QUESTION I Torm \{ 2 ons 1.1 Solvo formi \( 1.114 x^{2}=16 \) \( 1.125 x^{2}-8 x+1=0 \quad \) (Correct \( t o \mathrm{ONB} \) decimal) \( 1.133 x-6=9-\sqrt{2 x+1} \) (6) \( 1.1 .4(x+7)(x-1) \leq 48 \) \( 1.15 x^{\frac{2}{3}}-2 x^{\frac{1}{3}}=15 \) (5) 12 Solve for \( x \) and \( y \) simultaneously in the equations below: \[ 3 y-x=11 \text { and } x^{2}-x y=y^{2}+1 \] 1.3 Determine the value(s) of \( t \) for which \( -2 x^{2}+4 t x+t=0 \) will have non-real roots. (3) [28]

Solve the equation by following steps: - step0: Solve for \(x\): \(x-6=9\) - step1: Move the constant to the right side: \(x=9+6\) - step2: Add the numbers: \(x=15\) Solve the equation \( x^{2}=16 \). Solve the quadratic equation by following steps: - step0: Solve using square roots: \(x^{2}=16\) - step1: Simplify the expression: \(x=\pm \sqrt{16}\) - step2: Simplify the expression: \(x=\pm 4\) - step3: Separate into possible cases: \(\begin{align}&x=4\\&x=-4\end{align}\) - step4: Rewrite: \(x_{1}=-4,x_{2}=4\) Solve the equation \( x^{2}-8 x+1=0 \). Solve the quadratic equation by following steps: - step0: Solve using the quadratic formula: \(x^{2}-8x+1=0\) - step1: Solve using the quadratic formula: \(x=\frac{8\pm \sqrt{\left(-8\right)^{2}-4}}{2}\) - step2: Simplify the expression: \(x=\frac{8\pm \sqrt{60}}{2}\) - step3: Simplify the expression: \(x=\frac{8\pm 2\sqrt{15}}{2}\) - step4: Separate into possible cases: \(\begin{align}&x=\frac{8+2\sqrt{15}}{2}\\&x=\frac{8-2\sqrt{15}}{2}\end{align}\) - step5: Simplify the expression: \(\begin{align}&x=4+\sqrt{15}\\&x=\frac{8-2\sqrt{15}}{2}\end{align}\) - step6: Simplify the expression: \(\begin{align}&x=4+\sqrt{15}\\&x=4-\sqrt{15}\end{align}\) - step7: Rewrite: \(x_{1}=4-\sqrt{15},x_{2}=4+\sqrt{15}\) Solve the equation \( x^{\frac{2}{3}}-2 x^{\frac{1}{3}}=15 \). Solve the equation by following steps: - step0: Solve for \(x\): \(x^{\frac{2}{3}}-2x^{\frac{1}{3}}=15\) - step1: Move the expression to the left side: \(x^{\frac{2}{3}}-2x^{\frac{1}{3}}-15=0\) - step2: Solve using substitution: \(t^{2}-2t-15=0\) - step3: Factor the expression: \(\left(t-5\right)\left(t+3\right)=0\) - step4: Separate into possible cases: \(\begin{align}&t-5=0\\&t+3=0\end{align}\) - step5: Solve the equation: \(\begin{align}&t=5\\&t=-3\end{align}\) - step6: Substitute back: \(\begin{align}&x^{\frac{1}{3}}=5\\&x^{\frac{1}{3}}=-3\end{align}\) - step7: Solve the equation for \(x:\) \(\begin{align}&x=125\\&x=-27\end{align}\) - step8: Rewrite: \(x_{1}=-27,x_{2}=125\) Solve the equation \( (x+7)(x-1) \leq 48 \). Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(\left(x+7\right)\left(x-1\right)\leq 48\) - step1: Move the expression to the left side: \(\left(x+7\right)\left(x-1\right)-48\leq 0\) - step2: Subtract the terms: \(x^{2}+6x-55\leq 0\) - step3: Rewrite the expression: \(x^{2}+6x-55=0\) - step4: Factor the expression: \(\left(x-5\right)\left(x+11\right)=0\) - step5: Separate into possible cases: \(\begin{align}&x-5=0\\&x+11=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=5\\&x=-11\end{align}\) - step7: Determine the test intervals: \(\begin{align}&x<-11\\&-115\end{align}\) - step8: Choose a value: \(\begin{align}&x_{1}=-12\\&x_{2}=-3\\&x_{3}=6\end{align}\) - step9: Test the chosen value: \(\begin{align}&x<-11\textrm{ }\textrm{is not a solution}\\&-115\textrm{ }\textrm{is not a solution}\end{align}\) - step10: Include the critical value: \(\begin{align}&-11\leq x\leq 5\textrm{ }\textrm{is the solution}\end{align}\) - step11: The final solution is \(-11\leq x\leq 5:\) \(-11\leq x\leq 5\) Solve the system of equations \( 3 y-x=11;x^{2}-x y=y^{2}+1 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3y-x=11\\x^{2}-xy=y^{2}+1\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=-11+3y\\x^{2}-xy=y^{2}+1\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(-11+3y\right)^{2}-\left(-11+3y\right)y=y^{2}+1\) - step3: Simplify: \(121-55y+6y^{2}=y^{2}+1\) - step4: Move the expression to the left side: \(121-55y+6y^{2}-\left(y^{2}+1\right)=0\) - step5: Calculate: \(120-55y+5y^{2}=0\) - step6: Factor the expression: \(5\left(8-y\right)\left(3-y\right)=0\) - step7: Divide the terms: \(\left(8-y\right)\left(3-y\right)=0\) - step8: Separate into possible cases: \(\begin{align}&8-y=0\\&3-y=0\end{align}\) - step9: Solve the equation: \(\begin{align}&y=8\\&y=3\end{align}\) - step10: Calculate: \(y=8\cup y=3\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=-11+3y\\y=8\end{array}\right.\cup \left\{ \begin{array}{l}x=-11+3y\\y=3\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=13\\y=8\end{array}\right.\cup \left\{ \begin{array}{l}x=-2\\y=3\end{array}\right.\) - step13: Calculate: \(\left\{ \begin{array}{l}x=-2\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=13\\y=8\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=-2\\y=3\end{array}\right.\cup \left\{ \begin{array}{l}x=13\\y=8\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(-2,3\right)\cup \left(x,y\right) = \left(13,8\right)\) Solve the equation \( -2 x^{2}+4 t x+t=0 \). Solve the equation by following steps: - step0: Solve for \(t\): \(-2x^{2}+4tx+t=0\) - step1: Rewrite the expression: \(-2x^{2}+4xt+t=0\) - step2: Collect like terms: \(-2x^{2}+\left(4x+1\right)t=0\) - step3: Move the constant to the right side: \(\left(4x+1\right)t=0+2x^{2}\) - step4: Remove 0: \(\left(4x+1\right)t=2x^{2}\) - step5: Divide both sides: \(\frac{\left(4x+1\right)t}{4x+1}=\frac{2x^{2}}{4x+1}\) - step6: Divide the numbers: \(t=\frac{2x^{2}}{4x+1}\) Let's solve the problems step by step. ### 1. Solve for \( x \) #### 1.1 Solve \( x^{2} = 16 \) To solve this equation, we take the square root of both sides: \[ x = \pm 4 \] Thus, the solutions are: \[ x_{1} = -4, \quad x_{2} = 4 \] #### 1.2 Solve \( x^{2} - 8x + 1 = 0 \) Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 1, b = -8, c = 1 \). Calculating the discriminant: \[ D = b^2 - 4ac = (-8)^2 - 4 \cdot 1 \cdot 1 = 64 - 4 = 60 \] Now applying the quadratic formula: \[ x = \frac{8 \pm \sqrt{60}}{2} = \frac{8 \pm 2\sqrt{15}}{2} = 4 \pm \sqrt{15} \] Thus, the solutions are: \[ x_{1} = 4 - \sqrt{15}, \quad x_{2} = 4 + \sqrt{15} \quad (\text{approximately } x_{1} \approx 0.127, x_{2} \approx 7.873) \] #### 1.3 Solve \( x - 6 = 9 - \sqrt{2x + 1} \) Rearranging gives: \[ \sqrt{2x + 1} = 9 - (x - 6) = 15 - x \] Squaring both sides: \[ 2x + 1 = (15 - x)^2 \] \[ 2x + 1 = 225 - 30x + x^2 \] \[ x^2 - 32x + 224 = 0 \] Using the quadratic formula: \[ D = (-32)^2 - 4 \cdot 1 \cdot 224 = 1024 - 896 = 128 \] \[ x = \frac{32 \pm \sqrt{128}}{2} = \frac{32 \pm 8\sqrt{2}}{2} = 16 \pm 4\sqrt{2} \] Thus, the solutions are: \[ x_{1} = 16 - 4\sqrt{2}, \quad x_{2} = 16 + 4\sqrt{2} \] #### 1.4 Solve \( (x + 7)(x - 1) \leq 48 \) First, we expand and rearrange: \[ x^2 + 6x - 48 \leq 0 \] Finding the roots: \[ x^2 + 6x - 48 = 0 \] Using the quadratic formula: \[ D = 6^2 - 4 \cdot 1 \cdot (-48) = 36 + 192 = 228 \] \[ x = \frac{-6 \pm \sqrt{228}}{2} = \frac{-6 \pm 2\sqrt{57}}{2} = -3 \pm \sqrt{57} \] Thus, the critical points are: \[ x_{1} = -3 - \sqrt{57}, \quad x_{2} = -3 + \sqrt{57} \] The solution to the inequality is: \[ -11 \leq x \leq 5 \] #### 1.5 Solve \( x^{\frac{2}{3}} - 2x^{\frac{1}{3}} = 15 \) Let \( y = x^{\frac{1}{3}} \), then: \[ y^2 - 2y - 15 = 0 \] Using the quadratic formula: \[ D = (-2)^2 - 4 \cdot 1 \cdot (-15) = 4 + 60 = 64 \] \[ y = \frac{2 \pm 8}{2} = 5 \quad \text{or} \quad -3 \] Thus, \( x = 5^3 = 125 \) (since \( y = -3 \) is not valid for real \( x \)): \[ x_{1} = 125 \] ### 2. Solve for \( x \) and \( y \) simultaneously Given the equations: \[ 3y - x = 11 \quad \text{and} \quad x^2 - xy = y^2 + 1 \] From the first equation, we can express \( x \): \[ x = 3y - 11 \] Substituting into the second equation: \[ (3y - 11)^2 - (3y - 11)y = y^2 + 1 \] Expanding and simplifying leads to: \[ (9y^2 - 66y + 121) - (3y^2 - 11y) = y^2 + 1 \] \[ 6y^2 - 55y + 120 = 0 \] Using the quadratic formula: \[ D = (-55)^2 - 4 \cdot 6 \cdot 120 = 3025 - 2880 = 145 \] \[ y = \frac{55 \pm \sqrt{145}}{12} \] Substituting back to find \( x \): \[ x = 3y - 11 \] Thus, the solutions are: \[ (x,y) = (-2,3) \cup (13,8) \] ### 3. Determine the value(s) of \( t \) for which \( -2x^{2} + 4tx + t = 0 \) will have non-real roots. For non-real roots, the discriminant must be less than zero: \[ D = (4t)^2 - 4(-2)(t) < 0 \] \[ 16t^2 + 8t < 0 \] Factoring gives: \[ 8t(2t + 1) < 0 \] The critical points are \( t = 0 \) and \( t = -\frac{1}{2} \). Testing intervals, we find: \[ -\frac{1}{2} < t < 0 \] ### Summary of Results 1. \( x^{2} = 16 \): \( x_{1} = -4, x_{2} = 4 \) 2. \( x^{2} - 8x + 1 = 0 \): \( x_{1} = 4 - \sqrt{15},