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Solve Quadratic Equations Using the Square Root Property

Solve Quadratic Equations Using the Square Root Property

By the end of this section, you will be able to: solve quadratic equations of the form ax2=kax^2=k using the Square Root Property and solve quadratic equations of the form a(x−h)2=ka(x-h)^2=k using the Square Root Property.

A quadratic equation is an equation of the form ax2+bx+c=0ax^2+bx+c=0, where a≠0a\ne0. Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form ax2ax^2. We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.

We have seen that some quadratic equations can be solved by factoring. In this chapter, we will learn three other methods to use in case a quadratic equation cannot be factored.

Solve Quadratic Equations of the form ax2=kax^2=k using the Square Root Property

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation x2=9x^2=9.

x2=9Put the equation in standard form.x2−9=0Factor the difference of squares.(x−3)(x+3)=0Use the Zero Product Property.x−3=0orx+3=0Solve each equation.x=3orx=−3 \begin{array}{lrcl} &&& x^2=9 \\[4pt] \text{Put the equation in standard form.} & x^2-9 &=& 0 \\[4pt] \text{Factor the difference of squares.} & (x-3)(x+3) &=& 0 \\[4pt] \text{Use the Zero Product Property.} & x-3 &=& 0\quad\text{or}\quad x+3=0 \\[4pt] \text{Solve each equation.} & x &=& 3\quad\text{or}\quad x=-3 \end{array}

We can easily use factoring to find the solutions of similar equations, like x2=16x^2=16 and x2=25x^2=25, because 16 and 25 are perfect squares. In each case, we would get two solutions, x=4x=4, x=−4x=-4 and x=5x=5, x=−5x=-5. But what happens when we have an equation like x2=7x^2=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(-13)^2=169, so −13-13 is also a square root of 169. Therefore, both 13 and −13-13 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:

If n2=m, then n is a square root of m.\text{If }n^2=m,\text{ then }n\text{ is a square root of }m.

Since these equations are all of the form x2=kx^2=k, the square root definition tells us the solutions are the two square roots of kk. This leads to the Square Root Property.

Square Root Property. If x2=kx^2=k, then

x=korx=−korx=±k.x=\sqrt{k}\quad\text{or}\quad x=-\sqrt{k}\quad\text{or}\quad x=\pm\sqrt{k}.

Notice that the Square Root Property gives two solutions to an equation of the form x2=kx^2=k, the principal square root of kk and its opposite. We could also write the solution as x=±kx=\pm\sqrt{k}. We read this as xx equals positive or negative the square root of kk.

Now we will solve the equation x2=9x^2=9 again, this time using the Square Root Property.

x2=9Use the Square Root Property.x=±9x=±3So x=3 or x=−3. \begin{array}{lrcl} &&& x^2=9 \\[4pt] \text{Use the Square Root Property.} & x &=& \pm\sqrt{9} \\[4pt] &&& x=\pm3 \\[4pt] &&& \text{So }x=3\text{ or }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 x2=7x^2=7.

x2=7Use the Square Root Property.x=7,x=−7 \begin{array}{lrcl} &&& x^2=7 \\[4pt] \text{Use the Square Root Property.} & x &=& \sqrt{7},\quad x=-\sqrt{7} \end{array}

We cannot simplify 7\sqrt{7}, so we leave the answer as a radical.

Example. Solve x2−50=0x^2-50=0.

x2−50=0Isolate the quadratic term and make its coefficient one.x2=50Use the Square Root Property. Remember to write the ± symbol.x=±50Simplify the radical.x=±25⋅2x=±52Rewrite to show two solutions.x=52,x=−52 \begin{array}{lrcl} &&& x^2-50=0 \\[4pt] \text{Isolate the quadratic term and make its coefficient one.} & x^2 &=& 50 \\[4pt] \text{Use the Square Root Property. Remember to write the }\pm\text{ symbol.} & x &=& \pm\sqrt{50} \\[4pt] \text{Simplify the radical.} & x &=& \pm\sqrt{25}\cdot\sqrt{2} \\[4pt] &&& x=\pm5\sqrt{2} \\[4pt] \text{Rewrite to show two solutions.} & x &=& 5\sqrt{2},\quad x=-5\sqrt{2} \end{array}

Check the solutions:

