Lesson 9.1: Solve Quadratic Equations Using the Square Root Property

New Concept

The Square Root Property is a powerful tool for solving quadratic equations where you can isolate the squared term. This lesson shows how to solve equations like $ax^2=k$ and $a(x-h)^2=k$, providing a direct path to the solution.

What’s next

This is just the beginning. Next, you'll tackle interactive examples and practice problems to master isolating the variable and simplifying radicals.

Property

If $x^2 = k$, then

This is also written as $x = \pm \sqrt{k}$.
Notice that the Square Root Property gives two solutions to an equation of the form $x^2 = k$: the principal square root of $k$ and its opposite.

Examples

• To solve $x^2 = 64$, we apply the Square Root Property to get $x = \pm\sqrt{64}$. The two solutions are $x = 8$ and $x = -8$.
• To solve $p^2 = 19$, we use the property to find $p = \pm\sqrt{19}$. Since 19 is not a perfect square, we leave the answer in radical form.
• To solve $m^2 = -36$, we find $m = \pm\sqrt{-36}$. This simplifies to $m = \pm 6i$, giving two complex solutions.

Explanation

This property works because squaring a positive number and its negative opposite both result in the same positive value. This rule helps you find both original numbers by taking the square root and considering both the positive and negative results.

Property

To solve a quadratic equation using the square root property:
Step 1. Isolate the quadratic term and make its coefficient one. For an equation like $ax^2 = k$, this means rewriting it as $x^2 = \frac{k}{a}$.
Step 2. Use the Square Root Property: $x = \pm\sqrt{\frac{k}{a}}$.
Step 3. Simplify the radical.
Step 4. Check the solutions.

Examples

• Solve $5x^2 = 125$. First, divide by 5 to isolate $x^2$, giving $x^2 = 25$. Then, use the Square Root Property to find $x = \pm\sqrt{25}$, so the solutions are $x=5$ and $x=-5$.
• Solve $\frac{3}{4}u^2 + 2 = 29$. Subtract 2 to get $\frac{3}{4}u^2 = 27$. Multiply by $\frac{4}{3}$ to get $u^2 = 36$. The solutions are $u = \pm\sqrt{36}$, so $u=6$ and $u=-6$.
• Solve $3x^2 - 9 = 54$. Add 9 to get $3x^2 = 63$. Divide by 3 to get $x^2 = 21$. The solutions are $x = \pm\sqrt{21}$.

Explanation

Before you can use the Square Root Property, the $x^2$ term must be by itself. This method is about clearing away any coefficients or constants first. Once you have $x^2$ isolated, you can solve for the two possible values of $x$.

Property

We can use the Square Root Property to solve an equation of the form $a(x-h)^2 = k$ as well. The first step is to isolate the term that has the variable squared, which in this case is the binomial $(x-h)$.

1. Isolate the squared binomial: $(x-h)^2 = \frac{k}{a}$.
2. Use the Square Root Property: $x-h = \pm\sqrt{\frac{k}{a}}$.
3. Solve for $x$ by adding $h$ to both sides: $x = h \pm\sqrt{\frac{k}{a}}$.

Examples

• Solve $2(y-4)^2 = 98$. Divide by 2 to get $(y-4)^2 = 49$. Take the square root: $y-4 = \pm 7$. This gives two equations: $y-4=7$ (so $y=11$) and $y-4=-7$ (so $y=-3$).
• Solve $(x - \frac{1}{4})^2 = \frac{11}{16}$. Use the property: $x - \frac{1}{4} = \pm\sqrt{\frac{11}{16}} = \pm\frac{\sqrt{11}}{4}$. The solutions are $x = \frac{1}{4} \pm \frac{\sqrt{11}}{4}$.
• Solve $(3x-1)^2 = -27$. Use the property: $3x-1 = \pm\sqrt{-27} = \pm 3\sqrt{3}i$. Add 1 to get $3x = 1 \pm 3\sqrt{3}i$. Divide by 3 to get $x = \frac{1}{3} \pm \sqrt{3}i$.

Explanation

This method extends the Square Root Property to expressions in parentheses. Treat the entire binomial, like $(x-h)$, as a single squared item. Get that squared group alone, take the square root of both sides, and then solve for $x$.

Property

If one side of an equation is a perfect square trinomial, it can be factored to create an equation in the form $(ax \pm b)^2 = k$.

1. Factor the perfect square trinomial. For example, $4n^2 + 4n + 1$ factors to $(2n+1)^2$.
2. Rewrite the equation in the form $(ax \pm b)^2 = k$.
3. Solve by isolating the binomial and using the Square Root Property.

Examples

• Solve $x^2 + 12x + 36 = 81$. The left side factors to $(x+6)^2 = 81$. Take the square root: $x+6 = \pm 9$. The two solutions are $x=3$ and $x=-15$.
• Solve $9m^2 - 30m + 25 = 4$. Factor the left side to get $(3m-5)^2 = 4$. Then, $3m-5 = \pm 2$. This leads to $3m=7$ (so $m=\frac{7}{3}$) and $3m=3$ (so $m=1$).
• Solve $y^2 + 8y + 16 = 20$. Factor the trinomial to get $(y+4)^2 = 20$. Take the square root: $y+4 = \pm\sqrt{20} = \pm 2\sqrt{5}$. The solutions are $y = -4 \pm 2\sqrt{5}$.

Explanation

This is a smart way to simplify a complex equation. By recognizing and factoring a perfect square trinomial, you can turn an equation like $ax^2+bx+c=k$ into the much simpler $(x-h)^2=k$ form, which is easy to solve.