We are asked to write a radical equation that has x= 3 and x= 4 as its solutions. To do so, let's first rewrite these equations by setting them equal to zero.
ccc
x= 3 &and& x= 4
(x- 3)=0 &and& (x- 4)=0
Now, we will use the expressions on the left-hand sides of these equations as the factors of a quadratic equation in factored form. Since both of them are equal to zero, the product of these factors will also be equal to zero.
(x- 3)=0 and (x- 4)=0
⇓
(x- 3)(x- 4)=0Next, we need to rewrite this quadratic equation as a radical equation. To do so, we will take square root of both sides.
sqrt((x- 3)(x- 4))= sqrt()
Let's simplify our equation!
Great! We found a radical equation that has x= 3 and x= 4 as its solutions. Note that this is just one possible equation that fits the given requirements.
Further Exploration
For further exploration, let's examine some more examples that have the same solutions. We will write a general equation that has x= 3 and x= 4 as its solutions.
y= asqrt((x- 3)(x- 4)- h)+k
We can choose arbitrary values for the variables.
y
a
h
k
Obtained Equation
Simplified Equation
0
1
0
0= 1sqrt((x- 3)(x- 4)- 0)+
0=sqrt((x-3)(x-4))
0
2
0
0= 2sqrt((x- 3)(x- 4)- 0)+
0=2sqrt((x-3)(x-4))
0
-5
0
0= -5sqrt((x- 3)(x- 4)- 0)+
0=-5sqrt((x-3)(x-4))
7
1
-4
5
7= 1sqrt((x- 3)(x- 4)-( -4))+5
7=sqrt((x-3)(x-4)+8)+5
1
8
0
1
1= 8sqrt((x- 3)(x- 4)- 0)+1
1=8sqrt((x-3)(x-4))+1
We can notice a pattern that, when we multiply the square root expression by a real number without changing the other variables, the solution remains always the same. However, when we change h or k, we need to adjust the remaining variables to be sure that x= 3 and x= 4 are the solutions.