(52)2−50=?025⋅2−50=?00=0 ✓(−52)2−50=?025⋅2−50=?00=0 ✓ \begin{array}{rcl} (5\sqrt{2})^2-50 &\overset{?}{=}& 0 \\[4pt] 25\cdot2-50 &\overset{?}{=}& 0 \\[4pt] 0&=&0\ \checkmark \end{array} \qquad \begin{array}{rcl} (-5\sqrt{2})^2-50 &\overset{?}{=}& 0 \\[4pt] 25\cdot2-50 &\overset{?}{=}& 0 \\[4pt] 0&=&0\ \checkmark \end{array}

Solve x2−48=0x^2-48=0. Enter both solutions, separated by a comma.

Solve y2−27=0y^2-27=0. Enter both solutions, separated by a comma.

The steps to take to use the Square Root Property to solve a quadratic equation are listed here.

Solve a quadratic equation using the Square Root Property.

  1. Isolate the quadratic term and make its coefficient one.
  2. Use the Square Root Property.
  3. Simplify the radical.
  4. Check the solutions.

In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient 3 before using the Square Root Property.

Example. Solve 3z2=1083z^2=108.

3z2=108The quadratic term is isolated. Divide by 3 to make its coefficient 1.3z23=1083Simplify.z2=36Use the Square Root Property.z=±36Simplify the radical.z=±6Rewrite to show two solutions.z=6,z=−6 \begin{array}{lrcl} &&& 3z^2=108 \\[4pt] \text{The quadratic term is isolated. Divide by 3 to make its coefficient 1.} & \tfrac{3z^2}{3} &=& \tfrac{108}{3} \\[10pt] \text{Simplify.} & z^2 &=& 36 \\[4pt] \text{Use the Square Root Property.} & z &=& \pm\sqrt{36} \\[4pt] \text{Simplify the radical.} & z &=& \pm6 \\[4pt] \text{Rewrite to show two solutions.} & z &=& 6,\quad z=-6 \end{array}

Check: 3(6)2=1083(6)^2=108 and 3(−6)2=1083(-6)^2=108.

Solve 2x2=982x^2=98. Enter both solutions, separated by a comma.

Solve 5m2=805m^2=80. Enter both solutions, separated by a comma.

The Square Root Property states “If x2=kx^2=k.” What will happen if k<0k<0? This will be the case in the next example.

Example. Solve x2+72=0x^2+72=0.

x2+72=0Isolate the quadratic term.x2=−72Use the Square Root Property.x=±−72Simplify using complex numbers.x=±72 iSimplify the radical.x=±62 iRewrite to show two solutions.x=62 i,x=−62 i \begin{array}{lrcl} &&& x^2+72=0 \\[4pt] \text{Isolate the quadratic term.} & x^2 &=& -72 \\[4pt] \text{Use the Square Root Property.} & x &=& \pm\sqrt{-72} \\[4pt] \text{Simplify using complex numbers.} & x &=& \pm\sqrt{72}\,i \\[4pt] \text{Simplify the radical.} & x &=& \pm6\sqrt{2}\,i \\[4pt] \text{Rewrite to show two solutions.} & x &=& 6\sqrt{2}\,i,\quad x=-6\sqrt{2}\,i \end{array}

Check the solutions:

(62 i)2+72=?062(2)2i2+72=?036⋅2⋅(−1)+72=?00=0 ✓(−62 i)2+72=?0(−6)2(2)2i2+72=?036⋅2⋅(−1)+72=?00=0 ✓ \begin{array}{rcl} (6\sqrt{2}\,i)^2+72 &\overset{?}{=}& 0 \\[4pt] 6^2(\sqrt{2})^2i^2+72 &\overset{?}{=}& 0 \\[4pt] 36\cdot2\cdot(-1)+72 &\overset{?}{=}& 0 \\[4pt] 0&=&0\ \checkmark \end{array} \qquad \begin{array}{rcl} (-6\sqrt{2}\,i)^2+72 &\overset{?}{=}& 0 \\[4pt] (-6)^2(\sqrt{2})^2i^2+72 &\overset{?}{=}& 0 \\[4pt] 36\cdot2\cdot(-1)+72 &\overset{?}{=}& 0 \\[4pt] 0&=&0\ \checkmark \end{array}

Solve c2+12=0c^2+12=0. Enter both solutions, separated by a comma.

