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Solve Quadratic Equations Using the Square Root Property

Solve quadratic equations of the form

Solve Quadratic Equations of the form

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation x² = 9.

	\({x}^{2}=9\)
Put the equation in standard form.	\({x}^{2}-9=0\)
Factor the difference of squares.	\((x-3)(x+3)=0\)
Use the Zero Product Property.	\(x-3=0\ x-3=0\)
Solve each equation.	\(x=3\ x=-3\)

We can easily use factoring to find the solutions of similar equations, like x² = 16 and x² = 25, because 16 and 25 are perfect squares. In each case, we would get two solutions, \(x=4,x=-4\) and \(x=5,x=-5.\)

But what happens when we have an equation like x² = 7? Since 7 is not a perfect square, we cannot solve the equation by factoring.

Previously we learned that since 169 is the square of 13, we can also say that 13 is a square root of 169. Also, (−13)² = 169, so −13 is also a square root of 169. Therefore, both 13 and −13 are square roots of 169. So, every positive number has two square roots, one positive and one negative. We earlier defined the square root of a number in this way:

\[\text{If}\ {n}^{2}=m,\ \text{then}\ n\ \text{is a square root of}\ m.\]

Since these equations are all of the form x² = k, the square root definition tells us the solutions are the two square roots of k. This leads to the Square Root Property.

Notice that the Square Root Property gives two solutions to an equation of the form x² = k, the principal square root of \(k\) and its opposite. We could also write the solution as \(x=\pm \sqrt{k}.\) We read this as x equals positive or negative the square root of k.

Now we will solve the equation x² = 9 again, this time using the Square Root Property.

	\(\ {x}^{2}=9\)
Use the Square Root Property.	\(\ x=\pm \sqrt{9}\)
	\(\ x=\pm 3\)
	\(\text{So}\ x=3\ \text{or}\ x=-3.\)

	\({x}^{2}=7\)
Use the Square Root Property.	\(x=\sqrt{7},\ x=\text{-}\sqrt{7}\)

How to solve a Quadratic Equation of the form

Try it.

Solve: \({x}^{2}-50=0.\)

Solution

Example

Try it.

Solve: \(3{z}^{2}=108.\)

Solution

	\(3{z}^{2}=108\)
The quadratic term is isolated. Divide by 3 to make its coefficient 1.	\(\frac{3{z}^{2}}{3}=\frac{108}{3}\)
Simplify.	\({z}^{2}=36\)
Use the Square Root Property.	\(\ z=\pm \sqrt{36}\)
Simplify the radical.	\(\ z=\pm 6\)
Rewrite to show two solutions.	\(z=6,\ z=-6\)
Check the solutions:

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Solve Quadratic Equations of the Form

We can use the Square Root Property to solve an equation of the form a(x − h)² = k as well. Notice that the quadratic term, x, in the original form ax² = k is replaced with (x − h).

The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of a, then the Square Root Property can be used on (x − h)².

Example

Try it.

Solve: \(4{(y-7)}^{2}=48.\)

Solution

	\(4{(y-7)}^{2}=48\)
Divide both sides by the coefficient 4.	\(\ {(y-7)}^{2}=12\)
Use the Square Root Property on the binomial	\(\ y-7=\pm \sqrt{12}\)
Simplify the radical.	\(\ y-7=\pm 2\sqrt{3}\)
Solve for \(y.\)	\(\ y=7\pm 2\sqrt{3}\)
Rewrite to show two solutions.	\(y=7+2\sqrt{3},\) \(y=7-2\sqrt{3}\)
Check:

Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

Example

Try it.

Solve: \({(x-\frac{1}{3})}^{2}=\frac{5}{9}.\)

Solution

	\(\ {(x-\frac{1}{3})}^{2}=\frac{5}{9}\)
Use the Square Root Property.	\(\ x-\frac{1}{3}=\pm \sqrt{\frac{5}{9}}\)
Rewrite the radical as a fraction of square roots.	\(\ x-\frac{1}{3}=\pm \frac{\sqrt{5}}{\sqrt{9}}\)
Simplify the radical.	\(\ x-\frac{1}{3}=\pm \frac{\sqrt{5}}{3}\)
Solve for \(x\).	\(\ x=\frac{1}{3}\pm \frac{\sqrt{5}}{3}\)
Rewrite to show two solutions.	\(x=\frac{1}{3}+\frac{\sqrt{5}}{3},\ x=\frac{1}{3}-\frac{\sqrt{5}}{3}\)
Check: We leave the check for you.

