A graded collection to help you learn how to solve equations involving radicals. In each exercise we isolate the radical, square both sides carefully, and always check the solutions we obtain. This checking step is essential because squaring can introduce extraneous solutions: values that come out correctly from the algebraic manipulations but fail to satisfy the original equation, and that must therefore be discarded.
Exercise 1 — level ★☆☆☆☆
\[ \sqrt{x}=3 \]
Result
\[ x=9 \]
Solution
Solution strategy
The radical is already isolated. To remove it, we square both sides.
Step 1
\[ \sqrt{x}=3 \]
The square root is isolated on the left-hand side.
\[ (\sqrt{x})^2=3^2 \]
We square both sides to eliminate the radical.
\[ x=9 \]
Squaring the radical gives back the radicand.
Check
\[ \sqrt{9}=3 \]
The equality holds, so the solution is valid.
Result
\[ \boxed{x=9} \]
Exercise 2 — level ★☆☆☆☆
\[ \sqrt{x+1}=4 \]
Result
\[ x=15 \]
Solution
Solution strategy
The radical is already isolated. We square both sides and then solve the resulting linear equation.
Step 1
\[ \sqrt{x+1}=4 \]
The radical contains \(x+1\), so we need to remove the square root.
\[ (\sqrt{x+1})^2=4^2 \]
We square both sides.
\[ x+1=16 \]
After squaring we are left with a first-degree equation.
Step 2
\[ x=16-1 \]
We subtract \(1\) from both sides to isolate \(x\).
\[ x=15 \]
Check
\[ \sqrt{15+1}=\sqrt{16}=4 \]
The solution satisfies the original equation.
Result
\[ \boxed{x=15} \]
Exercise 3 — level ★☆☆☆☆
\[ \sqrt{2x-1}=3 \]
Result
\[ x=5 \]
Solution
Solution strategy
The radical is isolated. We square both sides and then solve the linear equation.
Step 1
\[ \sqrt{2x-1}=3 \]
The radicand is \(2x-1\). To free it from the radical, we square both sides.
\[ 2x-1=9 \]
The right-hand side becomes \(3^2=9\).
Step 2
\[ 2x=10 \]
We add \(1\) to both sides.
\[ x=5 \]
We divide both sides by \(2\).
Check
\[ \sqrt{2\cdot5-1}=\sqrt{9}=3 \]
The solution is correct.
Result
\[ \boxed{x=5} \]
Exercise 4 — level ★★☆☆☆
\[ \sqrt{x}=x-2 \]
Result
\[ x=4 \]
Solution
Solution strategy
The radical is isolated. Since \(\sqrt{x}\ge0\), the right-hand side must also be non-negative: \(x-2\ge0\), that is, \(x\ge2\).
Step 1
\[ \sqrt{x}=x-2 \]
We can square both sides, since the radical is already isolated.
\[ x=(x-2)^2 \]
The left-hand side becomes \(x\), while the right-hand side is expanded as the square of a binomial.
Step 2
\[ x=x^2-4x+4 \]
We have expanded \((x-2)^2=x^2-4x+4\).
\[ x^2-5x+4=0 \]
We move all terms to one side to obtain a quadratic equation.
Step 3
\[ x^2-5x+4=(x-1)(x-4) \]
We factor the trinomial by looking for two numbers whose product is \(4\) and whose sum is \(-5\): they are \(-1\) and \(-4\).
\[ (x-1)(x-4)=0 \]
A product is zero when at least one of its factors is zero.
\[ x=1 \quad \text{or} \quad x=4 \]
Check
For \(x=1\): \[ \sqrt{1}=1,\qquad 1-2=-1 \]
The two sides are not equal, so \(x=1\) is an extraneous solution.
For \(x=4\): \[ \sqrt{4}=2,\qquad 4-2=2 \]
The two sides match, so \(x=4\) is valid.
Result
\[ \boxed{x=4} \]
Exercise 5 — level ★★☆☆☆
\[ \sqrt{x+2}=x \]
Result
\[ x=2 \]
Solution
Solution strategy
The right-hand side is \(x\). Since a square root is always non-negative, we must have \(x\ge0\).
Step 1
\[ \sqrt{x+2}=x \]
The radical is isolated, so we can square both sides.
\[ x+2=x^2 \]
Squaring the radical removes the radical sign.
