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gladu [14]
3 years ago
11

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

Mathematics
1 answer:
mafiozo [28]3 years ago
5 0

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

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