We will start the solution to the next example by isolating the binomial term.

Example

Try it.

Solve: \(2{(x-2)}^{2}+3=57.\)

Solution

	\(2{(x-2)}^{2}+3=57\)
Subtract 3 from both sides to isolate the binomial term.	\(2{(x-2)}^{2}=54\)
Divide both sides by 2.	\({(x-2)}^{2}=27\)
Use the Square Root Property.	\(x-2=\pm \sqrt{27}\)
Simplify the radical.	\(x-2=\pm 3\sqrt{3}\)
Solve for \(x\).	\(x=2\pm 3\sqrt{3}\)
Rewrite to show two solutions.	\(x=2+3\sqrt{3},\ x=2-3\sqrt{3}\)
Check: We leave the check for you.

Sometimes the solutions are complex numbers.

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Key Concepts

Square Root Property
- If \({x}^{2}=k\), then \(x=\sqrt{k}\ \text{or}\ x=\text{-}\sqrt{k}\) or \(x=\pm \sqrt{k}\)
How to solve a quadratic equation using the square root property.
1. Isolate the quadratic term and make its coefficient one.
2. Use Square Root Property.
3. Simplify the radical.
4. Check the solutions.

Solve Quadratic Equations Using the Square Root Property

Solve Quadratic Equations of the Form ax² = k Using the Square Root Property

In the following exercises, solve each equation.

Try it.

\({a}^{2}=49\)

Solution

\(a=\pm 7\)

Try it.

\({b}^{2}=144\)

Try it.

\({r}^{2}-24=0\)

Solution

\(r=\pm 2\sqrt{6}\)

Try it.

\({t}^{2}-75=0\)

Try it.

\({u}^{2}-300=0\)

Solution

\(u=\pm 10\sqrt{3}\)

Try it.

\({v}^{2}-80=0\)

Try it.

\(4{m}^{2}=36\)

Solution

\(m=\pm 3\)

Try it.

\(3{n}^{2}=48\)

Try it.

\(\frac{4}{3}{x}^{2}=48\)

Solution

\(x=\pm 6\)

Try it.

\(\frac{5}{3}{y}^{2}=60\)

Try it.

\({x}^{2}+25=0\)

Solution

\(x=\pm 5i\)

Try it.

\({y}^{2}+64=0\)

Try it.

\({x}^{2}+63=0\)

Solution

\(x=\pm 3\sqrt{7}i\)

Try it.

\({y}^{2}+45=0\)

Try it.

\(\frac{4}{3}{x}^{2}+2=110\)

Solution

\(x=\pm 9\)

Try it.

\(\frac{2}{3}{y}^{2}-8=-2\)

Try it.

\(\frac{2}{5}{a}^{2}+3=11\)

Solution

\(a=\pm 2\sqrt{5}\)

Try it.

\(\frac{3}{2}{b}^{2}-7=41\)

Try it.

\(7{p}^{2}+10=26\)

Solution

\(p=\pm \frac{4\sqrt{7}}{7}\)

Try it.

\(2{q}^{2}+5=30\)

Try it.

\(5{y}^{2}-7=25\)

Solution

\(y=\pm \frac{4\sqrt{10}}{5}\)

Try it.

\(3{x}^{2}-8=46\)

Solve Quadratic Equations of the Form a(x − h)² = k Using the Square Root Property

In the following exercises, solve each equation.

Try it.

\({(u-6)}^{2}=64\)

Solution

\(u=14,u=-2\)

Try it.

\({(v+10)}^{2}=121\)

Try it.

\({(m-6)}^{2}=20\)

Solution

\(m=6\pm 2\sqrt{5}\)

Try it.

\({(n+5)}^{2}=32\)

Try it.

\({(r-\frac{1}{2})}^{2}=\frac{3}{4}\)

Solution

\(r=\frac{1}{2}\pm \frac{\sqrt{3}}{2}\)

Try it.

\({(x+\frac{1}{5})}^{2}=\frac{7}{25}\)

Try it.

\({(y+\frac{2}{3})}^{2}=\frac{8}{81}\)

Solution

\(y=-\frac{2}{3}\pm \frac{2\sqrt{2}}{9}\)

Try it.