Solve q2+24=0q^2+24=0. Enter both solutions, separated by a comma.

Our method also works when fractions occur in the equation; we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.

Example. Solve 23u2+5=17\tfrac{2}{3}u^2+5=17.

23u2+5=17Isolate the quadratic term.23u2=12Multiply by 32 to make the coefficient 1.32⋅23u2=32⋅12Simplify.u2=18Use the Square Root Property.u=±18Simplify the radical.u=±9⋅2u=±32Rewrite to show two solutions.u=32,u=−32 \begin{array}{lrcl} &&& \tfrac{2}{3}u^2+5=17 \\[10pt] \text{Isolate the quadratic term.} & \tfrac{2}{3}u^2 &=& 12 \\[10pt] \text{Multiply by }\tfrac{3}{2}\text{ to make the coefficient 1.} & \tfrac{3}{2}\cdot\tfrac{2}{3}u^2 &=& \tfrac{3}{2}\cdot12 \\[10pt] \text{Simplify.} & u^2 &=& 18 \\[4pt] \text{Use the Square Root Property.} & u &=& \pm\sqrt{18} \\[4pt] \text{Simplify the radical.} & u &=& \pm\sqrt{9}\cdot\sqrt{2} \\[4pt] &&& u=\pm3\sqrt{2} \\[4pt] \text{Rewrite to show two solutions.} & u &=& 3\sqrt{2},\quad u=-3\sqrt{2} \end{array}

Check: 23(32)2+5=17\tfrac{2}{3}(3\sqrt{2})^2+5=17 and 23(−32)2+5=17\tfrac{2}{3}(-3\sqrt{2})^2+5=17.

Solve 12x2+4=24\tfrac{1}{2}x^2+4=24. Enter both solutions, separated by a comma.

Solve 34y2−3=18\tfrac{3}{4}y^2-3=18. Enter both solutions, separated by a comma.

The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator.

Example. Solve 2x2−8=412x^2-8=41.

2x2−8=41Isolate the quadratic term.2x2=49Divide by 2 to make the coefficient 1.2x22=492Simplify.x2=492Use the Square Root Property.x=±492Rewrite the radical as a fraction of square roots.x=±492Rationalize the denominator.x=±49⋅22⋅2Simplify.x=±722Rewrite to show two solutions.x=722,x=−722 \begin{array}{lrcl} &&& 2x^2-8=41 \\[4pt] \text{Isolate the quadratic term.} & 2x^2 &=& 49 \\[4pt] \text{Divide by 2 to make the coefficient 1.} & \tfrac{2x^2}{2} &=& \tfrac{49}{2} \\[10pt] \text{Simplify.} & x^2 &=& \tfrac{49}{2} \\[10pt] \text{Use the Square Root Property.} & x &=& \pm\sqrt{\tfrac{49}{2}} \\[10pt] \text{Rewrite the radical as a fraction of square roots.} & x &=& \pm\tfrac{\sqrt{49}}{\sqrt{2}} \\[10pt] \text{Rationalize the denominator.} & x &=& \pm\tfrac{\sqrt{49}\cdot\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}} \\[10pt] \text{Simplify.} & x &=& \pm\tfrac{7\sqrt{2}}{2} \\[10pt] \text{Rewrite to show two solutions.} & x &=& \tfrac{7\sqrt{2}}{2},\quad x=-\tfrac{7\sqrt{2}}{2} \end{array}

Check: We leave the check for you.

Solve 5r2−2=345r^2-2=34. Enter both solutions, separated by a comma.

Solve 3t2+6=703t^2+6=70. Enter both solutions, separated by a comma.

Solve Quadratic Equations of the form a(x−h)2=ka(x-h)^2=k using the Square Root Property

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

ax2=ka(x−h)2=kax^2=k\qquad a(x-h)^2=k

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 aa, then the Square Root Property can be used on (x−h)2(x-h)^2.

Example. Solve 4(y−7)2=484(y-7)^2=48.