Step 2
\[ x^2-x-2=0 \]
We move all terms to one side to obtain a quadratic in standard form.
\[ (x-2)(x+1)=0 \]
We factor the trinomial: the product is \(-2\) and the sum is \(-1\).
\[ x=2 \quad \text{or} \quad x=-1 \]
We apply the zero-product property.
Check
\(x=-1\) cannot be accepted because it does not satisfy the condition \(x\ge0\).
For \(x=2\): \[ \sqrt{2+2}=\sqrt{4}=2 \]
The solution satisfies the original equation.
Result
\[ \boxed{x=2} \]
Exercise 6 — level ★★☆☆☆
\[ \sqrt{x+4}=x-2 \]
Result
\[ x=5 \]
Solution
Solution strategy
The radical is already isolated. Since a square root is always non-negative, we must also have \(x-2\ge0\), that is, \(x\ge2\).
Step 1
\[ \sqrt{x+4}=x-2 \]
We can square both sides to eliminate the radical.
\[ x+4=(x-2)^2 \]
The left-hand side becomes the radicand \(x+4\), while the right-hand side is the square of a binomial.
Step 2
\[ x+4=x^2-4x+4 \]
We expand \((x-2)^2\) using the formula \((a-b)^2=a^2-2ab+b^2\).
\[ x^2-5x=0 \]
We move all terms to one side and cancel \(+4\) with \(+4\).
Step 3
\[ x(x-5)=0 \]
We factor out the common factor \(x\).
\[ x=0 \quad \text{or} \quad x=5 \]
A product is zero if at least one of its factors is zero.
Check
\(x=0\) does not satisfy the condition \(x\ge2\), so we discard it.
For \(x=5\): \[ \sqrt{5+4}=\sqrt{9}=3 \]
The right-hand side is: \[ 5-2=3 \]
The two sides match, so \(x=5\) is valid.
Result
\[ \boxed{x=5} \]
Exercise 7 — level ★★★☆☆
\[ \sqrt{2x+3}=x \]
Result
\[ x=3 \]
Solution
Solution strategy
The right-hand side is \(x\). Since the left-hand side is a square root, we must have \(x\ge0\). After squaring, we check the solutions.
Step 1
\[ \sqrt{2x+3}=x \]
The radical is isolated, so we can square both sides.
\[ 2x+3=x^2 \]
Squaring the radical removes the radical.
Step 2
\[ x^2-2x-3=0 \]
We move all terms to one side to obtain a quadratic in standard form.
\[ (x-3)(x+1)=0 \]
We factor the trinomial: we need two numbers whose product is \(-3\) and whose sum is \(-2\), namely \(-3\) and \(+1\).
Step 3
\[ x-3=0 \quad \text{or} \quad x+1=0 \]
We apply the zero-product property.
\[ x=3 \quad \text{or} \quad x=-1 \]
Check
\(x=-1\) does not satisfy the condition \(x\ge0\), so we discard it.
For \(x=3\): \[ \sqrt{2\cdot3+3}=\sqrt{9}=3 \]
The right-hand side equals \(x=3\), so the equation is satisfied.
Result
\[ \boxed{x=3} \]
Exercise 8 — level ★★★☆☆
\[ \sqrt{x+5}=x-1 \]
Result
\[ x=4 \]
Solution
Solution strategy
The right-hand side must be non-negative, since it equals a square root. Hence \(x-1\ge0\), that is, \(x\ge1\).
Step 1
\[ \sqrt{x+5}=x-1 \]
The radical is isolated: we square both sides.
\[ x+5=(x-1)^2 \]
The radical disappears and the right-hand side becomes the square of a binomial.
Step 2
\[ x+5=x^2-2x+1 \]
We expand \((x-1)^2=x^2-2x+1\).
\[ x^2-3x-4=0 \]
We move all terms to one side to obtain a quadratic in standard form.
Step 3
\[ x^2-3x-4=(x-4)(x+1) \]
We factor the trinomial: the product must be \(-4\) and the sum \(-3\), so the numbers are \(-4\) and \(+1\).
\[ (x-4)(x+1)=0 \]
A product is zero when at least one of its factors is zero.
\[ x=4 \quad \text{or} \quad x=-1 \]
Check
\(x=-1\) does not satisfy the condition \(x\ge1\), so we discard it.
For \(x=4\): \[ \sqrt{4+5}=\sqrt{9}=3 \]
The right-hand side is: \[ 4-1=3 \]
The two sides match, so \(x=4\) is valid.
