Question

Find two positive even consecutive integers such that the square of the smaller integer is 10 more than the larger integer

Answer 1

Answer:

the numbers are $4,6$

Step-by-step explanation:

Let

x-------> the smaller even consecutive integer

x+2-------> the larger even consecutive integer

we know that

$x^{2}=(x+2)+10$

solve for x

$x^{2}-x-12=0$

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2}-x-12=0$

so

$a=1\\b=-1\\c=-12$

substitute in the formula

$x=\frac{1(+/-)\sqrt{-1^{2}-4(1)(-12)}} {2(1)}$

$x=\frac{1(+/-)\sqrt{49}} {2}$

$x=\frac{1(+/-)7} {2}$

$x=\frac{1(+)7} {2}=4$  ------> the solution (must be positive)

$x=\frac{1(-)7} {2}=-3$

therefore

the numbers are $4,6$