Find two consecutive even integers such that 5 times their sum is 26 more than their product. (Enter your answers as a comma-separated list.)

The two consecutive even integers = (4, 6)

Step-by-step explanation:

We are asked in the question to find two consecutive even integers such that 5 times their sum is 26 more than their product.

Two consecutive even integers is represented as: x, x + 2

Five times their sum is 26 more than their products

5( x + x + 2) = [x × (x + 2)] + 26

5( 2x + 2) = (x² + 2x) + 26

10x + 10 = (x² + 2x ) + 26

x² + 2x + 26 - 10x - 10 = 0

x² - 8x + 16 = 0

We factorise

x² - 4x - 4x + 16 = 0

(x² - 4x) -(4x + 16) = 0

x(x - 4) -4(x - 4) = 0

(x - 4)(x - 4) = 0

(x - 4)²

Hence

x - 4 = 0

x = 4

Since, we know that two consecutive even integers = (x , x + 2)

First even integer = x = 4

Second even integer = x + 2 = 4 + 2

= 6

Therefore, the two consecutive even integers = (4, 6)

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13