Question

Given that 10-square root of 18/square root of 2 where a and b are integers find the values of a and b

Answer 1

a = -3

b = 5

Explanation:

Prime factorize 18 and simplify the irrational number.

`\sf √(18)=√(3*3*2)=3√(2)`

`\sf (10-√(18))/(√(2))=(10-3√(2))/(√(2))`

Now, rationalize the denominator by multiplying the numerator and denominator by √2.

`\s = ((10-3√(2))*√(2))/(√(2))\\\\\\=(√(2)*10-3√(2)*√(2))/(√(2)*√(2))\\\\\\=(10√(2)-3*2)/(2)\\\\\\=(10√(2)-6)/(2)\\\\\\=(10√(2))/(2)+(-6)/(2)\\\\\\=5√(2)+(-3)\\\\= (-3) + 5√(2)`

Answer:

a = -3

b = 5

Answer 2

a = - 3 , b = 5

Explanation:

using the rules of radicals

•
`√(a)` ×
`√(b)` =
`√(ab)`

•
`√(a)` ×
`√(a)` = a

given

`(10-√(18) )/(√(2) )`

rationalise the denominator by multiplying the numerator and denominator by
`√(2)`

=
`(√(2)(10-√(18)) )/(√(2)(√(2)) )` ← distribute parenthesis on numerator using the rules

=
`(10√(2)-√(36) )/(2)`

=
`(10√(2))-6 )/(2)`

=
`(10√(2) )/(2)` -
`(6)/(2)`

= 5
`√(2)` - 3

= - 3 + 5
`√(2)` ← in the form a + b
`√(2)`

with a = - 3 and b = 5