Result
\[ \boxed{x=4} \]
Exercise 9 — level ★★★☆☆
\[ \sqrt{x+1}+\sqrt{x}=3 \]
Result
\[ x=\dfrac{16}{9} \]
Solution
Solution strategy
The domain requires \(x\ge0\). Since there are two radicals, we isolate one of them and then square.
Step 1
\[ \sqrt{x+1}+\sqrt{x}=3 \]
We isolate \(\sqrt{x+1}\) by moving \(\sqrt{x}\) to the right-hand side.
\[ \sqrt{x+1}=3-\sqrt{x} \]
One radical is now isolated, and we can square.
Step 2
\[ x+1=(3-\sqrt{x})^2 \]
We square both sides. On the right we get the square of a binomial.
\[ x+1=9-6\sqrt{x}+x \]
Indeed, \((3-\sqrt{x})^2=9-6\sqrt{x}+x\).
Step 3
\[ 1=9-6\sqrt{x} \]
We subtract \(x\) from both sides.
\[ 6\sqrt{x}=8 \]
We move the radical term to the left and the constant to the right.
\[ \sqrt{x}=\dfrac{4}{3} \]
We divide both sides by \(6\).
Step 4
\[ x=\left(\dfrac{4}{3}\right)^2 \]
We square once more to remove the last radical.
\[ x=\dfrac{16}{9} \]
Check
We substitute \(x=\dfrac{16}{9}\) into the original equation: \[ \sqrt{\dfrac{16}{9}+1}+\sqrt{\dfrac{16}{9}} \]
We add inside the first radical: \[ \sqrt{\dfrac{25}{9}}+\sqrt{\dfrac{16}{9}} \]
We compute the two roots: \[ \dfrac{5}{3}+\dfrac{4}{3}=3 \]
The equality holds, so the solution is valid.
Result
\[ \boxed{x=\dfrac{16}{9}} \]
Exercise 10 — level ★★★☆☆
\[ \sqrt{x+6}-\sqrt{x+1}=1 \]
Result
\[ x=3 \]
Solution
Solution strategy
The two radicands require \(x+6\ge0\) and \(x+1\ge0\), so the domain is \(x\ge-1\). We isolate one radical and then square.
Step 1
\[ \sqrt{x+6}-\sqrt{x+1}=1 \]
We move \(-\sqrt{x+1}\) to the right-hand side.
\[ \sqrt{x+6}=1+\sqrt{x+1} \]
Now the radical \(\sqrt{x+6}\) is isolated.
Step 2
\[ x+6=(1+\sqrt{x+1})^2 \]
We square both sides.
\[ x+6=1+2\sqrt{x+1}+x+1 \]
We expand the square of the binomial \((1+\sqrt{x+1})^2\).
Step 3
\[ x+6=x+2+2\sqrt{x+1} \]
We add the constants \(1+1=2\).
\[ 4=2\sqrt{x+1} \]
We subtract \(x+2\) from both sides.
\[ \sqrt{x+1}=2 \]
We divide both sides by \(2\).
Step 4
\[ x+1=4 \]
We square to remove the last radical.
\[ x=3 \]
We subtract \(1\) from both sides.
Check
We substitute \(x=3\): \[ \sqrt{3+6}-\sqrt{3+1} \]
We compute the radicands: \[ \sqrt{9}-\sqrt{4} \]
We compute the roots: \[ 3-2=1 \]
The equality holds, so the solution is valid.
Result
\[ \boxed{x=3} \]
Exercise 11 — level ★★★★☆
\[ \sqrt{x+1}+\sqrt{x-2}=3 \]
Result
\[ x=3 \]
Solution
Solution strategy
The domain requires \(x\ge2\). We isolate one radical and square twice.
Step 1
\[ \sqrt{x+1}+\sqrt{x-2}=3 \]
We move \(\sqrt{x-2}\) to the right-hand side to isolate one radical.
\[ \sqrt{x+1}=3-\sqrt{x-2} \]
Step 2
\[ x+1=(3-\sqrt{x-2})^2 \]
We square to remove the first radical.
\[ x+1=9-6\sqrt{x-2}+x-2 \]
Step 3
\[ x+1=x+7-6\sqrt{x-2} \]
We combine like terms.
\[ 6\sqrt{x-2}=6 \]
\[ \sqrt{x-2}=1 \]
Step 4
\[ x-2=1 \]
We square to remove the radical.
\[ x=3 \]
Check
\[ \sqrt{3+1}+\sqrt{3-2}=2+1=3 \]
The solution is correct.
