Question

(2x+5)(x-6)+2(3x+2)(3x-2)=5(2x+1)^2+11

Answer 1

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Answer 2

To determine whether the given equation is true or false, we'll simplify both sides of the equation and check if they are equal.

Given equation: (2x+5)(x-6) + 2(3x+2)(3x-2) = 5(2x+1)^2 + 11

Step 1: Expand both sides of the equation.

Left-hand side (LHS): (2x+5)(x-6) + 2(3x+2)(3x-2) Expand the first part: (2x+5)(x-6) = 2x(x) + 2x(-6) + 5(x) + 5(-6) = 2x^2 - 12x + 5x - 30 = 2x^2 - 7x - 30 Expand the second part: 2(3x+2)(3x-2) = 2(9x^2 - 4) = 18x^2 - 8

So the left-hand side (LHS) of the equation becomes: 2x^2 - 7x - 30 + 18x^2 - 8 = 20x^2 - 7x - 38

Right-hand side (RHS): 5(2x+1)^2 + 11 (2x+1)^2 = (2x+1)(2x+1) = 4x^2 + 2x + 2x + 1 = 4x^2 + 4x + 1 5(2x+1)^2 = 5(4x^2 + 4x + 1) = 20x^2 + 20x + 5 5(2x+1)^2 + 11 = 20x^2 + 20x + 5 + 11 = 20x^2 + 20x + 16

Step 2: Check if LHS equals RHS.

Comparing both sides, we have: LHS: 20x^2 - 7x - 38 RHS: 20x^2 + 20x + 16

Since the coefficients of x^2 are the same (20x^2), but the coefficients of x and constants are different, the given equation is false.

Thus, the equation (2x+5)(x-6) + 2(3x+2)(3x-2) = 5(2x+1)^2 + 11 is not true for all values of x.