Quadratic Equations | Quantitative Aptitude

 Quadratic Equations

If f(x) is a polynomial of degree 2 i.e f(x) =ax2+bx+c=0$a{x}^2+bx+c=0$,  where a, b and c ∈ R and a ≠ 0, then the equation f(x) = 0 is termed as a quadratic equation. That implies quadratic equations are in general the polynomial equations with degrees equal to 2 and one variable.

In this article, we will learn about the quadratic equations definition, formula, nature of roots, how to solve the equation using different methods with solved examples and many such concepts. Quadratic equations in algebra are the most frequently used topic. It is a very important topic from the perspective of competitive and entrance examinations. Questions from such a topic are frequently asked in bank exams like SBI PO, IBPS PO, SBI Clerk, IBPS Clerk, followed by exams like SSC CGL, RBI Grade B, etc.

What are Quadratic Equations?

The equation ax2+bx+c=0$a{x}^2+bx+c=0$ is the general or standard form of quadratic equation. For example: x2+5x+12=0$x^2+5x+12=0$ and 5x2−4x−16=0$5x^2-4x-16=0$ are quadratic equations in the variable x. In competitive examinations, you must simplify a given equation before deciding whether it is quadratic or not.

In the equation above ‘a’ is termed the leading coefficient and ‘c’ is named the absolute term of f (x). The values of x that satisfy the quadratic equation are called the roots of the quadratic equation (denoted by;α,β). Quadratic equations are the basics of algebra and one must be familiar with them to be able to score well in the algebra section of any exam. Once the basics are clear, these equations become very easy to work with.

Quadratic Equation Formula

As per the definition, the standard form of a quadratic equation is ax2+bx+c=0$a{x}^2+bx+c=0$, therefore the name quadratic comes as “quad” which means square because the variable gets squared(x2$x^2$) here.

For the equation: ax2+bx+c=0$a{x}^2+bx+c=0$

• ‘a’ denotes the coefficient of x2$x^2$.
• ‘b’ denotes the coefficient of x.
• ‘c’ is the constant.

These are also identified as polynomial equations of degree 2(due to the presence of x2$x^2$) in one variable. The important formula for quadratic equations in determining the roots is: x=−b±b2−4ac√2a$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The quadratic equation in mathematics will always possess two roots and the nature of roots may be either real or imaginary depending upon the equation. Always remember that quadratic equations are second-degree polynomial i.e if a given equation appears to be a third-degree equation that is not quadratic. You will learn more about the above formula in the coming headings.

Formulas for Solving Quadratic Equations

In the previous heading, we saw the formula to find the roots. Let us learn some other important formulas for quadratic equations for solving various types of questions in different examinations.

For a standard quadratic equation of the form ax2+bx+c=0$a{x}^2+bx+c=0$.

• The roots are obtained by the formula: x=−b±b2−4ac√2a$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.
• The discriminant is calculated by the formula:b2−4ac$b^2-4ac$
•  If we consider α and β as the roots of a given quadratic equation then:
• The sum of roots of quadratic equation=S =α + β
• α + β = −ba=−Coefficient of x Coefficient of x2$-\frac{b}{a}=-\frac{\text{Coefficient of }x}{\text{ Coefficient of }{x}^2}$
• The product of the roots= P=αβ
• αβ= ca=Constant term Coefficient of x2$\frac{c}{a}=\frac{\text{Constant term}}{\text{ Coefficient of }{x}^2}$
• The quadratic equation in the form of the above obtained roots: x2–(α+β)x+(αβ)=0$x^2–(\alpha +\beta )x+(\alpha \beta )=0$
• For a cubic equation ax3+bx2+cx+d=0$a{x}^3+b{x}^2+cx+d=0$, If α, β and γ are the roots then:
• Sum of the three roots= α+β+γ=−ba$\alpha +\beta +\gamma =-\frac{b}{a}$
• Product of the combination of two roots=αβ+βγ+λα=ca$\alpha \beta +\beta \gamma +\lambda \alpha =\frac{c}{a}$ , and
• Product of all the three roots= αβγ=−da$\alpha \beta \gamma =-\frac{d}{a}$

Roots of Quadratic Equation

The standard quadratic equations ax2+bx+c=0$a{x}^2+bx+c=0$ being second-degree equations in x hold two answers. These two answers for x are termed as the roots of the quadratic equations specified with the symbol (α, β). The question that comes here is how to find the roots of a quadratic equation.