Result
\[ \boxed{x=3} \]
Exercise 12 — level ★★★★☆
\[ \sqrt{x+5}+\sqrt{x}=5 \]
Result
\[ x=4 \]
Solution
Solution strategy
The domain is \(x\ge0\). We isolate one radical and then square.
Step 1
\[ \sqrt{x+5}=5-\sqrt{x} \]
We isolate one radical so we can remove it.
Step 2
\[ x+5=(5-\sqrt{x})^2 \]
We square both sides.
\[ x+5=25-10\sqrt{x}+x \]
Step 3
\[ 5=25-10\sqrt{x} \]
We simplify by subtracting \(x\) from both sides.
\[ 10\sqrt{x}=20 \]
\[ \sqrt{x}=2 \]
Step 4
\[ x=4 \]
We square to remove the radical.
Check
\[ \sqrt{4+5}+\sqrt{4}=3+2=5 \]
Result
\[ \boxed{x=4} \]
Exercise 13 — level ★★★★☆
\[ \sqrt{3x+4}=x+2 \]
Result
\[ x=-1 \quad \text{or} \quad x=0 \]
Solution
Solution strategy
Since the right-hand side is \(x+2\), we must have \(x+2\ge0\). The radical is already isolated.
Step 1
\[ 3x+4=(x+2)^2 \]
We square both sides.
Step 2
\[ 3x+4=x^2+4x+4 \]
\[ x^2+x=0 \]
We move all terms to one side.
Step 3
\[ x(x+1)=0 \]
\[ x=0 \quad \text{or} \quad x=-1 \]
Check
Both solutions satisfy the original equation.
Result
\[ \boxed{x=-1 \quad \text{or} \quad x=0} \]
Exercise 14 — level ★★★★☆
\[ \sqrt{x+2}+\sqrt{x-2}=4 \]
Result
\[ x=\dfrac{17}{4} \]
Solution
Solution strategy
The domain requires \(x\ge2\). We isolate one radical and then square.
Step 1
\[ \sqrt{x+2}=4-\sqrt{x-2} \]
Step 2
\[ x+2=(4-\sqrt{x-2})^2 \]
\[ x+2=16-8\sqrt{x-2}+x-2 \]
Step 3
\[ x+2=x+14-8\sqrt{x-2} \]
\[ 8\sqrt{x-2}=12 \]
\[ \sqrt{x-2}=\dfrac{3}{2} \]
Step 4
\[ x-2=\dfrac{9}{4} \]
\[ x=\dfrac{17}{4} \]
Check
\[ \dfrac{5}{2}+\dfrac{3}{2}=4 \]
Result
\[ \boxed{x=\dfrac{17}{4}} \]
Exercise 15 — level ★★★★☆
\[ \sqrt{x+9}-\sqrt{x}=1 \]
Result
\[ x=16 \]
Solution
Solution strategy
The domain requires \(x\ge0\). We isolate one radical and then square.
Step 1
\[ \sqrt{x+9}=1+\sqrt{x} \]
Step 2
\[ x+9=(1+\sqrt{x})^2 \]
\[ x+9=1+2\sqrt{x}+x \]
Step 3
\[ 9=1+2\sqrt{x} \]
\[ 2\sqrt{x}=8 \]
\[ \sqrt{x}=4 \]
Step 4
\[ x=16 \]
Check
\[ 5-4=1 \]
Result
\[ \boxed{x=16} \]
Exercise 16 — level ★★★★☆
\[ \sqrt{x+3}+\sqrt{x-1}=4 \]
Result
\[ x=\dfrac{13}{4} \]
Solution
Solution strategy
The domain requires \(x-1\ge0\), that is, \(x\ge1\). We isolate one of the two radicals and then square.
Step 1
\[ \sqrt{x+3}+\sqrt{x-1}=4 \]
We move \(\sqrt{x-1}\) to the right-hand side to isolate \(\sqrt{x+3}\).
\[ \sqrt{x+3}=4-\sqrt{x-1} \]
Now one radical is isolated, so we can square.
Step 2
\[ x+3=(4-\sqrt{x-1})^2 \]
We square both sides. On the right we have the square of a binomial.
\[ x+3=16-8\sqrt{x-1}+x-1 \]
We expand \((4-\sqrt{x-1})^2\): the cross term is \(-8\sqrt{x-1}\).
