The problem states that twice the square of the larger part, which is 2x², exceeds the square of the smaller part, which is (16 - x)², by 164. Setting up their difference gives 2x² - (16 - x)² = 164.

The problem states that twice the square of the larger part, which is 2x², exceeds the square of the smaller part, which is (16 - x)², by 164. Setting up their difference gives 2x² - (16 - x)² = 164.

Expanding (16 - x)² yields 256 - 32x + x². Subtracting this from 2x² gives 2x² - 256 + 32x - x² = 164. Simplifying and bringing all terms to the left side results in x² + 32x - 420 = 0.

divide 16 into 2 parts such that the twice the square of the largest part exceeds the square of the smaller part by 164

Answer: The two parts are 10 and 6.

Step-by-step solution

Step 1: Set up the algebraic expressions for the two parts

Let the larger part be $x$ . Since the sum of both parts is $16$ , the smaller part is $16 - x$ . For $x$ to be the larger part, we require $x > 8$ .

Step 2: Formulate the equation from the given condition

The problem states that twice the square of the larger part, which is $2x^2$ , exceeds the square of the smaller part, which is $(16 - x)^2$ , by $164$ . Setting up their difference gives $2x^2 - (16 - x)^2 = 164$ .

Step 3: Expand and simplify into standard quadratic form

Expanding $(16 - x)^2$ yields $256 - 32x + x^2$ . Subtracting this from $2x^2$ gives $2x^2 - 256 + 32x - x^2 = 164$ . Simplifying and bringing all terms to the left side results in $x^2 + 32x - 420 = 0$ .

Step 4: Factor the quadratic equation

We look for two numbers whose product is $-420$ and whose sum is $32$ . These numbers are $42$ and $-10$ . Splitting the middle term and factoring by grouping yields $(x + 42)(x - 10) = 0$ .

Step 5: Solve for the two parts

Since $x$ represents a part of the positive number $16$ , we reject the negative root $x = -42$ . Thus, the larger part is $x = 10$ , and the smaller part is $16 - 10 = 6$ .

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