One can find the roots of a quadratic equation using the formula; −b±b2−4ac√2a$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Nature of Roots of Quadratic Equation

Now that you know what are the roots of a given quadratic equation and how to find them. Let us step forward and learn about the nature of the root and how to find the nature of the roots of a quadratic equation.

The nature of the roots can be found by finding the roots using the formula. Another approach to determining nature is to find the value of b2–4ac$b^2–4ac$. Here, b2–4ac$b^2–4ac$ is the part of the formula used to solve a quadratic equation. Let us understand how the nature of roots depends on this equation.

The general formula to obtain the roots is α=−b−b2−4ac√2a and β=−b+b2−4ac√2a$\alpha =\frac{-b-\sqrt{b^2-4ac}}{2a}\text{ and }\beta =\frac{-b+\sqrt{b^2-4ac}}{2a}$

If b2–4ac=0$b^2–4ac=0$, then the nature of the roots α and β are real and equal.

If b2−4ac<0$b^2-4ac<0$, then the roots α and β are not real or the roots are imaginary.

If b2−4ac>0$b^2-4ac>0$ then the nature of the roots α and β are real and different/unequal.

If b2−4ac>0$b^2-4ac>0$ + a perfect square then the roots α and β are real, rational and unequal.

If b2−4ac>0$b^2-4ac>0$ + not a perfect square, then the roots α and β are real, irrational and unequal.

What is Discriminant?

To compose the standard form of a quadratic equation, the x2$x^2$ term is penned first, followed by the x term, and eventually, the constant term is written. The value b2–4ac$b^2–4ac$ denotes the discriminant of a quadratic equation, and is expressed by the symbol ‘D’.

• If D > 0, the roots obtained are real and unequal.
• If D = 0, the roots fetched are real and equal.
• If D < 0, existing roots are imaginary and unequal.

Relationship between Coefficient and Roots of Quadratic Equation

For the quadratic equation ax2+bx+c=0$a{x}^2+bx+c=0$, the terms a,b and c specify the coefficient of x2$x^2$, x, and the constant term respectively. On the other hand, the roots of a quadratic equation are denoted by the symbols alpha (α), and beta (β). Now is there any relationship between the coefficient and roots of a quadratic equation? The answer would be yes.

The sum and product of roots of a quadratic equation, can be computed using these coefficients as shown:

The sum of the roots=α + β = −ba=−Coefficient of x Coefficient of x2$-\frac{b}{a}=-\frac{\text{Coefficient of }x}{\text{ Coefficient of }{x}^2}$

The product of roots of quadratic equation= αβ= ca=Constant term Coefficient of x2$\frac{c}{a}=\frac{\text{Constant term}}{\text{ Coefficient of }{x}^2}$

Also, if α and β, are the roots, then the quadratic equation can be formed using these roots. The equation is as follows:

x2–(α+β)x+(αβ)=0$x^2–(\alpha +\beta )x+(\alpha \beta )=0$

How to Solve Quadratic Equations?

Now that you know what a quadratic equation is with the definition and formula for solving such questions followed by information on the roots, their nature and roots of the quadratic equation formula. Let us now understand the different methods of solving quadratic equations.

Factorizing Quadratic Equation

Factoring quadratic equations is an approach where the equation ax2+bx+c=0$a{x}^2+bx+c=0$ is factorised as (x – ∝)(x – β) = 0 and then equation is solved to get x = ∝ or x = β. A standard form of quadratic equation in one variable can be solved by factorising the equation, i.e., by making factors out of the equation.

Consider the equation x2−7x+12=0$x^2-7x+12=0$ , it can also be written as (x-3)(x-4)=0. Now this factorised equation can be very easily solved using a property of real numbers called the zero-product rule. We will cover the detailed procedure in the separate article.