\({(t-\frac{5}{6})}^{2}=\frac{11}{25}\)

Try it.

\({(a-7)}^{2}+5=55\)

Solution

\(a=7\pm 5\sqrt{2}\)

Try it.

\({(b-1)}^{2}-9=39\)

Try it.

\(4{(x+3)}^{2}-5=27\)

Solution

\(x=-3\pm 2\sqrt{2}\)

Try it.

\(5{(x+3)}^{2}-7=68\)

Try it.

\({(5c+1)}^{2}=-27\)

Solution

\(c=-\frac{1}{5}\pm \frac{3\sqrt{3}}{5}i\)

Try it.

\({(8d-6)}^{2}=-24\)

Try it.

\({(4x-3)}^{2}+11=-17\)

Solution

\(x=\frac{3}{4}\pm \frac{\sqrt{7}}{2}i\)

Try it.

\({(2y+1)}^{2}-5=-23\)

Try it.

\({m}^{2}-4m+4=8\)

Solution

\(m=2\pm 2\sqrt{2}\)

Try it.

\({n}^{2}+8n+16=27\)

Try it.

\({x}^{2}-6x+9=12\)

Solution

\(x=3\pm 2\sqrt{3}\)

Try it.

\({y}^{2}+12y+36=32\)

Try it.

\(25{x}^{2}-30x+9=36\)

Solution

\(x=-\frac{3}{5},x=\frac{9}{5}\)

Try it.

\(9{y}^{2}+12y+4=9\)

Try it.

\(36{x}^{2}-24x+4=81\)

Solution

\(x=-\frac{7}{6},x=\frac{11}{6}\)

Try it.

\(64{x}^{2}+144x+81=25\)

Try it.

In your own words, explain the Square Root Property.

Solution

Answers will vary.

Try it.

In your own words, explain how to use the Square Root Property to solve the quadratic equation \({(x+2)}^{2}=16\).

Condensed: the full section is in OpenStax Intermediate Algebra 2e.

Solve Quadratic Equations of the Form

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation \({x}^{2}=9\).

\[\begin{array}{llllll} & & & {x}^{2} & = & 9 \\ \text{Put the equation in standard form.} & & & {x}^{2}-9 & = & 0 \\ \text{Factor the left side.} & & & (x-3)(x+3) & = & 0 \\ \text{Use the Zero Product Property.} & & & (x-3)=0,\ (x+3) & = & 0 \\ \text{Solve each equation.} & & & x=3,\ x & = & -3 \\ \text{Combine the two solutions into}\ \pm \ \text{form.} & & & x & = & \pm \ 3 \\ \text{(The solution is read}\ ‘x\ \text{is equal to positive or negative 3.’)} & & \end{array}\]

We can easily use factoring to find the solutions of similar equations, like \({x}^{2}=16\) and \({x}^{2}=25\), because 16 and 25 are perfect squares. But what happens when we have an equation like \({x}^{2}=7\)? Since 7 is not a perfect square, we cannot solve the equation by factoring.

These equations are all of the form \({x}^{2}=k\).
We defined the square root of a number in this way:

\[\text{If}\ {n}^{2}=m,\ \text{then}\ n\ \text{is a square root of}\ m.\]

This leads to the Square Root Property.

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. We could also write the solution as \(x=\pm \ \sqrt{k}\).

Now, we will solve the equation \({x}^{2}=9\) again, this time using the Square Root Property.

\[\begin{array}{llllll} & & & {x}^{2} & = & 9 \\ \text{Use the Square Root Property.} & & & x & = & \pm \ \sqrt{9} \\ \text{Simplify the radical.} & & & x & = & \pm \ 3 \\ \text{Rewrite to show the two solutions.} & & & x=3,x & = & -3\end{array}\]

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation \({x}^{2}=7\).

\[\begin{array}{llllllll}\begin{array}{l} \\ \\ \\ \text{Use the Square Root Property.}\end{array} & & & \begin{array}{lll}{x}^{2} & = & 7 \\ x & = & \pm \ \sqrt{7}\end{array} \\ \text{Rewrite to show two solutions.} & & & x=\sqrt{7},\ x=\text{-}\sqrt{7} \\ \text{We cannot simplify}\ \sqrt{7},\ \text{so we leave the answer as a radical.} & & \end{array}\]

Example

Try it.

