maths.free › Algebra › Quadratic Equations and Functions › Solve Quadratic Equations Using the Square Root Property
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.- Isolate the quadratic term and make its coefficient one.
- Use Square Root Property.
- Simplify the radical.
- 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:
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.
Nu jij. Geen rekenmachine regelt deze, maar de stukken ervan zijn berekenbaar. Probeer er een hieronder, of typ je eigen.
Praktijk (40)
Probeer ze eerst op papier. Onthul het antwoord om te controleren; geverifieerde kunnen worden geopend in de oplossing voor elke stap.
-
Simplify: \(\sqrt{128}.\)
Onthul het antwoord
\(8\sqrt{2}\)
-
Simplify: \(\sqrt{\frac{32}{5}}\).
Onthul het antwoord
\(\frac{4\sqrt{10}}{5}\)
-
Factor: \(9{x}^{2}-12x+4\).
Onthul het antwoord
\({\left(3x-2\right)}^{2}\)
-
Solve: \({x}^{2}-50=0.\)
-
Solve: \({x}^{2}-48=0.\)
Onthul het antwoord
\(x=4\sqrt{3},x=-4\sqrt{3}\)
-
Solve: \({y}^{2}-27=0.\)
Onthul het antwoord
\(y=3\sqrt{3},y=-3\sqrt{3}\)
-
Solve: \(3{z}^{2}=108.\)
Onthul het antwoord
\(3{z}^{2}=108\) 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: \(2{x}^{2}=98.\)
Onthul het antwoord
\(x=7,x=-7\)
-
Solve: \(5{m}^{2}=80.\)
Onthul het antwoord
\(m=4,m=-4\)
-
Solve: \({x}^{2}+72=0\).
Onthul het antwoord
\({x}^{2}+72=0\) 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: \({c}^{2}+12=0.\)
Onthul het antwoord
\(c=2\sqrt{3}i,\ \ c=-2\sqrt{3}i\)
-
Solve: \({q}^{2}+24=0.\)
Onthul het antwoord
\(c=2\sqrt{6}i,\ \ c=-2\sqrt{6}i\)
-
Solve: \(\frac{2}{3}{u}^{2}+5=17.\)
Onthul het antwoord
\(\frac{2}{3}{u}^{2}+5=17\) 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: \(\frac{1}{2}{x}^{2}+4=24.\)
Onthul het antwoord
\(x=2\sqrt{10},\ x=-2\sqrt{10}\)
-
Solve: \(\frac{3}{4}{y}^{2}-3=18.\)
Onthul het antwoord
\(y=2\sqrt{7},\ y=-2\sqrt{7}\)
-
Solve: \(2{x}^{2}-8=41.\)
Onthul het antwoord
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: \(5{r}^{2}-2=34.\)
Onthul het antwoord
\(r=\frac{6\sqrt{5}}{5},\ r=-\frac{6\sqrt{5}}{5}\)
-
Solve: \(3{t}^{2}+6=70.\)
Onthul het antwoord
\(t=\frac{8\sqrt{3}}{3},\ t=-\frac{8\sqrt{3}}{3}\)
-
Solve: \(4{(y-7)}^{2}=48.\)
Onthul het antwoord
\(4{(y-7)}^{2}=48\) 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: \(3{(a-3)}^{2}=54.\)
Onthul het antwoord
\(a=3+3\sqrt{2},\ a=3-3\sqrt{2}\)
-
Solve: \(2{(b+2)}^{2}=80.\)
Onthul het antwoord
\(b=-2+2\sqrt{10},\ b=-2-2\sqrt{10}\)
-
Solve: \({(x-\frac{1}{3})}^{2}=\frac{5}{9}.\)
Onthul het antwoord
\(\ {(x-\frac{1}{3})}^{2}=\frac{5}{9}\) 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: \({(x-\frac{1}{2})}^{2}=\frac{5}{4}.\)
Onthul het antwoord
\(x=\frac{1}{2}+\frac{\sqrt{5}}{2}\),\(x=\frac{1}{2}-\frac{\sqrt{5}}{2}\)
-
Solve: \({(y+\frac{3}{4})}^{2}=\frac{7}{16}.\)
Onthul het antwoord
\(y=-\frac{3}{4}+\frac{\sqrt{7}}{4},\ y=-\frac{3}{4}-\frac{\sqrt{7}}{4}\)
-
Solve: \(2{(x-2)}^{2}+3=57.\)
Onthul het antwoord
\(2{(x-2)}^{2}+3=57\) 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: \(5{(a-5)}^{2}+4=104.\)
Onthul het antwoord
\(a=5+2\sqrt{5},\ a=5-2\sqrt{5}\)
-
Solve: \(3{(b+3)}^{2}-8=88.\)
Onthul het antwoord
\(b=-3+4\sqrt{2},\ b=-3-4\sqrt{2}\)
-
Solve: \({(2x-3)}^{2}=-12.\)
Onthul het antwoord
\(\ {(2x-3)}^{2}=-12\) 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: \({(3r+4)}^{2}=-8.\)
Onthul het antwoord
\(r=-\frac{4}{3}+\frac{2\sqrt{2}i}{3},\ \text{r}=-\frac{4}{3}-\frac{2\sqrt{2}i}{3}\)
-
Solve: \({(2t-8)}^{2}=-10.\)
Onthul het antwoord
\(t=4+\frac{\sqrt{10}i}{2},\ \text{t}=4-\frac{\sqrt{10}i}{2}\)
-
Solve: \(4{n}^{2}+4n+1=16.\)
Onthul het antwoord
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: \(9{m}^{2}-12m+4=25.\)
Onthul het antwoord
\(m=\frac{7}{3},\ m=-1\)
-
Solve: \(16{n}^{2}+40n+25=4.\)
Onthul het antwoord
\(n=-\frac{3}{4},\ n=-\frac{7}{4}\)
-
\({a}^{2}=49\)
Onthul het antwoord
\(a=\pm 7\)
-
\({b}^{2}=144\)
-
\({r}^{2}-24=0\)
Onthul het antwoord
\(r=\pm 2\sqrt{6}\)
-
\({t}^{2}-75=0\)
-
\({u}^{2}-300=0\)
Onthul het antwoord
\(u=\pm 10\sqrt{3}\)
-
\({v}^{2}-80=0\)
-
\(4{m}^{2}=36\)
Onthul het antwoord
\(m=\pm 3\)
Hier gebruikte symbolen
Tik op elk symbool voor de volledige definitie, een afbeelding en wat elke letter erin betekent.
Hoe: 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
Vragen die mensen stellen
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.
Delen van deze pagina zijn aangepast van OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0), OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Gecondenseerd en opnieuw verklaard, fouten zijn van ons. Omdat de bron ervan een licentie heeft voor niet-commercieel gebruik, is deze les gratis, verkoopt niets en wordt gedeeld onder dezelfde licentie.
Meer in Algebra
Linear equationsQuadratic equationsSystems of equationsInequalitiesFactoringExpandingSimplifying expressionsFunctions and graphsExponential and logarithmic equationsPolynomial equationsAbsolute value