Solve: x- √ (x+5) = 7?
I have no idea how to work this out, if someone could give me the steps, that would be great. It reads x minus sq. root of x plus 5 equals 7.
Update:Thank you for the help, last question, how do you get the 14x?
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x - √(x + 5) = 7
x - 7 = √(x + 5)
x^2 - 14x + 49 = x + 5
x^2 - 15x + 44 = 0
(x - 4)(x - 11) = 0
x = 4 or x = 11
x = 4 is extraneous, so x = 11 is the only solution.
Move stuff around until the square root is alone on one side:
x - 7 = â(x+5)
Then, square both sides:
(x - 7)^2 = (x+5)
x^2 - 14x + 49 = x + 5
Next, move stuff around until you have a typical quadratic, in the standard format of
ax^2 + bx + c = 0
x^2 - 15x + 44 = 0
Now you can solve it with any of the methods used to find the roots of quadratics, including the quadratic formula.
Once you find possible values, you MUST go back and check it they work in the original equation.
Hint: the original equation has a square root, and you cannot have a negative number under the square root (it you are working with real numbers).
Everybody else is right except for the exclusion of 4 as an answer.
x - sqrt(x + 5) = 7
4 - sqrt(4 + 5) = 7
4 - sqrt(9) = 7
sqrt(9) can be 3 or -3
4 - (-3) = 7
4 + 3 = 7
4 is a valid solution.
â(x+5)=x-7
x+5=x^2-14x+49
x^2-15x+44=0
(x-11)(x-4)=0
xâ{11, 4}
11-â16=7 check.
4-â9=1 no.
x=11