Solve: \({x}^{2}=169\).

Solution

\(\begin{array}{llllllllll}\begin{array}{l} \\ \\ \\ \text{Use the Square Root Property.} \\ \text{Simplify the radical.}\end{array} & & & \ \begin{array}{lll}{x}^{2} & = & 169 \\ x & = & \pm \ \sqrt{169} \\ x & = & \pm \ 13\end{array} \\ \text{Rewrite to show two solutions.} & & & \ x=13,\ x=-13\end{array}\)

Condensed: the full section is in OpenStax Elementary Algebra 2e.

Solve Quadratic Equations of the Form

We can use the Square Root Property to solve an equation like \({(x-3)}^{2}=16\), too. We will treat the whole binomial, \((x-3)\), as the quadratic term.

Example

Try it.

Solve: \({(x-3)}^{2}=16\).

Solution

	\({(x-3)}^{2}=16\)
Use the Square Root Property.	\(\ x-3=\pm \ \sqrt{16}\)
Simplify.	\(\ x-3=\pm \ 4\)
Write as two equations.	\(x-3=4,x-3=\text{-}4\)
Solve.	\(\ x=7,x=\text{-}1\)
Check.

Example

Try it.

Solve: \({(y-7)}^{2}=12\).

Solution

	\({(y-7)}^{2}=12\)
Use the Square Root Property.	\(\ y-7=\pm \ \sqrt{12}\)
Simplify the radical.	\(\ y-7=\pm \ 2\sqrt{3}\)
Solve for y.	\(\ y=7\pm 2\sqrt{3}\)
Rewrite to show two solutions.	\(y=7+2\sqrt{3},y=7-2\sqrt{3}\)
Check.

Remember, when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

Example

Try it.

Solve: \({(x-\frac{1}{2})}^{2}=\frac{5}{4}.\)

Solution

Use the Square Root Property. Rewrite the radical as a fraction of square roots. Simplify the radical. Solve for x.	\(\begin{array}{lll} \\ {(x-\frac{1}{2})}^{2} & = & \frac{5}{4} \\ x-\frac{1}{2} & = & \pm \ \sqrt{\frac{5}{4}} \\ x-\frac{1}{2} & = & \pm \ \frac{\sqrt{5}}{\sqrt{4}} \\ x-\frac{1}{2} & = & \pm \ \frac{\sqrt{5}}{2} \\ x & = & \frac{1}{2}\pm \frac{\sqrt{5}}{2}\end{array}\)
Rewrite to show two solutions.	\(x=\frac{1}{2}+\frac{\sqrt{5}}{2},\ x=\frac{1}{2}-\frac{\sqrt{5}}{2}\)
Check. We leave the check for you.

We will start the solution to the next example by isolating the binomial.

Example

Try it.

Solve: \({(x-2)}^{2}+3=30\).

Solution

	\(\ \begin{array}{lll} \\ {(x-2)}^{2}+3 & = & 30\end{array}\)
Isolate the binomial term.	\(\ \begin{array}{lll} \\ {(x-2)}^{2} & = & 27\end{array}\)
Use the Square Root Property.	\(\ \begin{array}{lll} \\ x-2 & = & \pm \ \sqrt{27}\end{array}\)
Simplify the radical.	\(\ \begin{array}{lll} \\ x-2 & = & \pm \ 3\sqrt{3}\end{array}\)
Solve for x.	\(\ \begin{array}{lll} \\ x & = & 2\pm 3\sqrt{3}\end{array}\)
Rewrite to show two solutions.	\(\ x=2+3\sqrt{3},\ x=2-3\sqrt{3}\)
Check. We leave the check for you.

Condensed: the full section is in OpenStax Elementary Algebra 2e.

Solve Quadratic Equations Using the Square Root Property

Solve Quadratic Equations of the form \(a{x}^{2}=k\) Using the Square Root Property

In the following exercises, solve the following quadratic equations.

Try it.

\({a}^{2}=49\)

Solution

\(a=\pm \ 7\)

Try it.

\({b}^{2}=144\)

Try it.

\({r}^{2}-24=0\)

Solution

\(r=\pm \ 2\sqrt{6}\)

Try it.

\({t}^{2}-75=0\)

Try it.

\({u}^{2}-300=0\)

Solution

\(u=\pm \ 10\sqrt{3}\)

Try it.

