Solving Quadratic Equations by Square Root Property Show
Quadratic Equations by Square Root Property
To solve an equation by using the square root property, you will first isolate the term that contains the squared variable. You can then take the square root of both sides and solve for the variable. Make sure to write the final answer in simplified form. Note that there is always the possibility of two roots for every square root: one positive and one negative. Placing a ± sign in front of the side containing the constant after you take the square root will ensure that the final answer will include both possible roots. Example Solve: 2x2 + 3 = 27 Solution First, isolate the portion of the equation that's actually being squared. 2x2 + 3 − 3 = 27 − 3 2x2 = 24 x2 = 12 Now square root both sides and simplify. The square root method can be used for solving quadratic equations in the form "x² = b." This method can yield two answers, as the square root of a number can be a negative or a positive number. If an equation can be expressed in this form, it can be solved by finding the square roots of x. Put the Equation Into the Proper FormIn the equation x² - 49 = 0, the second element on the left side (-49) must be removed to isolate x². This is easily accomplished by adding 49 to both sides of the equation. It is important to remember to always apply changes like this to both sides of the equal sign or you will get an incorrect answer. x² - 49 (+ 49) = 0 (+ 49) yields an equation in the proper form for the square root method: x² = 49. Find the Rootsx² is made up of an element (x) which has been squared, or multiplied by itself (x · x). In other words, finding the square root is finding the number (x or -x) that is the root of the squared number. In the equation x² = 49, √49 = +/- 7, yielding the final answer x = +/- 7. Isolate the SquareSometimes you may be given an equation to solve by this method that is in the form ax² = b. In this case, you can isolate x² by multiplying both sides of the equation by the reciprocal of "a." The reciprocal of "a" is 1/a, and the product of these terms equals 1. If you have a fraction, such as 3/4, simply turn the fraction upside down to get its reciprocal: 4/3. Example With ReciprocalIn the equation 6x² = 72, multiplying both sides of the equation by the reciprocal of 6, or 1/6, will convert it to the proper form for solving by this method. The equation (1/6)6x² = 72(1/6) works out to x² = 12. X then is equal to √12. You can then factor 12: 12 = 2 · 2 · 3, or 2² · 3. Remembering that either the positive or negative square root could be the answer yields the final answer: x = +/- 2√3. By the end of this section, you will be able to: Before you get started, take this readiness quiz.
A quadratic equation is an equation of the form \(a x^{2}+b x+c=0\), where \(a≠0\). Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form \(ax^{2}\). 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 \(ax^{2}=k\) using the Square Root PropertyWe have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation \(x^{2}=9\). \(x^{2}=9\) Put the equation in standard form. \(x^{2}-9=0\) Factor the difference of squares. \((x-3)(x+3)=0\) Use the Zero Produce Property. \(x-3=0 \quad x-3=0\) Solve each equation. \(x=3 \quad x=-3\) 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. 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 \(x^{2}=7\)? 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, \((-13)^{2}=169\), 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: If \(n^{2}=m\), then \(n\) is a square root of \(m\). Since these equations are all of the form \(x^{2}=k\), the square root definition tells us the solutions are the two square roots of \(k\). This leads to the Square Root Property. Definition \(\PageIndex{1}\)Square Root Property If \(x^{2}=k\), then \(x=\sqrt{k} \quad\) or \(\quad x=-\sqrt{k} \quad\) or \(\quad x=\pm \sqrt{k}\) 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. 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 \(x^{2}=9\) again, this time using the Square Root Property. \(\begin{aligned} &x^{2} =9 \\ \text { Use the Square Root Property. } \quad& x=\pm \sqrt{9} \\& x =\pm 3 \end{aligned}\) So \(x=3\) or \(x=-3\) What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation \(x^{2}=7\). \(x^{2}=7\) Use the Square Root Property. \(x=\sqrt{7}, \quad x=-\sqrt{7}\) We cannot simplify \(\sqrt{7}\), so we leave the answer as a radical. Example \(\PageIndex{1}\) How to Solve a Quadratic Equation of the form \(ax^{2}\k\) Using the Square Root PropertySolve: \(x^{2}-50=0\). Solution: Table 9.1.1
