Solve the quadratic equations using the quadratic formula and completing the square. Simplify the surds.

Welcome back — missed you this week. Quadratic Equations 1. Solve: 3x^2 - 7x + 2 = 0 Step 1: Identify coefficients a=3, b=-7, c=2. Step 2: Use the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a. x = -(-7) ± sqrt((-7)^2 - 4(3)(2))2(3) x = 7 ± sqrt(49 - 24)6 x = 7 ± sqrt(25)6 x = (7 ± 5)/(6) Step 3: Calculate the two possible values for x. x_1 = (7 + 5)/(6) = (12)/(6) = 2 x_2 = (7 - 5)/(6) = (2)/(6) = (1)/(3) The solutions are x=2 and x=(1)/(3). x = 2, x = (1)/(3) 2. Solve using completing the square: x^2 + 6x + 5 = 0 Step 1: Move the constant term to the right side. x^2 + 6x = -5 Step 2: Complete the square on the left side by adding ((6)/(2))^2 = 3^2 = 9 to both sides. x^2 + 6x + 9 = -5 + 9 Step 3: Factor the left side as a perfect square. (x+3)^2 = 4 Step 4: Take the square root of both sides. x+3 = ±sqrt(4) x+3 = ± 2 Step 5: Solve for x. x = -3 ± 2 x_1 = -3 + 2 = -1 x_2 = -3 - 2 = -5 The solutions are x=-1 and x=-5. x = -1, x = -5 Surds and Indices 1. Simplify: sqrt(50) + 3sqrt(8) Step 1: Simplify each surd by finding the largest perfect square factor. sqrt(50) = sqrt(25 × 2) = sqrt(25) × sqrt(2) = 5sqrt(2) sqrt(8) = sqrt(4 × 2) = sqrt(4) × sqrt(2) = 2sqrt(2) Step 2: Substitute the simplified surds back into the expression. 5sqrt(2) + 3(2sqrt(2)) 5sqrt(2) + 6sqrt(2) Step 3: Combine the like terms. 11sqrt(2) The simplified expression is 11sqrt(2). 11sqrt(2) 2. Simplify: (2^3 × 2^-5) Step 1: Use the rule of indices a^m × a^n = a^m+n. 2^3 + (-5) 2^3 - 5 2^-2 Step 2: Use the rule a^-n = (1)/(a^n). (1)/(2^2) (1)/(4) The simplified expression is (1)/(4). (1)/(4) Logarithms 1. Simplify: _10(100) + _10(0.001) Step 1: Evaluate each logarithm. _10(100) = _10(10^2) = 2 _10(0.001) = _10(10^-3) = -3 Step 2: Add the results. 2 + (-3) = -1 The simplified expression is -1. -1 2. Solve: _2(x) = 5 Step 1: Convert the logarithmic equation to an exponential equation using the definition _b(a) = c b^c = a. x = 2^5 Step 2: Calculate the value. x = 32 The solution is x=32. x = 32 Trigonometry 1. Solve: x = 1/2 for 0^ ≤ x ≤ 360^ Step 1: Find the principal value. x = ((1)/(2)) = 30^ Step 2: Identify the quadrants where x is positive. Sine is positive in the first and second quadrants. Step 3: Find the solution in the second quadrant. x = 180^ - 30^ = 150^ The solutions in the given range are 30^ and 150^. x = 30^, 150^ 2. Prove: ^2 x + ^2 x = 1 Step 1: Consider a right-angled triangle with angle x. Let the opposite side be y, the adjacent side be a, and the hypotenuse be h. Step 2: Write the definitions of x and x. x = oppositehypotenuse = (y)/(h) x = adjacenthypotenuse = (a)/(h) Step 3: Substitute these into the expression ^2 x + ^2 x. ^2 x + ^2 x = ((y)/(h))^2 + ((a)/(h))^2 ^2 x + ^2 x = (y^2)/(h^2) + (a^2)/(h^2) ^2 x + ^2 x = (y^2 + a^2)/(h^2) Step 4: Apply the Pythagorean theorem, which states that in a right-angled triangle, y^2 + a^2 = h^2. ^2 x + ^2 x = (h^2)/(h^2) ^2 x + ^2 x = 1 The identity is proven. ^2 x + ^2 x = 1 (Proven) Calculus (Differentiation) 1. Differentiate: y = 4x^3 - 3x^2 + 2 Step 1: Apply the power rule for differentiation, (d)/(dx)(ax^n) = nax^n-1, and the rule that the derivative of a constant is 0. (dy)/(dx) = (d)/(dx)(4x^3) - (d)/(dx)(3x^2) + (d)/(dx)(2) (dy)/(dx) = 4(3)x^3-1 - 3(2)x^2-1 + 0 (dy)/(dx) = 12x^2 - 6x The derivative is 12x^2 - 6x. (dy)/(dx) = 12x^2 - 6x 2. Find gradient at x = 1 for y = x^2 + 3x Step 1: Differentiate the function to find the gradient function. (dy)/(dx) = (d)/(dx)(x^2) + (d)/(dx)(3x) (dy)/(dx) = 2x^2-1 + 3x^1-1 (dy)/(dx) = 2x + 3 Step 2: Substitute x=1 into the gradient function. (dy)/(dx)|_x=1 = 2(1) + 3 (dy)/(dx)|_x=1 = 2 + 3 (dy)/(dx)|_x=1 = 5 The gradient at x=1 is 5. Gradient = 5 Calculus (Integration) 1. Integrate: (3x^2 + 2x)\,dx Step 1: Apply the power rule for integration, x^n\,dx = x^n+1n+1 + C, to each term. 3x^2\,dx + 2x\,dx 3(x^2+12+1) + 2(x^1+11+1) + C 3((x^3)/(3)) + 2((x^2)/(2)) + C Step 2: Simplify the expression. x^3 + x^2 + C The integral is x^3 + x^2 + C. x^3 + x^2 + C 2. Find area under y = x from 0 to 5 Step 1: Set up the definite integral for the area. Area = _0^5 x\,dx Step 2: Integrate the function. x\,dx = x^1+11+1 = (x^2)/(2) Step 3: Evaluate the definite integral using the limits of integration. Area = [(x^2)/(2)]_0^5 Area = (5^2)/(2) - (0^2)/(2) Area = (25)/(2) - 0 Area = (25)/(2) The area under y=x from 0 to 5 is (25)/(2) square units. Area = (25)/(2) square units Send me the next one 📸