A two-digit number has digit product 18. Adding 63 interchanges its digits, so the units digit exceeds the tens digit by 7. Letting the tens digit be x, this gives x^2 + 7x - 18 = 0. Using the quadratic formula, find the original number.

Question

Accepted Answer

29. Let digits be tens a, units b with ab = 18 and (10a + b) + 63 = 10b + a => 9b - 9a = 63 => b - a = 7. So b = a + 7, a(a+7) = 18 => a^2 + 7a - 18 = 0 => a = -7 ± √49 + 722 = -7 ± 11/2 = 2. Then b = 9, number = 29 (29 + 63 = 92).

A two-digit number has digit product 18. Adding 63 interchanges its digits, so the units digit exceeds the tens digit by 7.
Letting the tens digit be $x$ , this gives $x^{2} + 7 x - 18 = 0$ .
Using the quadratic formula, find the original number.

Step-by-Step Solution

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More Class 10 The Quadratic Formula practice

A two-digit number has digit product 18. Adding 63 interchanges its digits, so the units digit exceeds the tens digit by 7.Letting the tens digit be x, this gives x2+7x−18=0.Using the quadratic formula, find the original number.

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 The Quadratic Formula practice

A two-digit number has digit product 18. Adding 63 interchanges its digits, so the units digit exceeds the tens digit by 7.
Letting the tens digit be $x$ , this gives $x^{2} + 7 x - 18 = 0$ .
Using the quadratic formula, find the original number.