Step 3
\[ x+3=x+15-8\sqrt{x-1} \]
We combine the constants \(16-1=15\).
\[ 3=15-8\sqrt{x-1} \]
We subtract \(x\) from both sides.
\[ 8\sqrt{x-1}=12 \]
We move the radical term to the left and the constant to the right.
\[ \sqrt{x-1}=\dfrac{3}{2} \]
We divide both sides by \(8\).
Step 4
\[ x-1=\left(\dfrac{3}{2}\right)^2 \]
We square to remove the last radical.
\[ x-1=\dfrac{9}{4} \]
We compute the square of \(\dfrac{3}{2}\).
\[ x=1+\dfrac{9}{4}=\dfrac{13}{4} \]
We add \(1\) to both sides.
Check
We substitute \(x=\dfrac{13}{4}\): \[ \sqrt{\dfrac{13}{4}+3}+\sqrt{\dfrac{13}{4}-1} \]
We compute the radicands: \[ \sqrt{\dfrac{25}{4}}+\sqrt{\dfrac{9}{4}} \]
We compute the roots: \[ \dfrac{5}{2}+\dfrac{3}{2}=4 \]
The equality holds, so the solution is valid.
Result
\[ \boxed{x=\dfrac{13}{4}} \]
Exercise 17 — level ★★★★★
\[ \sqrt{2x+1}+\sqrt{x-3}=4 \]
Result
\[ x=4 \]
Solution
Solution strategy
The domain requires \(x-3\ge0\), that is, \(x\ge3\). We isolate one radical: after the first squaring a radical will still remain, so we will need to square a second time.
Step 1
\[ \sqrt{2x+1}+\sqrt{x-3}=4 \]
We move \(\sqrt{x-3}\) to the right-hand side.
\[ \sqrt{2x+1}=4-\sqrt{x-3} \]
In this way \(\sqrt{2x+1}\) is isolated.
Step 2
\[ 2x+1=(4-\sqrt{x-3})^2 \]
We square both sides.
\[ 2x+1=16-8\sqrt{x-3}+x-3 \]
We expand the square of the binomial.
Step 3
\[ 2x+1=x+13-8\sqrt{x-3} \]
We combine the constants \(16-3=13\).
\[ 8\sqrt{x-3}=12-x \]
We isolate the remaining radical.
Step 4
\[ 64(x-3)=(12-x)^2 \]
We square both sides to eliminate the second radical.
Step 5
\[ 64x-192=x^2-24x+144 \]
We expand both sides: on the left we distribute \(64\), and on the right we expand \((12-x)^2\).
\[ x^2-88x+336=0 \]
We move all terms to one side and combine like terms.
Step 6
\[ x^2-88x+336=(x-4)(x-84) \]
We factor the trinomial: \(4\cdot84=336\) and \(4+84=88\).
\[ (x-4)(x-84)=0 \]
We apply the zero-product property.
\[ x=4 \quad \text{or} \quad x=84 \]
Check
For \(x=4\): \[ \sqrt{2\cdot4+1}+\sqrt{4-3} = \sqrt{9}+\sqrt{1} = 3+1=4 \]
Hence \(x=4\) is valid.
For \(x=84\): \[ \sqrt{2\cdot84+1}+\sqrt{84-3} = \sqrt{169}+\sqrt{81} = 13+9=22 \]
The result is not \(4\), so \(x=84\) is an extraneous solution.
Result
\[ \boxed{x=4} \]
Exercise 18 — level ★★★★★
\[ \sqrt{x+13}-\sqrt{x+4}=1 \]
Result
\[ x=12 \]
Solution
Solution strategy
The domain requires \(x+4\ge0\), that is, \(x\ge-4\). We isolate one radical and then square.
Step 1
\[ \sqrt{x+13}-\sqrt{x+4}=1 \]
We move \(-\sqrt{x+4}\) to the right-hand side.
\[ \sqrt{x+13}=1+\sqrt{x+4} \]
Now \(\sqrt{x+13}\) is isolated.
Step 2
\[ x+13=(1+\sqrt{x+4})^2 \]
We square both sides.
\[ x+13=1+2\sqrt{x+4}+x+4 \]
We expand the square of the binomial \((1+\sqrt{x+4})^2\).
Step 3
\[ x+13=x+5+2\sqrt{x+4} \]
We combine the constants \(1+4=5\).
\[ 8=2\sqrt{x+4} \]
We subtract \(x+5\) from both sides.
\[ \sqrt{x+4}=4 \]
We divide both sides by \(2\).
Step 4
\[ x+4=16 \]
We square to remove the last radical.
\[ x=12 \]
We subtract \(4\) from both sides.