Exercise \(\PageIndex{1}\)Solve: \(x^{2}-48=0\). Answer\(x=4 \sqrt{3}, x=-4 \sqrt{3}\) Exercise \(\PageIndex{2}\)Solve: \(y^{2}-27=0\). Answer\(y=3 \sqrt{3}, y=-3 \sqrt{3}\) 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
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 \(\PageIndex{2}\)Solve: \(3 z^{2}=108\). Solution: Table 9.1.2
Exercise \(\PageIndex{3}\)Solve: \(2x^{2}=98\). Answer\(x=7, x=-7\) Exercise \(\PageIndex{4}\)Solve: \(5m^{2}=80\). Answer\(m=4, m=-4\) The Square Root Property states ‘If \(x^{2}=k\),’ What will happen if \(k<0\)? This will be the case in the next example. Example \(\PageIndex{3}\)Solve: \(x^{2}+72=0\). Solution: Table 9.1.3
Exercise \(\PageIndex{5}\)Solve: \(c^{2}+12=0\). Answer\(c=2 \sqrt{3} i, \quad c=-2 \sqrt{3} i\) Exercise \(\PageIndex{6}\)Solve: \(q^{2}+24=0\). Answer\(c=2 \sqrt{6} i, \quad c=-2 \sqrt{6} i\) 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 \(\PageIndex{4}\)Solve: \(\frac{2}{3} u^{2}+5=17\). Solution: Table 9.1.4
Exercise \(\PageIndex{7}\)Solve: \(\frac{1}{2} x^{2}+4=24\). Answer\(x=2 \sqrt{10}, x=-2 \sqrt{10}\) Exercise \(\PageIndex{8}\)Solve: \(\frac{3}{4} y^{2}-3=18\). Answer\(y=2 \sqrt{7}, y=-2 \sqrt{7}\) The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator. Example \(\PageIndex{5}\)Solve: \(2 x^{2}-8=41\). Solution: Table 9.1.5
Exercise \(\PageIndex{9}\)Solve: \(5 r^{2}-2=34\). Answer\(r=\frac{6 \sqrt{5}}{5}, \quad r=-\frac{6 \sqrt{5}}{5}\) Exercise \(\PageIndex{10}\)Solve: \(3 t^{2}+6=70\). Answer\(t=\frac{8 \sqrt{3}}{3}, \quad t=-\frac{8 \sqrt{3}}{3}\) Solve Quadratic Equation of the Form \(a(x-h)^{2}=k\) Using the Square Root PropertyWe 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 \(ax^{2}=k\) 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 \(\PageIndex{6}\)Solve: \(4(y-7)^{2}=48\). Solution: Table 9.1.6
Exercise \(\PageIndex{11}\)Solve: \(3(a-3)^{2}=54\). Answer\(a=3+3 \sqrt{2}, \quad a=3-3 \sqrt{2}\) Exercise \(\PageIndex{12}\)Solve: \(2(b+2)^{2}=80\). Answer\(b=-2+2 \sqrt{10}, \quad b=-2-2 \sqrt{10}\) Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately. Example \(\PageIndex{7}\)Solve: \(\left(x-\frac{1}{3}\right)^{2}=\frac{5}{9}\). Solution: \(\left(x-\frac{1}{3}\right)^{2}=\frac{5}{9}\) 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. Exercise \(\PageIndex{13}\)Solve: \(\left(x-\frac{1}{2}\right)^{2}=\frac{5}{4}\). Answer\(x=\frac{1}{2}+\frac{\sqrt{5}}{2}, x=\frac{1}{2}-\frac{\sqrt{5}}{2}\) Exercise \(\PageIndex{14}\)Solve: \(\left(y+\frac{3}{4}\right)^{2}=\frac{7}{16}\). Answer\(y=-\frac{3}{4}+\frac{\sqrt{7}}{4}, y=-\frac{3}{4}-\frac{\sqrt{7}}{4}\) We will start the solution to the next example by isolating the binomial term. Example \(\PageIndex{8}\)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. Exercise \(\PageIndex{15}\)Solve: \(5(a-5)^{2}+4=104\). Answer\(a=5+2 \sqrt{5}, a=5-2 \sqrt{5}\) Exercise \(\PageIndex{16}\)Solve: \(3(b+3)^{2}-8=88\). \(b=-3+4 \sqrt{2}, \quad b=-3-4 \sqrt{2}\) Sometimes the solutions are complex numbers. Example \(\PageIndex{9}\)Solve: \((2 x-3)^{2}=-12\). Solution: \((2 x-3)^{2}=-12\) Use the Square Root Property. \(2 x-3=\pm \sqrt{-12}\) Simplify the radical. \(2 x-3=\pm 2 \sqrt{3} i\) Add \(3\) to both sides. \(2 x=3 \pm 2 \sqrt{3} i\) 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. Exercise \(\PageIndex{17}\)Solve: \((3 r+4)^{2}=-8\). Answer\(r=-\frac{4}{3}+\frac{2 \sqrt{2} i}{3}, r=-\frac{4}{3}-\frac{2 \sqrt{2} i}{3}\) Exercise \(\PageIndex{18}\)Solve: \((2 t-8)^{2}=-10\). Answer\(t=4+\frac{\sqrt{10} i}{2}, t=4-\frac{\sqrt{10 i}}{2}\) The left sides of the equations in the next two examples do not seem to be of the form \(a(x-h)^{2}\). But they are perfect square trinomials, so we will factor to put them in the form we need. Example \(\PageIndex{10}\)Solve: \(4 n^{2}+4 n+1=16\). Solution: We notice the left side of the equation is a perfect square trinomial. We will factor it first. Table 9.1.7
Exercise \(\PageIndex{19}\)Solve: \(9 m^{2}-12 m+4=25\). Answer\(m=\frac{7}{3}, \quad m=-1\) Exercise \(\PageIndex{20}\)Solve: \(16 n^{2}+40 n+25=4\). Answer\(n=-\frac{3}{4}, \quad n=-\frac{7}{4}\) Access this online resource for additional instruction and practice with using the Square Root Property to solve quadratic equations.
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Solving quadratic equation by square root method
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