Question

The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers.

Answer 1

$$\begin{gathered} \text{Let the smaller number be}x. \\ \therefore \mathrm{L}\text{arger number =}(2x-5) \\ \mathrm{Also}, \\ {(2x-5)}^2-x^2=88 \\ \Rightarrow 4x^2-20x+25-x^2=88 \\ \Rightarrow 3x^2-20x-63=0 \\ \Rightarrow 3x^2-(27-7)x-63=0 \\ \Rightarrow 3x^2-27x+7x-63=0 \\ \Rightarrow 3x(x-9)+7(x-9)=0 \\ \Rightarrow (x-9)(3x+7)=0 \\ \Rightarrow x=9\text{or}x=\frac{7}{3} \\ \mathrm{The}\mathrm{number}\mathrm{cannot}\mathrm{be}\mathrm{a}\mathrm{fraction};\mathrm{therefore},\mathrm{we}\mathrm{have}: \\ \text{Smaller number = 9} \\ \text{Larger number = (2 × 9 - 5) = 13} \end{gathered}$$