\({v}^{2}-80=0\)

Try it.

\(4{m}^{2}=36\)

Solution

\(m=\pm \ 3\)

Try it.

\(3{n}^{2}=48\)

Try it.

\({x}^{2}+20=0\)

Solution

no real solution

Try it.

\({y}^{2}+64=0\)

Try it.

\(\frac{2}{5}{a}^{2}+3=11\)

Solution

\(a=\pm \ 2\sqrt{5}\)

Try it.

\(\frac{3}{2}{b}^{2}-7=41\)

Try it.

\(7{p}^{2}+10=26\)

Solution

\(p=\pm \ \frac{4\sqrt{7}}{7}\)

Try it.

\(2{q}^{2}+5=30\)

Solve Quadratic Equations of the Form \(a{(x-h)}^{2}=k\) Using the Square Root Property

In the following exercises, solve the following quadratic equations.

Try it.

\({(x+2)}^{2}=9\)

Solution

\(x=1,x=-5\)

Try it.

\({(y-5)}^{2}=36\)

Try it.

\({(u-6)}^{2}=64\)

Solution

\(u=14,u=-2\)

Try it.

\({(v+10)}^{2}=121\)

Try it.

\({(m-6)}^{2}=20\)

Solution

\(m=6\pm 2\sqrt{5}\)

Try it.

\({(n+5)}^{2}=32\)

Try it.

\({(r-\frac{1}{2})}^{2}=\frac{3}{4}\)

Solution

\(r=\frac{1}{2}\pm \frac{\sqrt{3}}{2}\)

Try it.

\({(t-\frac{5}{6})}^{2}=\frac{11}{25}\)

Try it.

\({(a-7)}^{2}+5=55\)

Solution

\(a=7\pm 5\sqrt{2}\)

Try it.

\({(b-1)}^{2}-9=39\)

Try it.

\({(5c+1)}^{2}=-27\)

Solution

no real solution

Try it.

\({(8d-6)}^{2}=-24\)

Try it.

\({m}^{2}-4m+4=8\)

Solution

\(m=2\pm 2\sqrt{2}\)

Try it.

\({n}^{2}+8n+16=27\)

Try it.

\(25{x}^{2}-30x+9=36\)

Solution

\(x=-\frac{3}{5},x=\frac{9}{5}\)

Try it.

\(9{y}^{2}+12y+4=9\)

Mixed Practice

In the following exercises, solve using the Square Root Property.

Try it.

\(2{r}^{2}=32\)

Solution

\(r=\pm \ 4\)

Try it.

\(4{t}^{2}=16\)

Try it.

\({(a-4)}^{2}=28\)

Solution

\(a=4\pm 2\sqrt{7}\)

Try it.

\({(b+7)}^{2}=8\)

Try it.

\(9{w}^{2}-24w+16=1\)

Solution

\(w=1,w=\frac{5}{3}\)

Try it.

\(4{z}^{2}+4z+1=49\)

Try it.

\({a}^{2}-18=0\)

Solution

\(a=\pm \ 3\sqrt{2}\)

Try it.

\({b}^{2}-108=0\)

Try it.

\({(p-\frac{1}{3})}^{2}=\frac{7}{9}\)

Solution

\(p=\frac{1}{3}\pm \frac{\sqrt{7}}{3}\)

Try it.

\({(q-\frac{3}{5})}^{2}=\frac{3}{4}\)

Try it.

\({m}^{2}+12=0\)

Solution

no real solution

Try it.

\({n}^{2}+48=0\)

Try it.

\({u}^{2}-14u+49=72\)

Solution

\(u=7\pm 6\sqrt{2}\)

Try it.

\({v}^{2}+18v+81=50\)

Try it.

\({(m-4)}^{2}+3=15\)

Solution

\(m=4\pm 2\sqrt{3}\)

Try it.

\({(n-7)}^{2}-8=64\)

Try it.

\({(x+5)}^{2}=4\)

Solution

\(x=-3,x=-7\)

Try it.

\({(y-4)}^{2}=64\)

Try it.

\(6{c}^{2}+4=29\)

Solution

\(c=\pm \ \frac{5\sqrt{6}}{6}\)

Try it.

\(2{d}^{2}-4=77\)

Try it.

\({(x-6)}^{2}+7=3\)

Solution

no real solution

Try it.

