#### Question

The sum of the squares of two numbers as 233 and one of the numbers as 3 less than twice the other number find the numbers.

#### Solution

Let the numbers be integers. One of the numbers be *x*. So, the other will be (2x - 3).

Then according to question,

x^{2} + (2x - 3)^{2} = 233

x^{2} + 4x^{2} - 12x + 9 = 233

5x^{2} - 12x + 9 - 233 = 0

5x^{2} - 12x - 224 = 0

5x^{2} - 40x + 28x - 224 = 0

5x(x - 8) + 28(x - 8) = 0

(x - 8)(5x + 28) = 0

x - 8 = 0

x = 8

Or

5x + 28 = 0

5x = -28

x = -28/5

Since, we have assumed the numbers to be integers, so *x* cannot be a rational number/fraction.

Therefore, for *x* = 8

Other number = (2x - 3) = 2(8) - 3 = 16 - 3 = 13

Thus, whole numbers be 8, 13.

Is there an error in this question or solution?

#### APPEARS IN

Solution The Sum of the Squares of Two Numbers as 233 and One of the Numbers as 3 Less than Twice the Other Number Find the Numbers. Concept: Solutions of Quadratic Equations by Factorization.