Formula Method of Finding Roots

Another approach for finding the roots is using the formula to find the roots of a quadratic equation. That is, solving the equation using the Sridharacharya formula. This can be a tricky method in terms of calculations but it is comparatively faster. This approach is generally used when the solution cannot be obtained using the factorisation method. Here the roots for the standard equation are obtained using the below formula:

x=−b±b2−4ac√2a$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Using the above formula you will get two roots of x one with a positive sign and the other with a negative sign as shown below:

−b+b2−4ac√2a,−b−b2−4ac√2a$\frac{-b+\sqrt{b^2-4ac}}{2a},\frac{-b-\sqrt{b^2-4ac}}{2a}$

Method of Completing the Square

In this method, you will learn how to find the roots of quadratic equations by the method of completing the squares. The steps involved in solving are:

For the equation; ax2+bx+c=0$a{x}^2+bx+c=0$

• First, divide all terms of the equation by the coefficient of x2$x^2$ i.e by ‘a’.
• The equation becomes: x2+bxa+ca=0$x^2+\frac{bx}{a}+\frac{c}{a}=0$
• Next, keep the terms of x2$x^2$ and x on one side and shift the term c/a to the right side of the equation; x2+bxa=−ca$x^2+\frac{bx}{a}=-\frac{c}{a}$
• Moreover, complete the square on the left side of the equation and counterbalance this by adding the identical value to the right.
• An equation similar to (x+s)2=w$(x+s{)}^2=w$ is obtained.
• In the next step, take the square root for both sides of the equation.
• Lastly, shift the numeral present on the left side of the equation to the right side to find x.

Graphing Method to Find the Roots

Lastly, the solution can also be obtained by graphing quadratic equations. Consider the general form of a quadratic equation as a function of y, the equation becomes y=ax2+bx+c$y=a{x}^2+bx+c$.

Next, substitute different values for x and obtain the respective values of y and plot the graph accordingly. The graph in general is a parabola-shaped graph for the quadratic equation.

Quadratic Equations having Common Roots

For the given two quadratic equations a1x2+b1x+c1=0 and a2x2+b2x+c2=0$a_1{x}^2+b_{1}x+c_1=0\text{ and }{a}_2{x}^2+b_{2}x+c_2=0$. The condition for the two equations has a common root can be obtained as shown below.

x2(b1c2−b2c1)=−x(a1c2−a2c1)=1(a1b2−a2b1)$\frac{x^2}{\left(b_1{c}_2-b_2{c}_1\right)}=\frac{-x}{\left(a_1{c}_2-a_2{c}_1\right)}=\frac{1}{\left(a_1{b}_2-a_2{b}_1\right)}$

The above equations are obtained using the determinant method.

Solving the equation we get:

x2=(b1c2−b2c1)(a1b2−a2b1) and x=(a2c1−a1c2)(a1b2−a2b1)$x^2=\frac{\left(b_1{c}_2-b_2{c}_1\right)}{\left(a_1{b}_2-a_2{b}_1\right)}\text{ and}x=\frac{\left(a_2{c}_1-a_1{c}_2\right)}{\left(a_1{b}_2-a_2{b}_1\right)}$

Now squaring x and equating with x2$x^2$ we get;

x2=(a2c1−a1c2)2(a1b2−a2b1)2$x^2=\frac{{\left(a_2{c}_1-a_1{c}_2\right)}^2}{{\left(a_1{b}_2-a_2{b}_1\right)}^2}$

(a2c1−a1c2)2(a1b2−a2b1)2=(b1c2−b2c1)(a1b2−a2b1)$\frac{{\left(a_2{c}_1-a_1{c}_2\right)}^2}{{\left(a_1{b}_2-a_2{b}_1\right)}^2}=\frac{\left(b_1{c}_2-b_2{c}_1\right)}{\left(a_1{b}_2-a_2{b}_1\right)}$

(a2c1−a1c2)2=(b1c2−b2c1)(a1b2−a2b1)${\left(a_2{c}_1-a_1{c}_2\right)}^2=\left(b_1{c}_2-b_2{c}_1\right)\left(a_1{b}_2-a_2{b}_1\right)$ is required condition for the equation to have one common root.

However, if both the roots of quadratic equations, a1x2+b1x+c1=0 and a2x2+b2x+c2=0$a_1{x}^2+b_{1}x+c_1=0\text{ and }{a}_2{x}^2+b_{2}x+c_2=0$ are common then:

a1a2=b1b2=c1c2$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

How to Solve Quadratic Equations with Common Roots?

In the previous heading, we saw the condition for a standard form of a quadratic equation to have one and two common roots. For a single common root the condition is;

x2(b1c2−b2c1)=−x(a1c2−a2c1)=1(a1b2−a2b1)$\frac{x^2}{\left(b_1{c}_2-b_2{c}_1\right)}=\frac{-x}{\left(a_1{c}_2-a_2{c}_1\right)}=\frac{1}{\left(a_1{b}_2-a_2{b}_1\right)}$ and for two common roots the condition is; a1a2=b1b2=c1c2$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$. Let us now understand how to solve such equations having common roots with an example.

