Question

Solve the radical equation. If there is no real solution, choose no solution.

Answer 1

square both sides to get

x^2 = 6 - x

now get it into standard form

x^2 = 6 - x

x^2 +x = 6

x^2 +x - 6 = 0

what's next?

Answer 2

you do the quadratic fromula?

Answer 3

you could if you want to, or you can factor

Answer 4

either way works

Answer 5

ahh ok well factoring is much faster

Answer 6

keep in mind that factoring doesn't always work for every quadratic, so the quadratic formula is a good plan B

Answer 7

i got -3 and 2

Answer 8

now check those possible solutions

Answer 9

one of those possible solutions will not work

Answer 10

the negative one

Answer 11

good, plugging in x = -3 produces a false equation

however, plugging in x = 2 produces a true equation

Answer 12

so x = 2 is the only solution

Answer 13

so 2! omg i feel smart lol Thanx!!!

Answer 14

that's because you are smart

Answer 15

you're welcome

Answer 16

how do i do this?

Answer 17

hmm one sec, let me think

Answer 18

the best way would probably be to do this

Answer 19

x <= sqrt(x)

x - sqrt(x) <= 0

now we need to find the roots to f(x) = x - sqrt(x)

f(x) = x - sqrt(x)

0 = x - sqrt(x)

sqrt(x) = x

( sqrt(x) )^2 = x^2

x = x^2

0 = x^2 - x

x^2 - x = 0

x(x-1) = 0

x = 0 or x-1 = 0

x = 0 or x = 1

Answer 20

the two roots of f(x) = x - sqrt(x) are x = 0 or x = 1

let's plug in a value to the left of x = 0 and see what we get


plug in x = -1

x <= sqrt(x)

-1 <= sqrt(-1)

since you cannot take the square root of a negative number, this means that x >= 0

now let's plug in a number between 0 and 1

plug in 0.5

x <= sqrt(x)

0.5 <= sqrt(0.5)

0.5 <= 0.70710678118654

and that's true, so the solution set so far in interval notation is [0, 1]

Answer 21

now let's plug in a number to the right of x = 1, so plug in x = 4

x <= sqrt(x)

4 <= sqrt(4)

4 <= 2

that's false, so (1, infinity) is not part of the solution set

Answer 22

so in the end, the solution set of x <= sqrt(x) is [0, 1]