Question:
The difference of squares of two number is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers.
Solution:
Let the smaller numbers be $x$
Then according to question,
The larger number be $=2 x-5$, then
$(2 x-5)^{2}-x^{2}=88$
$4 x^{2}-20 x+25-x^{2}-88=0$
$3 x^{2}-20 x-63=0$
$3 x^{2}-20 x-63=0$
$3 x^{2}-27 x+7 x-63=0$
$3 x(x-9)+7(x-9)=0$
$(x-9)(3 x+7)=0$
$(x-9)=0$
$x=9$
Or
$(3 x+7)=0$
$x=\frac{-7}{3}$
Since, x being a positive integer so, x cannot be negative,
Therefore,
When $x=9$ then larger number be
$2 x-5=2 \times 9-5$
$=18-5$
$=13$
Thus, two consecutive number be either 9,13
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