The product of two consecutive even positive integers is 168. Let the smaller integer be x, so the next even integer is (x + 2). Form a quadratic from x(x + 2) = 168 and solve by factorisation. What is the smaller integer?

Question

Accepted Answer

12. x(x + 2) = 168 gives x^2 + 2x - 168 = 0 = (x - 12)(x + 14). x = 12 (reject -14). The integers are 12 and 14; the smaller is 12.

The product of two consecutive even positive integers is 168.
Let the smaller integer be $x$ , so the next even integer is $(x + 2)$ . Form a quadratic from $x (x + 2) = 168$ and solve by factorisation.
What is the smaller integer?

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More Class 10 Solving by Factorisation practice

The product of two consecutive even positive integers is 168.Let the smaller integer be x, so the next even integer is (x+2). Form a quadratic from x(x+2)=168 and solve by factorisation.What is the smaller integer?

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 Solving by Factorisation practice

The product of two consecutive even positive integers is 168.
Let the smaller integer be $x$ , so the next even integer is $(x + 2)$ . Form a quadratic from $x (x + 2) = 168$ and solve by factorisation.
What is the smaller integer?