Rearrange the radical equation so that one of the radical expressions is isolated. Then, raise both sides of the equation to a power equal to the index of the radicals.
x=5
Practice makes perfect
To solve equations with a variable expression inside a radical, we will first rearrange the radical equation so that one of the radical expressions is isolated. Then we can raise both sides of the equation to a power equal to the index of the radicals.
Since we still have a radical expression on one side of the equation, we will follow the same method as before — we will raise both sides of the equation to the second power. When we do this, we will have to expand the left-hand side of the equation using a perfect square trinomial.
Finally, all of the radicals have been removed from the equation. Because we have an x-variable with an exponent, we cannot directly solve for it. Therefore, let's move all of the non-zero terms to the left-hand side of the equation and see what remains.
We are left with a quadratic equation. We can solve it using the Quadratic Formula.
ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a
Let's identify the values of a, b, and c in our case.
4825-1210x+49x^2=0 ⇓
49x^2+( -1210)x+ 4825=0
We can see that a= 49, b= - 1210, and c= 4825. Let's substitute these values into the Quadratic Formula.
The solutions for this equation are x= 605± 36049.
x=605± 360/49
x_1=605-360/49
x_2=605+360/49
x_1=245/49
x_2=965/49
x_1=5
x_2=965/49
Using the Quadratic Formula, we found that the solutions of the given equation are x_1=5 and x_2= 96549. Let's check them to see if we have any extraneous solutions.
The substitution of 96549 resulted in a contradiction. This means that 96549 is an extraneous solution. Therefore, x=5 is the only real solution to the equation.
