Answer:
the numbers are [tex]4,6[/tex]
Step-by-step explanation:
Let
x-------> the smaller even consecutive integer
x+2-------> the larger even consecutive integer
we know that
[tex]x^{2}=(x+2)+10[/tex]
solve for x
[tex]x^{2}-x-12=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2}-x-12=0[/tex]
so
[tex]a=1\\b=-1\\c=-12[/tex]
substitute in the formula
[tex]x=\frac{1(+/-)\sqrt{-1^{2}-4(1)(-12)}} {2(1)}[/tex]
[tex]x=\frac{1(+/-)\sqrt{49}} {2}[/tex]
[tex]x=\frac{1(+/-)7} {2}[/tex]
[tex]x=\frac{1(+)7} {2}=4[/tex] ------> the solution (must be positive)
[tex]x=\frac{1(-)7} {2}=-3[/tex]
therefore
the numbers are [tex]4,6[/tex]
Two positive even consecutive numbers are [tex]4,6[/tex].
Let [tex]x,\;x+2[/tex] are two positive consecutive even number.
According to question,
[tex]x^2=(x+2)+10[/tex]
[tex]x^2-x-12=0[/tex]
Solve the quadratic equation,
[tex]x^2-4x+3x-12=0\\x(x-4)-3(x-4)=0\\(x-4)(x-3)=0\\\; x-4=0\\ \; x-3=0\\[/tex]
So [tex]x=3 , 4[/tex].
Positive even number is the requirement so [tex]3[/tex] is not a even number so eliminate.
Hence value of [tex]x[/tex] is [tex]4[/tex].
Two positive even consecutive numbers are [tex]4,6[/tex].
Learn more about quadratic equation here:
https://brainly.com/app/ask?q=quadratic
