Step-by-step explanation:
√(4-3x) = x
1. square both sides:
[√(4-3x)]²= (x) ²
4-3x = x²
2. shift all the terms to the right side
(minimise dealing with negative coefficient of x² terms to prevent careless mistakes)
0= x²+3x-4
no matter how you change the places of the terms, it will still be:
x²+3x-4=0
3. solve the quadratic equation, either by cross multiplication(and factorising) or the quadratic formula(if you're taught)
for me I will be using the cross multiplication table that my calculator can help me with(gives me the solution but better to show the factorised form to prevent missing out method marks)
as such:
x²+3x-4=0
(x+4)(x-1)=0
hence:
(x+4)=0 or (x-1)=0
finally:
x= -4 or x= 1
however we're not done yet as some solutions may not match with the question/ original equation provided, this we still need to check if there are any values that does not match
substitute x= -4 into the original equation:
√4-3(-4) = √4+12 = √16 = 4 > -1 , hence x= -1 is not the answer
you can substitute the values into a scientific calculator(that has the functions required) to quickly determine
then to double confirm, do the same for the other value of X that we have found:
√4-3(1) = √4-3 = √1 = 1
since the other value of x matches with the equation provided in the question, then we can now confirm that the value of x=1 is the correct value
reason for such situation to happen:
remember in step 1 when we squared both sides of the equation, that is the reason for having to check for the correct X value. let me explain further
when we square numbers, we are potentially introducing another possible answer other than the correct answer intended for us to find as well as to match with the question.
for example:
1 = 1² = (-1)²
likewise, I can even take any other number to compare, such as 100:
100=10²=(-10)²
and this scenario only happens for the even powers of x, such as x², x⁴, x⁶ so on and so forth.
hence whenever such questions come out, do remember to double check with the original question/equation
lastly, the checking portion will only apply for even powers of x in questions and/or when you happen to square them
hope this helps!
. If both coordinates are positive or both are negative, we get a maximum value of 
; otherwise, we get a minimum of 
.