How do you solve this? 'A rectangle has a length (x+16)cm and width x cm. The square has length(3x-2) cm. The area of the rectangle is 16cm^2 more than the area of the square. The area of the square is more than 1cm^2. Find the value of x.
Calculate area of rectangle, calculate area of square. Question says area of rectangle = area of square + 16 rearrange and put it into that double bracket quadratic thing to find x or use the quadratic equation init, ull probably get it with dis.
How do you solve this? 'A rectangle has a length (x+16)cm and width x cm. The square has length(3x-2) cm. The area of the rectangle is 16cm^2 more than the area of the square. The area of the square is more than 1cm^2. Find the value of x.
Area of Rectangle = (x + 16)(x) Area of Rectangle = x² + 16x
Area of Square = (3x-2)(3x-2) Area of Square = 9x² - 12x + 4