4(y−7)2=48Divide both sides by the coefficient 4.(y−7)2=12Use the Square Root Property on the binomial.y−7=±12Simplify the radical.y−7=±23Solve for y.y=7±23Rewrite to show two solutions.y=7+23,y=7−23 \begin{array}{lrcl} &&& 4(y-7)^2=48 \\[4pt] \text{Divide both sides by the coefficient 4.} & (y-7)^2 &=& 12 \\[4pt] \text{Use the Square Root Property on the binomial.} & y-7 &=& \pm\sqrt{12} \\[4pt] \text{Simplify the radical.} & y-7 &=& \pm2\sqrt{3} \\[4pt] \text{Solve for }y. & y &=& 7\pm2\sqrt{3} \\[4pt] \text{Rewrite to show two solutions.} & y &=& 7+2\sqrt{3},\quad y=7-2\sqrt{3} \end{array}

Check: 4(7+23−7)2=484(7+2\sqrt{3}-7)^2=48 and 4(7−23−7)2=484(7-2\sqrt{3}-7)^2=48.

Solve 3(a−3)2=543(a-3)^2=54. Enter both solutions, separated by a comma.

Solve 2(b+2)2=802(b+2)^2=80. Enter both solutions, separated by a comma.

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

Example. Solve (x−13)2=59(x-\tfrac{1}{3})^2=\tfrac{5}{9}.

(x−13)2=59Use the Square Root Property.x−13=±59Rewrite the radical as a fraction of square roots.x−13=±59Simplify the radical.x−13=±53Solve for x.x=13±53Rewrite to show two solutions.x=13+53,x=13−53 \begin{array}{lrcl} &&& (x-\tfrac{1}{3})^2=\tfrac{5}{9} \\[10pt] \text{Use the Square Root Property.} & x-\tfrac{1}{3} &=& \pm\sqrt{\tfrac{5}{9}} \\[10pt] \text{Rewrite the radical as a fraction of square roots.} & x-\tfrac{1}{3} &=& \pm\tfrac{\sqrt{5}}{\sqrt{9}} \\[10pt] \text{Simplify the radical.} & x-\tfrac{1}{3} &=& \pm\tfrac{\sqrt{5}}{3} \\[10pt] \text{Solve for }x. & x &=& \tfrac{1}{3}\pm\tfrac{\sqrt{5}}{3} \\[10pt] \text{Rewrite to show two solutions.} & x &=& \tfrac{1}{3}+\tfrac{\sqrt{5}}{3},\quad x=\tfrac{1}{3}-\tfrac{\sqrt{5}}{3} \end{array}

Check: We leave the check for you.

Solve (x−12)2=54(x-\tfrac{1}{2})^2=\tfrac{5}{4}. Enter both solutions, separated by a comma.

Solve (y+34)2=716(y+\tfrac{3}{4})^2=\tfrac{7}{16}. Enter both solutions, separated by a comma.

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

Example. Solve 2(x−2)2+3=572(x-2)^2+3=57.

2(x−2)2+3=57Subtract 3 from both sides to isolate the binomial term.2(x−2)2=54Divide both sides by 2.(x−2)2=27Use the Square Root Property.x−2=±27Simplify the radical.x−2=±33Solve for x.x=2±33Rewrite to show two solutions.x=2+33,x=2−33 \begin{array}{lrcl} &&& 2(x-2)^2+3=57 \\[4pt] \text{Subtract 3 from both sides to isolate the binomial term.} & 2(x-2)^2 &=& 54 \\[4pt] \text{Divide both sides by 2.} & (x-2)^2 &=& 27 \\[4pt] \text{Use the Square Root Property.} & x-2 &=& \pm\sqrt{27} \\[4pt] \text{Simplify the radical.} & x-2 &=& \pm3\sqrt{3} \\[4pt] \text{Solve for }x. & x &=& 2\pm3\sqrt{3} \\[4pt] \text{Rewrite to show two solutions.} & x &=& 2+3\sqrt{3},\quad x=2-3\sqrt{3} \end{array}

Check: We leave the check for you.

Solve 5(a−5)2+4=1045(a-5)^2+4=104. Enter both solutions, separated by a comma.

Solve 3(b+3)2−8=883(b+3)^2-8=88. Enter both solutions, separated by a comma.

Sometimes the solutions are complex numbers.

Example. Solve (2x−3)2=−12(2x-3)^2=-12.

