A transformation matrix P transforms an object of area 10 cm2 to an image of area 60 cm2. If P = [[x, 4], [3, x+3]], find the possible values of x.

You're on a roll — 4. Area Transformation The area of an object is transformed by a matrix P. The relationship between the area of the object and the area of its image is given by: Area of Image = |(P)| × Area of Object Step 1: Identify the given values. Area of Object A = 10 cm^2 Area of Image B = 60 cm^2 Transformation matrix P = x & 4 \\ 3 & x+3 Step 2: Calculate the determinant of the matrix P. For a 2 × 2 matrix a & b \\ c & d , the determinant is ad - bc. (P) = x(x+3) - (4)(3) (P) = x^2 + 3x - 12 Step 3: Substitute the values into the area transformation formula. 60 = |x^2 + 3x - 12| × 10 Step 4: Solve for |x^2 + 3x - 12|. |x^2 + 3x - 12| = (60)/(10) |x^2 + 3x - 12| = 6 Step 5: Solve the absolute value equation. This leads to two cases. Case 1: x^2 + 3x - 12 = 6 x^2 + 3x - 18 = 0 Factor the quadratic equation: (x+6)(x-3) = 0 This gives two possible values for x: x = -6 or x = 3 Case 2: x^2 + 3x - 12 = -6 x^2 + 3x - 6 = 0 Use the quadratic formula x = -b ± sqrt(b^2 - 4ac)2a where a=1, b=3, c=-6. x = -3 ± sqrt(3^2 - 4(1)(-6))2(1) x = -3 ± sqrt(9 + 24)2 x = -3 ± sqrt(33)2 This gives two possible values for x: x = -3 + sqrt(33)2 or x = -3 - sqrt(33)2 Step 6: Identify the positive values of x. From Case 1: x=3 is positive. x=-6 is negative. From Case 2: sqrt(33) ≈ 5.74. x = -3 + sqrt(33)2 ≈ (-3 + 5.74)/(2) = (2.74)/(2) = 1.37 (positive). x = -3 - sqrt(33)2 ≈ (-3 - 5.74)/(2) = (-8.74)/(2) = -4.37 (negative). The positive values of x are 3 and -3 + sqrt(33)2. 3 done, 2 left today. You're making progress.