Check
We substitute \(x=12\): \[ \sqrt{12+13}-\sqrt{12+4} \]
We compute the radicands: \[ \sqrt{25}-\sqrt{16} \]
We compute the roots: \[ 5-4=1 \]
The equality holds, so the solution is valid.
Result
\[ \boxed{x=12} \]
Exercise 19 — level ★★★★★
\[ \sqrt{2x+3}-\sqrt{x+1}=1 \]
Result
\[ x=-1 \quad \text{or} \quad x=3 \]
Solution
Solution strategy
The domain requires \(x+1\ge0\), that is, \(x\ge-1\). We isolate one radical and then square.
Step 1
\[ \sqrt{2x+3}-\sqrt{x+1}=1 \]
We move \(-\sqrt{x+1}\) to the right-hand side.
\[ \sqrt{2x+3}=1+\sqrt{x+1} \]
Now \(\sqrt{2x+3}\) is isolated.
Step 2
\[ 2x+3=(1+\sqrt{x+1})^2 \]
We square both sides.
\[ 2x+3=1+2\sqrt{x+1}+x+1 \]
We expand the square of the binomial.
Step 3
\[ 2x+3=x+2+2\sqrt{x+1} \]
We combine the constants.
\[ x+1=2\sqrt{x+1} \]
We subtract \(x+2\) from both sides and isolate the radical expression.
Step 4
Let us set: \[ t=\sqrt{x+1} \]
This substitution is helpful because the equation contains both \(x+1\) and \(\sqrt{x+1}\). Moreover, since \(t\) is a square root, we have \(t\ge0\).
Since: \[ t=\sqrt{x+1} \]
we obtain: \[ t^2=x+1 \]
We replace \(x+1\) by \(t^2\) and \(\sqrt{x+1}\) by \(t\).
\[ t^2=2t \]
Step 5
\[ t^2-2t=0 \]
We move all terms to one side.
\[ t(t-2)=0 \]
We factor out the common factor \(t\).
\[ t=0 \quad \text{or} \quad t=2 \]
We apply the zero-product property.
Step 6
If \(t=0\), then: \[ \sqrt{x+1}=0 \]
Squaring: \[ x+1=0 \implies x=-1 \]
If \(t=2\), then: \[ \sqrt{x+1}=2 \]
Squaring: \[ x+1=4 \implies x=3 \]
Check
For \(x=-1\): \[ \sqrt{2(-1)+3}-\sqrt{-1+1} = \sqrt{1}-\sqrt{0} = 1-0=1 \]
Hence \(x=-1\) is valid.
For \(x=3\): \[ \sqrt{2\cdot3+3}-\sqrt{3+1} = \sqrt{9}-\sqrt{4} = 3-2=1 \]
So \(x=3\) is also valid.
Result
\[ \boxed{x=-1 \quad \text{or} \quad x=3} \]
Exercise 20 — level ★★★★★
\[ \sqrt{x+9}+\sqrt{x-3}=6 \]
Result
\[ x=7 \]
Solution
Solution strategy
The domain requires \(x-3\ge0\), that is, \(x\ge3\). We isolate one radical and then square.
Step 1
\[ \sqrt{x+9}+\sqrt{x-3}=6 \]
We move \(\sqrt{x-3}\) to the right-hand side.
\[ \sqrt{x+9}=6-\sqrt{x-3} \]
Now \(\sqrt{x+9}\) is isolated.
Step 2
\[ x+9=(6-\sqrt{x-3})^2 \]
We square both sides.
\[ x+9=36-12\sqrt{x-3}+x-3 \]
We expand the square of the binomial \((6-\sqrt{x-3})^2\).
Step 3
\[ x+9=x+33-12\sqrt{x-3} \]
We combine the constants \(36-3=33\).
\[ 9=33-12\sqrt{x-3} \]
We subtract \(x\) from both sides.
\[ 12\sqrt{x-3}=24 \]
We isolate the radical term.
\[ \sqrt{x-3}=2 \]
We divide both sides by \(12\).
Step 4
\[ x-3=4 \]
We square to remove the last radical.
\[ x=7 \]
We add \(3\) to both sides.
Check
We substitute \(x=7\): \[ \sqrt{7+9}+\sqrt{7-3} \]
We compute the radicands: \[ \sqrt{16}+\sqrt{4} \]
We compute the roots: \[ 4+2=6 \]
The equality holds, so the solution is valid.
Result
\[ \boxed{x=7} \]