Example: For the given quadratic equation 4x2+24x+6=0$4x^2+24x+6=0$ and 7x2–6x+k=0$7x^2–6x+k=0$ what is the value of K for which the equations will have a common root?

Solution: (a2c1−a1c2)2=(b1c2−b2c1)(a1b2−a2b1)${\left(a_2{c}_1-a_1{c}_2\right)}^2=\left(b_1{c}_2-b_2{c}_1\right)\left(a_1{b}_2-a_2{b}_1\right)$ is required condition for the equation to have one common root.

Form the equation, a1$a_1$ = 4, b1$b_1$ = 24, c1$c_1$ = 6, a2$a_2$ = 7, b2$b_2$ = -6 and c2$c_2$ = k

Substituting the values in the equation we get:

[(24k)−(−6×6)]×[4(−6)−(7×24)]=(7×6−4k)2$[(24k)-(-6\times 6)]\times [4\left(-6\right)-(7\times 24)]=(7\times 6-4k{)}^2$

[24k+36]×[−24−168]=(42−4k)2$[24k+36]\times [-24-168]=(42-4k{)}^2$

−4608k−6912=1764−336k+16k2$-4608k-6912=1764-336k+16k^2$

16k2+4272k+8676=0$16k^2+4272k+8676=0$

Therefore, the values of k are -264.95, -2.04.

How to Find the Range of a Quadratic Equation?

The general form of a quadratic function is represented by the equation y=ax2+bx+c$y=a{x}^2+bx+c$. As per the definition of the domain, it includes all the input values. Hence while determining the domain the concern of the graph is the horizontal axis, i.e. all the values of x. therefore, the domain is all real values.

Now coming towards the range: The range of a function is the collection of all possible output values. This implies that while calculating the range the focus area is the vertical axis, that is, the values on the y axis. The value of the range also depends on the opening of the parabola. To identify whether the parabola opens upwards or downward, check for the sign of the coefficient of x2$x^2$.  If the sign of “a” is positive, then the parabola is concave upward and if “a” is negative, the parabola is concave downward.

Also, if the parabola is concave upward, the range will have all the real values that are greater than or equal to; y=f(−b2a)$y=f\left(-\frac{b}{2a}\right)$. Similarly, if the parabola is concave downward, the range will have all the real values less than or equal to; y=f(−b2a)$y=f\left(-\frac{b}{2a}\right)$

Solved Example: Find the range of the quadratic function given by the equation y=x2+8x+12$y=x^2+8x+12$.

Solution: On comparing the standard equation with the given one.

y=ax2+bx+c$y=a{x}^2+bx+c$

We get a = 1, b = 8 and c = 12

As the coefficient of x2$x^2$ i.e. “a” is positive, the parabola is concave upward. Hence the range will have all the real values that are greater than or equal to y=f(−b2a)$y=f\left(-\frac{b}{2a}\right)$

x = -b / 2a

Substitute the values we get;

x = -8/2(12)

x = -8/24

x = -0.33

Substitute -0.33 for x in the given equation to obtain the y-coordinate at the vertex.

y=(−0.33)2+8(−0.33)+12$y=(-0.33{)}^2+8(-0.33)+12$

y = 0.1089 – 2.64+ 12

y = 9.46

Therefore, the y-coordinate of the vertex is 9.46.

As the parabola is concave upward, the range will have all the real values that are greater than or equal to 9.46.

Maximum and Minimum Value of Quadratic Equation

The minimum, as well as the maximum value of the quadratic equation, depends on the nature of the graph, i.e. whether the graph opens upwards or downwards. When the graph is concave upwards i.e. a>0 then the expression holds a minimum value at x = -b/2. On the other hand when the graph is concave downwards i.e. a<0 then the expression holds a maximum value at x = -b/2a.

How to Find Maximum and Minimum values of Quadratic Functions?

In the previous heading, we use the two particular cases for the maximum and minimum values of the quadratic equations.

When the coefficient of x2$x^2$ i.e. a > 0, the values lie in the ranges of [ f(-b/2a), ∞)

When the coefficient of x2$x^2$ i.e. a < 0, the values lie in the ranges of (-∞, f(-b/2a)]

Solved Example: Locate the maximum or minimum value of the given quadratic equation −9(x–3)2+3$-9(x–3{)}^2+3$.

