This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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1.1.1 Solve for x: x(x-3)=0 Step 1: For the product of two factors to be zero, at least one of the factors must be zero. x = 0 or x-3 = 0 Step 2: Solve for x in the second equation. x = 0 or x = 3 The solutions for x are: x=0 or x=3 1.1.2 Solve for x: 2x^2+1=4x (correct to TWO decimal places) Step 1: Rearrange the equation into the standard quadratic form ax^2+bx+c=0. 2x^2 - 4x + 1 = 0 Step 2: Identify the coefficients a, b, c and apply the quadratic formula x = -b ± sqrt(b^2-4ac)2a. Here, a=2, b=-4, c=1. x = -(-4) ± sqrt((-4)^2 - 4(2)(1))2(2) x = 4 ± sqrt(16 - 8)4 x = 4 ± sqrt(8)4 Step 3: Simplify the square root and calculate the two possible values for x. x = 4 ± 2sqrt(2)4 x = 2 ± sqrt(2)2 Step 4: Calculate the numerical values and round to two decimal places. x_1 = 2 + sqrt(2)2 ≈ (2 + 1.41421)/(2) ≈ (3.41421)/(2) ≈ 1.7071 ≈ 1.71 x_2 = 2 - sqrt(2)2 ≈ (2 - 1.41421)/(2) ≈ (0.58579)/(2) ≈ 0.2929 ≈ 0.29 The solutions for x are: x ≈ 1.71 or x ≈ 0.29 1.1.3 Solve for x: x^2 - 2x - 3 > 0 Step 1: Find the critical points by solving the corresponding equality x^2 - 2x - 3 = 0. Factor the quadratic expression: (x-3)(x+1) = 0 This gives critical points x=3 and x=-1. Step 2: Determine the intervals where the inequality holds true. Since the parabola y=x^2-2x-3 opens upwards (coefficient of x^2 is positive), the expression is greater than zero outside its roots. We test values in the intervals (-, -1), (-1, 3), and (3, ). • For x < -1 (e.g., x=-2): (-2)^2 - 2(-2) - 3 = 4 + 4 - 3 = 5 > 0 (True) • For -1 < x < 3 (e.g., x=0): (0)^2 - 2(0) - 3 = -3 < 0 (False) • For x > 3 (e.g., x=4): (4)^2 - 2(4) - 3 = 16 - 8 - 3 = 5 > 0 (True) The solution for x is: x < -1 or x > 3 1.1.4 Solve for x: 2^2x - 2^x+2 - 32 = 0 Step 1: Rewrite the terms using exponent rules. 2^2x = (2^x)^2 2^x+2 = 2^x · 2^2 = 4 · 2^x Substitute these into the equation: (2^x)^2 - 4 · 2^x - 32 = 0 Step 2: Let k = 2^x. Substitute k into the equation to form a quadratic equation. k^2 - 4k - 32 = 0 Step 3: Solve the quadratic equation for k by factoring. (k-8)(k+4) = 0 This gives two possible values for k: k = 8 or k = -4 Step 4: Substitute back 2^x = k and solve for x. Case 1: 2^x = 8 2^x = 2^3 x = 3 Case 2: 2^x = -4 An exponential function with a positive base, like 2^x, can never be negative. Therefore, there is no real solution for x in this case. The solution for x is: x=3 1.1.5 Solve for x: sqrt(-2x+4) - x = 2 Step 1: Isolate the square root term on one side of the equation. sqrt(-2x+4) = x+2 Step 2: Square both sides of the equation to eliminate the square root. (sqrt(-2x+4))^2 = (x+2)^2 -2x+4 = x^2 + 4x + 4 Step 3: Rearrange the equation into a standard quadratic form ax^2+bx+c=0. 0 = x^2 + 4x + 2x + 4 - 4 0 = x^2 + 6x Step 4: Solve the quadratic equation for x by factoring. x(x+6) = 0 This gives two possible solutions: x = 0 or x = -6 Step 5: Check for extraneous solutions by substituting each value back into the original equation. For x=0: sqrt(-2(0)+4) - 0 = 2 sqrt(4) - 0 = 2 2 = 2 This solution is valid. For x=-6: sqrt(-2(-6)+4) - (-6) = 2 sqrt(12+4) + 6 = 2 sqrt(16) + 6 = 2 4 + 6 = 2 10 = 2 This statement is false, so x=-6 is an extraneous solution. The solution for x is: x=0 2 Solve for x and y simultaneously: 2x + y = 3 (Equation 1) y^2 + xy = 2 (Equation 2) Step 1: From Equation 1, express y in terms of x. y = 3 - 2x (Equation 3) Step 2: Substitute Equation 3 into Equation 2. (3-2x)^2 + x(3-2x) = 2 Step 3: Expand and simplify the equation. (9 - 12x + 4x^2) + (3x - 2x^2) = 2 9 - 12x + 4x^2 + 3x - 2x^2 = 2 2x^2 - 9x + 9 = 2 2x^2 - 9x + 7 = 0 Step 4: Solve the quadratic equation for x by factoring. (2x-7)(x-1) = 0 This gives two possible values for x: 2x-7=0 x = (7)/(2) x-1=0 x = 1 Step 5: Substitute each value of x back into Equation 3 to find the corresponding y values. Case 1: If x = (7)/(2) y = 3 - 2((7)/(2)) y = 3 - 7 y = -4 Case 2: If x = 1 y = 3 - 2(1) y = 3 - 2 y = 1 The pairs of solutions (x,y) are: ((7)/(2), -4) and (1, 1) Drop the next question! 📸