(2x−3)2=−12Use the Square Root Property.2x−3=±−12Simplify the radical.2x−3=±23 iAdd 3 to both sides.2x=3±23 iDivide both sides by 2.x=3±23 i2Rewrite in standard form.x=32±23 i2Simplify.x=32±3 iRewrite to show two solutions.x=32+3 i,x=32−3 i \begin{array}{lrcl} &&& (2x-3)^2=-12 \\[4pt] \text{Use the Square Root Property.} & 2x-3 &=& \pm\sqrt{-12} \\[4pt] \text{Simplify the radical.} & 2x-3 &=& \pm2\sqrt{3}\,i \\[4pt] \text{Add 3 to both sides.} & 2x &=& 3\pm2\sqrt{3}\,i \\[4pt] \text{Divide both sides by 2.} & x &=& \tfrac{3\pm2\sqrt{3}\,i}{2} \\[10pt] \text{Rewrite in standard form.} & x &=& \tfrac{3}{2}\pm\tfrac{2\sqrt{3}\,i}{2} \\[10pt] \text{Simplify.} & x &=& \tfrac{3}{2}\pm\sqrt{3}\,i \\[10pt] \text{Rewrite to show two solutions.} & x &=& \tfrac{3}{2}+\sqrt{3}\,i,\quad x=\tfrac{3}{2}-\sqrt{3}\,i \end{array}

Check: We leave the check for you.

Solve (3r+4)2=−8(3r+4)^2=-8. Enter both solutions, separated by a comma.

Solve (2t−8)2=−10(2t-8)^2=-10. Enter both solutions, separated by a comma.

The left sides of the equations in the next two examples do not seem to be of the form a(x−h)2a(x-h)^2. But they are perfect square trinomials, so we will factor to put them in the form we need.

Example. Solve 4n2+4n+1=164n^2+4n+1=16.

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

4n2+4n+1=16Factor the perfect square trinomial.(2n+1)2=16Use the Square Root Property.2n+1=±16Simplify the radical.2n+1=±4Solve for n.2n=−1±4Divide each side by 2.n=−1±42Rewrite to show two solutions.n=−1+42,n=−1−42Simplify each equation.n=32,n=−52 \begin{array}{lrcl} &&& 4n^2+4n+1=16 \\[4pt] \text{Factor the perfect square trinomial.} & (2n+1)^2 &=& 16 \\[4pt] \text{Use the Square Root Property.} & 2n+1 &=& \pm\sqrt{16} \\[4pt] \text{Simplify the radical.} & 2n+1 &=& \pm4 \\[4pt] \text{Solve for }n. & 2n &=& -1\pm4 \\[4pt] \text{Divide each side by 2.} & n &=& \tfrac{-1\pm4}{2} \\[10pt] \text{Rewrite to show two solutions.} & n &=& \tfrac{-1+4}{2},\quad n=\tfrac{-1-4}{2} \\[10pt] \text{Simplify each equation.} & n &=& \tfrac{3}{2},\quad n=-\tfrac{5}{2} \end{array}

Check: 4(32)2+4(32)+1=164(\tfrac{3}{2})^2+4(\tfrac{3}{2})+1=16 and 4(−52)2+4(−52)+1=164(-\tfrac{5}{2})^2+4(-\tfrac{5}{2})+1=16.

Solve 9m2−12m+4=259m^2-12m+4=25. Enter both solutions, separated by a comma.

Solve 16n2+40n+25=416n^2+40n+25=4. Enter both solutions, separated by a comma.

Key terms: The Square Root Property states that if x2=kx^2=k, then x=kx=\sqrt{k} or x=−kx=-\sqrt{k}, which can be written x=±kx=\pm\sqrt{k}.

This page adapts [OpenStax *Intermediate Algebra 2e*, Section 9.1](https://openstax.org/books/intermediate-algebra-2e/pages/9-1-solve-quadratic-equations-using-the-square-root-property), by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under [CC BY-NC-SA 4.0](https://creativecommons.org/licenses/by-nc-sa/4.0/). Access the original for free at [openstax.org](https://openstax.org/books/intermediate-algebra-2e/pages/9-1-solve-quadratic-equations-using-the-square-root-property). Changes: adapted the source for web presentation and converted the Try It exercises to interactive questions; omitted the Be Prepared questions, media links, and end-of-section exercise sets.
Quadratic Equations and FunctionsSolve Quadratic Equations by Completing the Square