Solution: Given the equation; −9(x–3)2+3$-9(x–3{)}^2+3$

Solving the equation we get:

−9(x−3)2+3=−9(x2−6x+9)+3=−9x2+54x−81+3$-9(x-3{)}^2+3=-9(x^2-6x+9)+3=-9x^2+54x-81+3$

=−9x2+54x−78$-9x^2+54x-78$

As, the coefficient of x2$x^2$ is less than zero i.e. a < 0, the expression will hold a maximum value at x = -b/2a

By substituting the values we get, x=3

Therefore, the maximum value of the quadratic equation−9(x–3)2+3$-9(x–3{)}^2+3$ is 3.

Word Problems on Quadratic Equation

A lot of daily life problems can be solved using quadratic equations. To solve any kind of word problem, first, you must translate the words into algebraic equations and then determine which method to apply to solve the equation so formed. Word problems are not specific to any one type. There can be innumerable types of word problems. Hence, there is no fixed technique or method to solve word problems. It’s all about logic and practice. However, if you keep the following points in mind, then it will be easy for you to solve any kind of word problem.

1. First, you must carefully read and try to understand what the question is asking for and what quantity is to be found.
2. Write whatever information is given in the question.
3. The quantity which is unknown, suppose it to be any variable, for example, x.
4. Now see which expressions are equal and form an equation using them.
5. Solve the resulting equation for the variable.
6. Lastly, check what the question demands, and then substitute the value of the variable you just found, to find the final answer.

Some practice problems for the are as follows:

Word Problem 1: Nine times a whole number is equal to five less than twice the square of the number. Find the number?

Solution: Let the required whole number be x.

According to the question,
⇒9x=2x2−5⇒2x2−9x−5=0⇒(x−5)(2x+1)=0⇒x−5=0 or 2x+1=0⇒x=5 or x=−12$\begin{array}{l}\Rightarrow 9x=2x^2-5\Rightarrow 2x^2-9x-5=0\\ \Rightarrow (x-5)(2x+1)=0\\ \Rightarrow x-5=0\text{ or }2x+1=0\\ \Rightarrow x=5\text{ or }x=-\frac{1}{2}\end{array}$

Since x is supposed to be a whole number, the answer, i.e., the required whole number is 5.

Word Problem 2: Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

Solution: Let the natural number be x.
According to the question,
⇒x2−84=3(x+8)$\Rightarrow x^2-84=3(x+8)$
⇒x2−3x−108=0⇒x2−12x+9x−108=0$\Rightarrow x^2-3x-108=0\Rightarrow x^2-12x+9x-108=0$
⇒x(x−12)+9(x−12)=0⇒(x−12)(x+9)=0$\Rightarrow x(x-12)+9(x-12)=0\Rightarrow (x-12)(x+9)=0$
⇒x−12=0$\Rightarrow x-12=0$ or x+9=0⇒x=12$x+9=0\Rightarrow x=12$ or x=−9$x=-9$.
Since x is supposed to be a natural number, we will not consider x=-9. Hence, the required natural number is 12.

Word Problem 3: The sum of two natural numbers is 8. If the sum of their reciprocals is 815$\frac{8}{15}$, then find the numbers?

Solution: Let the numbers be x and 8-x where x∈N$x\in N$ and x<8$x<8$.
According to the question ⇒1x+18−x=815$\Rightarrow \frac{1}{x}+\frac{1}{8-x}=\frac{8}{15}$
⇒8−x+xx(8−x)=815$\Rightarrow \frac{8-x+x}{x(8-x)}=\frac{8}{15}$
⇒8x(8−x)=815$\Rightarrow \frac{8}{x(8-x)}=\frac{8}{15}$
⇒15=x(8−x)$\Rightarrow 15=x(8-x)$
⇒15=8x−x2$\Rightarrow 15=8x-x^2$
⇒x2−8x+15=0$\Rightarrow x^2-8x+15=0$
⇒(x−3)(x−5)=0$\Rightarrow (x-3)(x-5)=0$
⇒x−3=0$\Rightarrow x-3=0$ or x−5=0$x-5=0$
⇒x=3$\Rightarrow x=3$ or x=5$x=5$.
When x=3, then the numbers are 3,8-3 i.e. 3,5.
When x=5, then the numbers are 5,8-5 i.e. 5,3.
Hence, the required numbers are 3,5.