\({(y-4)}^{2}+10=9\)

Try it.

Explain why the equation \({x}^{2}+12=8\) has no solution.

Solution

Answers will vary.

Try it.

Explain why the equation \({y}^{2}+8=12\) has two solutions.

Condensed: the full section is in OpenStax Elementary Algebra 2e.

Ankehitriny ianao Tsy misy milina fanisana afaka mamaha io olana io, saingy azo atao ny maminavina ny ampahany amin'io. Andramo ny iray eto ambany, na soraty ny anao manokana.

Fanazaran-tena (40)

Andramo amin'ny taratasy aloha ny tsirairay. Asehoy ny valinteny mba hijerena; azo sokafana ao amin'ny mpamolavola ny valinteny voaporofo ho an'ny dingana tsirairay.

Simplify: \(\sqrt{128}.\)

Asehoy ny valinteny

\(8\sqrt{2}\)
Simplify: \(\sqrt{\frac{32}{5}}\).

Asehoy ny valinteny

\(\frac{4\sqrt{10}}{5}\)
Factor: \(9{x}^{2}-12x+4\).

Asehoy ny valinteny

\({\left(3x-2\right)}^{2}\)
Solve: \({x}^{2}-50=0.\)
Solve: \({x}^{2}-48=0.\)

Asehoy ny valinteny

\(x=4\sqrt{3},x=-4\sqrt{3}\)
Solve: \({y}^{2}-27=0.\)

Asehoy ny valinteny

\(y=3\sqrt{3},y=-3\sqrt{3}\)

Solve: \(3{z}^{2}=108.\)

Asehoy ny valinteny

	\(3{z}^{2}=108\)
The quadratic term is isolated. Divide by 3 to make its coefficient 1.	\(\frac{3{z}^{2}}{3}=\frac{108}{3}\)
Simplify.	\({z}^{2}=36\)
Use the Square Root Property.	\(\ z=\pm \sqrt{36}\)
Simplify the radical.	\(\ z=\pm 6\)
Rewrite to show two solutions.	\(z=6,\ z=-6\)
Check the solutions:

Solve: \(2{x}^{2}=98.\)

Asehoy ny valinteny

\(x=7,x=-7\)
Solve: \(5{m}^{2}=80.\)

Asehoy ny valinteny

\(m=4,m=-4\)

Solve: \({x}^{2}+72=0\).

Asehoy ny valinteny

	\({x}^{2}+72=0\)
Isolate the quadratic term.	\(\ {x}^{2}=-72\)
Use the Square Root Property.	\(\ x=\pm \sqrt{-72}\)
Simplify using complex numbers.	\(\ x=\pm \sqrt{72}\ i\)
Simplify the radical.	\(\ x=\pm 6\sqrt{2}\ i\)
Rewrite to show two solutions.	\(x=6\sqrt{2}\ i,\ x=-6\sqrt{2}\ i\)
Check the solutions:

Solve: \({c}^{2}+12=0.\)

Asehoy ny valinteny

\(c=2\sqrt{3}i,\ \ c=-2\sqrt{3}i\)
Solve: \({q}^{2}+24=0.\)

Asehoy ny valinteny

\(c=2\sqrt{6}i,\ \ c=-2\sqrt{6}i\)

Solve: \(\frac{2}{3}{u}^{2}+5=17.\)

Asehoy ny valinteny

	\(\frac{2}{3}{u}^{2}+5=17\)
Isolate the quadratic term.
Multiply by \(\frac{3}{2}\) to make the coefficient 1.
Simplify.
Use the Square Root Property.
Simplify the radical.
Simplify.
Rewrite to show two solutions.
Check:

Solve: \(\frac{1}{2}{x}^{2}+4=24.\)

Asehoy ny valinteny

\(x=2\sqrt{10},\ x=-2\sqrt{10}\)
Solve: \(\frac{3}{4}{y}^{2}-3=18.\)

Asehoy ny valinteny

\(y=2\sqrt{7},\ y=-2\sqrt{7}\)

Solve: \(2{x}^{2}-8=41.\)

Asehoy ny valinteny


Isolate the quadratic term.
Divide by \(2\) to make the coefficient 1.
Simplify.
Use the Square Root Property.
Rewrite the radical as a fraction of square roots.
Rationalize the denominator.
Simplify.
Rewrite to show two solutions.
Check: We leave the check for you.

