x = -7 $$. ### 1.12 Solve for $$ x $$ in $$ x^{2}+3x-2=0 $$ We will use the quadratic formula: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ where $$ a = 1, b = 3, c = -2 $$. **Step 1: Calculate the discriminant** $$ b^2 - 4ac = 3^2 - 4(1)(-2) = 9 + 8 = 17 $$ **Step 2: Apply the quadratic formula** $$ x = \frac{-3 \pm \sqrt{17}}{2} $$ Calculating the two roots: 1. $$ x_1 = \frac{-3 + \sqrt{17}}{2} $$ 2. $$ x_2 = \frac{-3 - \sqrt{17}}{2} $$ Now, let's calculate these values to two decimal places. ### 1.13 Solve for $$ x $$ in $$ \sqrt{x-1}+3=x $$ **Step 1: Isolate the square root** $$ \sqrt{x-1} = x - 3 $$ **Step 2: Square both sides** $$ x - 1 = (x - 3)^2 $$ **Step 3: Expand and rearrange** $$ x - 1 = x^2 - 6x + 9 $$ $$ 0 = x^2 - 7x + 10 $$ **Step 4: Factor the quadratic** $$ 0 = (x - 5)(x - 2) $$ Thus, $$ x = 5 $$ or $$ x = 2 $$. **Step 5: Check for extraneous solutions** 1. For $$ x = 5 $$: $$ \sqrt{5-1} + 3 = 2 + 3 = 5 \quad \text{(valid)} $$ 2. For $$ x = 2 $$: $$ \sqrt{2-1} + 3 = 1 + 3 = 4 \quad \text{(not valid)} $$ Thus, the only solution is $$ x = 5 $$. ### 1.14 Solve the inequality $$ x^{2}>3(x+6) $$ **Step 1: Rearrange the inequality** $$ x^2 - 3x - 18 > 0 $$ **Step 2: Factor the quadratic** $$ (x - 6)(x + 3) > 0 $$ **Step 3: Determine the intervals** The critical points are $$ x = 6 $$ and $$ x = -3 $$. We test intervals: 1. $$ (-\infty, -3) $$: Choose $$ x = -4 $$ → $$ (-)(-) > 0 $$ (valid) 2. $$ (-3, 6) $$: Choose $$ x = 0 $$ → $$ (-)(+) < 0 $$ (not valid) 3. $$ (6, \infty) $$: Choose $$ x = 7 $$ → $$ (+)(+) > 0 $$ (valid) Thus, the solution is $$ x < -3 $$ or $$ x > 6 $$. ### 1.12 Solve for $$ x $$ and $$ y $$ simultaneously: $$ \begin{array}{l} 2y + x = 1 \ x^2 + y^2 + 3xy + y = 0 \end{array} $$ **Step 1: Solve the first equation for $$ x $$** $$ x = 1 - 2y $$ **Step 2: Substitute into the second equation** $$ (1 - 2y)^2 + y^2 + 3(1 - 2y)y + y = 0 $$ Expanding this: $$ (1 - 4y + 4y^2) + y^2 + (3y - 6y^2) + y = 0 $$ $$ 1 - 4y + 4y^2 + y^2 + 3y - 6y^2 + y = 0 $$ $$ 1 - 4y + 4y^2 + 3y - 6y^2 + y = 0 $$ $$ 1 - 4y + 3y - 2y^2 = 0 $$ $$ -2y^2 + 0y + 1 = 0 \implies 2y^2 = 1 \implies y^2 = \frac{1}{2} \implies y = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2} $$ **Step 3: Find corresponding $$ x $$ values** For $$ y = \frac{\sqrt{2}}{2} $$: $$ x = 1 - 2\left(\frac{\sqrt{2}}{2}\right) = 1 - \sqrt{2} $$ For $$ y = -\frac{\sqrt{2}}{2} $$: $$ x = 1 - 2\left(-\frac{\sqrt{2}}{2}\right) = 1 + \sqrt{2} $$ Thus, the solutions are: 1. $$ (1 - \sqrt{2}, \frac{\sqrt{2}}{2}) $$ 2. $$ (1 + \sqrt{2}, -\frac{\sqrt{2}}{2}) $$ ### 1.13 If $$ f(x)=0 $$ has roots $$ x=\frac{-5 \pm \sqrt{3-12 k^{2}}}{4} $$, for which values of $$ k $$ will the roots be equal? For the roots to be equal, the discriminant must be zero: $$ 3 - 12k^2 = 0 $$ **Step 1: Solve for $$ k $$** $$ 12k^2 = 3 \implies k^2 = To solve the equation $$ 12k^2 = 3 $$, divide both sides by 12: \[ k^2 = \frac{3}{12} = \frac{1}{4} $$ Taking the square root of both sides: $$ k = \pm \frac{1}{2} $$ Thus, the roots are equal when $$ k = \frac{1}{2} $$ or $$ k = -\frac{1}{2} $$.

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