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Solving Quadratic Equations with Square Roots

A quadratic equation is an equation where the highest degree is 2.2. Quadratic equations coincide with quadratic functions, and can be used to solve for various points on the graph, or parabola, of the function. There are many different ways to solve quadratic equations. One way is with the use of square roots.
Concept

Quadratic Equation

In standard form, quadratic equations take the form y=ax2+bx+c,y=ax^2+bx+c, and can be solved in various ways. In general, they are solved for the value(s) of xx that make the equation equal to 0.0. Thus, y=ax2+bx+cy=ax^2+bx+c becomes 0=ax2+bx+c. 0=ax^2+bx+c. Graphically, all points with a yy-coordinate of 00 are the xx-intercepts of the function — or the zeros of the parabola. That means, solving a quadratic equation leads to finding the zeros of the parabola. Since a parabola can have 0,1,0, 1, or 22 zeros, a quadratic equation can have 0,1,0, 1, or 22 solutions.

Theory

Simple Quadratic Equation

Simple quadratic equations take the form ax2+c=0 ax^2+c=0 and are solved using inverse operations. Once x2x^2 remains on the left-hand side, the equation can be written as x2=d,x^2=d, where d=-cad=\frac{\text{-} c}{a}. The value of dd gives the number of solutions the equation has.

d>0:2 real solutionsd=0:1 real solutiond<0:0 real solutions\begin{aligned} \mathbf{d>0} \quad &: \quad 2\text{ real solutions}\\ \mathbf{d=0} \quad &: \quad 1\text{ real solution}\\ \mathbf{d<0} \quad &: \quad 0\text{ real solutions} \end{aligned}
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Exercise

Without solving completely, determine the number of solutions the quadratic equation has. x2+75=50 x^2+75=50

Show Solution
Solution

A quadratic equation can have 0,1,0, 1, or 22 real solutions. Without solving completely, it's possible to determine how many just by isolating x2.x^2. Here, this means subtracting 7575 on both sides of the equation.

x2+75=50x2=-25 x^2+75=50 \quad \Rightarrow \quad x^2=\text{-} 25 Since -25<0,\text{-} 25 <0, the equation has 00 real solutions. Since the solutions to a quadratic equation give the zeros of the parabola, we can conclude that the parabola does not intersect the xx-axis.

Method

Solving Quadratic Equations with Square Roots

Quadratic equations of the form ax2+c=0 ax^2+c=0 can be solved using inverse operations. To undo the exponent of 22 on the x2x^2 term, square roots must be used. Consider 5x2500=0.5x^2-500=0.

1

Isolate x2x^2
To begin, isolate x2x^2 using inverse operations to move cc and then aa to the opposite side of the equation. Here, that means adding 500500 and then dividing by 5.5.
5x2500=05x^2-500=0
5x2=5005x^2=500
x2=100x^2=100

2

Square-root both sides of the equation
Now that x2x^2 has been isolated, it is necessary to undo the exponent. Exponents and radicals of the same index undo each other. Thus, square roots undo powers of 22 and cube roots undo powers of 3,3, etc. To finish isolating x2,x^2, square-root both sides of the equation.
x2=100x^2=100
x=±100x=\pm \sqrt{100}
Using a calculator to determine 100\sqrt{100} gives one value — 10.10. This is because calculators give the principal root of a square root. It is necessary to remember that there are two values that, when raised to the power of 2,2, equal 100.100. 1010=100and(-10)(-10)=100, 10 \cdot 10 = 100 \quad \text{and} \quad (\text{-} 10)(\text{-} 10)=100, Thus, x=10x=10 and x=-10.x=\text{-} 10.
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Exercise

The area of a square measures 81 cm2.81 \text{ cm}^2. Write a simple quadratic equation to represent the area. Then, determine the side length of the square.

Show Solution
Solution
To begin, recall that the area of a square can be found using the formula A=s2, A=s^2, where ss is the length of one of the sides. We can write an equation to represent the area of the given square by substituting A=81.A=81. This gives 81=s2. 81=s^2. Solving this equation for ss gives the side's length. Since ss is raised to the power of 2,2, we can square root both sides.
81=s281 = s^2
±81=s\pm \sqrt{81}= s
±9=s\pm 9 = s
s=±9s = \pm 9
The values of ss that satisfy the created equation are s=9s=9 and s=-9.s=\text{-} 9. Since ss represents a side length, it can only be positive. Thus, s=9s=9 is the only viable solution. The side length of the square measures 99 cm.
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