Solve: \(5{r}^{2}-2=34.\)

Asehoy ny valinteny

\(r=\frac{6\sqrt{5}}{5},\ r=-\frac{6\sqrt{5}}{5}\)
Solve: \(3{t}^{2}+6=70.\)

Asehoy ny valinteny

\(t=\frac{8\sqrt{3}}{3},\ t=-\frac{8\sqrt{3}}{3}\)

Solve: \(4{(y-7)}^{2}=48.\)

Asehoy ny valinteny

	\(4{(y-7)}^{2}=48\)
Divide both sides by the coefficient 4.	\(\ {(y-7)}^{2}=12\)
Use the Square Root Property on the binomial	\(\ y-7=\pm \sqrt{12}\)
Simplify the radical.	\(\ y-7=\pm 2\sqrt{3}\)
Solve for \(y.\)	\(\ y=7\pm 2\sqrt{3}\)
Rewrite to show two solutions.	\(y=7+2\sqrt{3},\) \(y=7-2\sqrt{3}\)
Check:

Solve: \(3{(a-3)}^{2}=54.\)

Asehoy ny valinteny

\(a=3+3\sqrt{2},\ a=3-3\sqrt{2}\)
Solve: \(2{(b+2)}^{2}=80.\)

Asehoy ny valinteny

\(b=-2+2\sqrt{10},\ b=-2-2\sqrt{10}\)

Solve: \({(x-\frac{1}{3})}^{2}=\frac{5}{9}.\)

Asehoy ny valinteny

	\(\ {(x-\frac{1}{3})}^{2}=\frac{5}{9}\)
Use the Square Root Property.	\(\ x-\frac{1}{3}=\pm \sqrt{\frac{5}{9}}\)
Rewrite the radical as a fraction of square roots.	\(\ x-\frac{1}{3}=\pm \frac{\sqrt{5}}{\sqrt{9}}\)
Simplify the radical.	\(\ x-\frac{1}{3}=\pm \frac{\sqrt{5}}{3}\)
Solve for \(x\).	\(\ x=\frac{1}{3}\pm \frac{\sqrt{5}}{3}\)
Rewrite to show two solutions.	\(x=\frac{1}{3}+\frac{\sqrt{5}}{3},\ x=\frac{1}{3}-\frac{\sqrt{5}}{3}\)
Check: We leave the check for you.

Solve: \({(x-\frac{1}{2})}^{2}=\frac{5}{4}.\)

Asehoy ny valinteny

\(x=\frac{1}{2}+\frac{\sqrt{5}}{2}\),\(x=\frac{1}{2}-\frac{\sqrt{5}}{2}\)
Solve: \({(y+\frac{3}{4})}^{2}=\frac{7}{16}.\)

Asehoy ny valinteny

\(y=-\frac{3}{4}+\frac{\sqrt{7}}{4},\ y=-\frac{3}{4}-\frac{\sqrt{7}}{4}\)

Solve: \(2{(x-2)}^{2}+3=57.\)

Asehoy ny valinteny

	\(2{(x-2)}^{2}+3=57\)
Subtract 3 from both sides to isolate the binomial term.	\(2{(x-2)}^{2}=54\)
Divide both sides by 2.	\({(x-2)}^{2}=27\)
Use the Square Root Property.	\(x-2=\pm \sqrt{27}\)
Simplify the radical.	\(x-2=\pm 3\sqrt{3}\)
Solve for \(x\).	\(x=2\pm 3\sqrt{3}\)
Rewrite to show two solutions.	\(x=2+3\sqrt{3},\ x=2-3\sqrt{3}\)
Check: We leave the check for you.