If you’ve learned Quadratic Equations, you can move on to learn about Average formulas in detail here!

Solved Examples on Quadratic Equation

By now you might be familiar with the definition, roots of quadratic equation formulas, nature of roots, discriminant, how to solve these equations using different methods and terms like range, maximum and minimum value, etc. Let us now practise some quadratic equation questions to understand all the discussed content from the exam viewpoint.

Solved Example 1: Find the roots of the given equation 3x2−4x−4=0$3x^2-4x-4=0$.

Solution: Comparing the given equation with ax2+bx+c=0$a{x}^2+bx+c=0$, we get
⇒a=3,b=−4,c=−4$\Rightarrow a=3,b=-4,c=-4$
Substituting the values in the formula; ⇒x=−b±b2−4ac√2a$\Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
⇒x=−(−4)=(−4)2−4.3⋅(−4)√2×3

=4±16+48√6=4±64√6=4±86$\Rightarrow x=\frac{-(-4)=\sqrt{(-4{)}^2-4.3\cdot (-4)}}{2\times 3}=\frac{4\pm \sqrt{16+48}}{6}=\frac{4\pm \sqrt{64}}{6}=\frac{4\pm 8}{6}$
⇒4+86,4−86=126,−46=2,−23$\Rightarrow \frac{4+8}{6},\frac{4-8}{6}=\frac{12}{6},\frac{-4}{6}=2,-\frac{2}{3}$.
Hence, the roots of the given equation are 2,−23$2,-\frac{2}{3}$.

Solved Example 2: Obtain the roots of the given equation 2x2+7–√x−7=0$2x^2+\sqrt{7}x-7=0$.

Solution: Comparing the given equation with ax2+bx+c=0$a{x}^2+bx+c=0$, we get
⇒a=2,b=7–√,c=−7$\Rightarrow a=2,b=\sqrt{7},c=-7$
Substituting the values in the formula; ⇒x=−b±b2−4ac√2a$\Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
⇒x=−7√±(7√)2−4.2⋅(7√)√2×2=7√±63√4$\Rightarrow x=\frac{-\sqrt{7}\pm \sqrt{(\sqrt{7}{)}^2-4.2\cdot (\sqrt{7})}}{2\times 2}=\frac{\sqrt{7}\pm \sqrt{63}}{4}$
⇒x=7√±37√4=27√4,−47√4=7√2,−7–√$\Rightarrow x=\frac{\sqrt{7}\pm 3\sqrt{7}}{4}=\frac{2\sqrt{7}}{4},\frac{-4\sqrt{7}}{4}=\frac{\sqrt{7}}{2},-\sqrt{7}$
Hence, the roots of the given equation are 7√2,−7–√$\frac{\sqrt{7}}{2},-\sqrt{7}$.

Solved Example 3: Determine the roots of the equation (x+3) (x-3)=40.

Solution: Given(x+3) (x-3)=40
x2−9=40$x^2-9=40$
x2−9−40=0$x^2-9-40=0$
x2−49=0$x^2-49=0$ writing as (ax2+bx+c=0)$(a{x}^2+bx+c=0)$

Using factorization method;
(x+7) (x-7)=0 (factorising left side)
x+7=0 or x-7=0 (zero-product rule)
x=-7 or x=7.
Hence, the roots of the given equation are -7, 7.

Solved Example 4: What are the roots of the equation 2x+9−−−−−√=13−x$\sqrt{2x+9}=13-x$?

Solution: Given 2x+9−−−−−√=13−x$\sqrt{2x+9}=13-x$
Squaring both sides, we get
2x+9=(13−x)2$2x+9=(13-x{)}^2$
⇒2x+9=169−26x+x2$\Rightarrow 2x+9=169-26x+x^2$
⇒−x2+28x−160=0$\Rightarrow -x^2+28x-160=0$
⇒x2−28x+160=0$\Rightarrow x^2-28x+160=0$
⇒(x−8)(x−20)=0$\Rightarrow (x-8)(x-20)=0$ (factorising left side)
⇒x−8=0orx−20=0$\Rightarrow x-8=0orx-20=0$ (zero-product rule)
⇒x=8orx=20$\Rightarrow x=8orx=20$
But x=20 does not satisfy the given equation, so it is rejected. Hence, the root of the given equation is 8.

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