1.1 Solve for $$ x $$ in each of the following: 1.1.1 $$ x(2 x+1)=0 $$ 1.1.2 $$ 5 x^{2}+2 x-6=0 $$ (correct to TWO decimal places) 1.1.3 $$ 2 x^{2}-2 \geq 3 x $$ 1.1.4 $$ \sqrt{2 x+5}-\frac{3}{\sqrt{2 x+5}}=-2 $$ 1.2 Solve for $$ x $$ and $$ y $$ simultancously: $$ y+x=2 \text { and } x^{2}+3 x y+8=0 $$ 1.3 The roots of the equation $$ f(x)=0 $$ are $$ x=\frac{4 \pm \sqrt{16-4 m(-m+5}}{2 m} $$ Determine the values of $$ m $$ for which the roots will be non-real 1.4 Show that the maximum value of $$ \sqrt{-x^{2}+4 x+12} $$ is 4 .

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**1.1.1** We have $$ x(2x+1)=0. $$ Using the zero‐product property, $$ x=0 \quad \text{or} \quad 2x+1=0. $$ Solving $$2x+1=0$$: $$ 2x=-1\quad \Longrightarrow\quad x=-\frac{1}{2}. $$ --- **1.1.2** We solve the quadratic $$ 5x^2+2x-6=0. $$ Using the quadratic formula $$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}, $$ with $$a=5$$, $$b=2$$, and $$c=-6$$: $$ \Delta = b^2-4ac=2^2-4(5)(-6)=4+120=124. $$ Thus, $$ x=\frac{-2\pm\sqrt{124}}{10}=\frac{-2\pm2\sqrt{31}}{10}=\frac{-1\pm\sqrt{31}}{5}. $$ Calculating approximate values (to two decimal places): $$ \sqrt{31}\approx5.57, $$ so $$ x\approx \frac{-1+5.57}{5}\approx\frac{4.57}{5}\approx0.91,\quad \text{and}\quad x\approx \frac{-1-5.57}{5}\approx\frac{-6.57}{5}\approx-1.31. $$ --- **1.1.3** We solve the inequality $$ 2x^2-2\geq3x. $$ Rearrange it by subtracting $$3x$$ from both sides: $$ 2x^2-3x-2\geq 0. $$ Find the roots of $$ 2x^2-3x-2=0. $$ Compute the discriminant: $$ \Delta=(-3)^2-4\cdot2\cdot(-2)=9+16=25. $$ Then, $$ x=\frac{3\pm 5}{4}. $$ This gives the roots $$ x=\frac{8}{4}=2\quad \text{and}\quad x=\frac{-2}{4}=-\frac{1}{2}. $$ Since the quadratic opens upward (leading coefficient $$2>0$$), the solution to the inequality is $$ x\leq -\frac{1}{2}\quad \text{or}\quad x\geq2. $$ --- **1.1.4** We solve $$ \sqrt{2x+5}-\frac{3}{\sqrt{2x+5}}=-2. $$ Let $$ u=\sqrt{2x+5}, $$ noting that $$u>0$$. Then the equation becomes $$ u-\frac{3}{u}=-2. $$ Multiply through by $$u$$: $$ u^2-3=-2u. $$ Rearrange to obtain a quadratic in $$u$$: $$ u^2+2u-3=0. $$ Factor or use the quadratic formula. Factoring, we look for two numbers whose product is $$-3$$ and sum is $$2$$: $$ (u+3)(u-1)=0. $$ The possible solutions are $$ u=-3\quad \text{or}\quad u=1. $$ Since $$u>0$$, we discard $$u=-3$$. Hence, $$ u=1. $$ Returning to the original variable: $$ \sqrt{2x+5}=1\quad \Longrightarrow\quad 2x+5=1, $$ so $$ 2x=-4\quad \Longrightarrow\quad x=-2. $$ --- **1.2** We solve for $$x$$ and $$y$$ simultaneously: $$ \begin{cases} x+y=2, \ x^2+3xy+8=0. \end{cases} $$ Express $$y$$ in terms of $$x$$ from the first equation: $$ y=2-x. $$ Substitute into the second equation: $$ x^2+3x(2-x)+8=0. $$ Expand: $$ x^2+6x-3x^2+8=0. $$ Combine like terms: $$ -2x^2+6x+8=0. $$ Multiply both sides by $$-1$$: $$ 2x^2-6x-8=0. $$ Divide by 2: $$ x^2-3x-4=0. $$ Factor the quadratic: $$ (x-4)(x+1)=0. $$ Thus, $$ x=4\quad \text{or}\quad x=-1. $$ Now substitute back for $$y$$: - If $$x=4$$, then $$y=2-4=-2$$. - If $$x=-1$$, then $$y=2-(-1)=3$$. --- **1.3** The roots of the equation are given by $$ x=\frac{4\pm\sqrt{16-4m(-m+5)}}{2m}. $$ First, simplify the expression under the square root: $$ 16-4m(-m+5)=16+4m(m-5). $$ For the roots to be non-real, the expression under the square root must be negative: $$ 16+4m(m-5)<0. $$ Divide both sides by 4: $$ 4+m(m-5)<0. $$ Expand $$m(m-5)$$: $$ m^2-5m+4<0. $$ Factor the quadratic: $$ m^2-5m+4=(m-1)(m-4)<0. $$ This inequality holds when \[ 1**1.1.1** $$ x = 0 $$ or $$ x = -\frac{1}{2} $$ **1.1.2** $$ x \approx 0.91 $$ or $$ x \approx -1.31 $$ **1.1.3** $$ x \leq -\frac{1}{2} $$ or $$ x \geq 2 $$ **1.1.4** $$ x = -2 $$ **1.2** $$ x = 4, y = -2 $$ $$ x = -1, y = 3 $$ **1.3** $$ m $$ values for non-real roots: $$ 1 < m < 4 $$ **1.4** Maximum value is 4

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