Solve: \(5{(a-5)}^{2}+4=104.\)

Asehoy ny valinteny

\(a=5+2\sqrt{5},\ a=5-2\sqrt{5}\)
Solve: \(3{(b+3)}^{2}-8=88.\)

Asehoy ny valinteny

\(b=-3+4\sqrt{2},\ b=-3-4\sqrt{2}\)

Solve: \({(2x-3)}^{2}=-12.\)

Asehoy ny valinteny

	\(\ {(2x-3)}^{2}=-12\)
Use the Square Root Property.	\(\ 2x-3=\pm \sqrt{-12}\)
Simplify the radical.	\(\ 2x-3=\pm 2\sqrt{3}\ i\)
Add 3 to both sides.	\(\ 2x=3\pm 2\sqrt{3}\ i\)
Divide both sides by 2.	\(\ x=\frac{3\pm 2\sqrt{3}\ i}{2}\)
Rewrite in standard form.	\(\ x=\frac{3}{2}\pm \frac{2\sqrt{3}\ i}{2}\)
Simplify.	\(\ x=\frac{3}{2}\pm \sqrt{3}\ i\)
Rewrite to show two solutions.	\(x=\frac{3}{2}+\sqrt{3}\ i,\ x=\frac{3}{2}-\sqrt{3}\ i\)
Check: We leave the check for you.

Solve: \({(3r+4)}^{2}=-8.\)

Asehoy ny valinteny

\(r=-\frac{4}{3}+\frac{2\sqrt{2}i}{3},\ \text{r}=-\frac{4}{3}-\frac{2\sqrt{2}i}{3}\)
Solve: \({(2t-8)}^{2}=-10.\)

Asehoy ny valinteny

\(t=4+\frac{\sqrt{10}i}{2},\ \text{t}=4-\frac{\sqrt{10}i}{2}\)

Solve: \(4{n}^{2}+4n+1=16.\)

Asehoy ny valinteny

We notice the left side of the equation is a perfect square trinomial. We will factor it first.

	\(\ 4{n}^{2}+4n+1=16\)
Factor the perfect square trinomial.	\(\ {(2n+1)}^{2}=16\)
Use the Square Root Property.	\(\ 2n+1=\pm \sqrt{16}\)
Simplify the radical.	\(\ 2n+1=\pm 4\)
Solve for \(n\).	\(\ 2n=-1\pm 4\)
Divide each side by 2.	\(\ \frac{2n}{2}=\frac{-1\pm 4}{2}\) \(\ n=\frac{-1\pm 4}{2}\)
Rewrite to show two solutions.	\(n=\frac{-1+4}{2}\), \(n=\frac{-1-4}{2}\)
Simplify each equation.	\(n=\frac{3}{2}\), \(\ n=-\frac{5}{2}\)
Check:

Solve: \(9{m}^{2}-12m+4=25.\)

Asehoy ny valinteny

\(m=\frac{7}{3},\ m=-1\)
Solve: \(16{n}^{2}+40n+25=4.\)

Asehoy ny valinteny

\(n=-\frac{3}{4},\ n=-\frac{7}{4}\)
\({a}^{2}=49\)

Asehoy ny valinteny

\(a=\pm 7\)
\({b}^{2}=144\)
\({r}^{2}-24=0\)

Asehoy ny valinteny

\(r=\pm 2\sqrt{6}\)
\({t}^{2}-75=0\)
\({u}^{2}-300=0\)

Asehoy ny valinteny

\(u=\pm 10\sqrt{3}\)
\({v}^{2}-80=0\)
\(4{m}^{2}=36\)

Asehoy ny valinteny

\(m=\pm 3\)

Aoka ho miasa ny anao

Ny kaonty maimaimpoana dia mampiditra ny fanamarihana amin'ny lesona tsirairay, ny tahirin'ny zavatra vitanao, ny olana voavahao ao amin'ny toerana iray, ary ny mpampianatra iray izay azonao anontanio momba ity pejy ity. misokatra ho an'ny rehetra ny matematika, na efa niditra na tsy niditra.

Misoratra anarana Fidirana

Marika ampiasaina eto

Tsindrio ny marika iray mba hahitana ny famaritana feno, sary, ary ny dikan'ny litera tsirairay ao aminy.

Ahoana ny fomba: Solve Quadratic Equations Using the Square Root Property

Solve quadratic equations of the form
Solve quadratic equations of the form
Isolate the quadratic term and make its coefficient one.
Use Square Root Property.
Simplify the radical.
Check the solutions.
Square Root Property
If

Fanontaniana napetrak'ireo olona

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Nalaina avy amin'ny OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0), OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Nofafana sy nanazavana indray eto; ny hadisoana dia antsika. Satria nahazo alalana ho an'ny fampiasana tsy ara-barotra ny loharanom-baovaony, maimaimpoana ity lesona ity, tsy mivarotra na inona na inona, ary mizara amin'ny alalana mitovy.

Mbola maro ao Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value