Introduction to Quadratic Equations and Finding the Roots

Introduction to Quadratic Equations and Finding the Roots: Quadratic equations are commonly used in everyday life, such as when calculating areas, a product’s profit, or estimating an object’s speed. It can be defined as an equation in one variable in which the maximum power of the variable is \(2\), with the most common form being \(a{x^2} + bx + c = 0\).

Here, \(x\) is the variable, with coefficients as \(a,b\), and \(c\), and the \(a \ne 0\). A quadratic equation can have a maximum of \(2\) roots, and these roots can be real or imaginary depending on the sign of the discriminant. Now, what is discriminant? Let us study the formula of discriminant, the nature of roots of the quadratic equation, and its graph in the following article.

What is the Degree of a Polynomial?

The degree of a polynomial is the highest power of a variable in a polynomial equation. For example: \(3{x^8} + 4{x^3} + 5x + 1\) is a polynomial of degree \(8\).

We also classify the polynomial on the basis of degrees as follows:

Learn About Quadratic Equations Here

Thus, a polynomial of degree \(2\) is known as a Quadratic Polynomial. A quadratic polynomial, when equated to zero, becomes a quadratic equation.

Roots of a Quadratic Equation

The roots of a quadratic equation are the values of variables that satisfy that equation. In other words, if \(f(\alpha ) = 0\), then \(x = \alpha \) is known as a root of quadratic equation \(f(x) = 0\).

Relationship Between Roots and Degree of a Polynomial Equation

According to the fundamental theorem of algebra, the maximum number of roots a polynomial can have equals the degree of the polynomial. For example, the number of roots of a linear polynomial is \(1\), the maximum number of roots in a quadratic polynomial and cubic polynomial is \(2\) and \(3\), respectively.

General Form of a Quadratic Equation

The general form of a quadratic equation is \(a{x^2} + bx + c = 0\)

Where, \(a \ne 0\) and \(a,b,c\) are numbers (real or complex), and \(x\) is the variable.

The following theorem suggests the number of roots of a quadratic equation.

Theorem: A quadratic equation cannot have more than two roots.

Proof: If possible, let \(\alpha ,\beta ,\gamma \) be three distinct roots of the quadratic equation \(a{x^2} + bx + c = 0\), where \(a,b,c \in R\) and \(a \ne 0\).

Then, each one of \(\alpha ,\beta ,\gamma \) will satisfy this equation.

\( \Rightarrow a{\alpha ^2} + b\alpha + c = 0\,\,\,\,…(i)\)

\(a{\beta ^2} + b\beta + c = 0\,\,….(ii)\)

\(a{\gamma ^2} + b\gamma + c = 0\,\,\,….(iii)\)

Subtracting \((ii)\) from \((i)\), we get,

\(a\left( {{\alpha ^2} – {\beta ^2}} \right) + b(\alpha – \beta ) = 0\)

\( \Rightarrow (\alpha – \beta )[ a(\alpha + \beta ) + b] = 0\)

\( \Rightarrow a(\alpha + \beta ) + b = 0\) [\(\because \alpha \) and \(\beta \) are distinct \(\therefore \,\alpha – \beta \ne 0\)] \(….(iv)\)

Subtracting \((iii)\) from \((ii)\) we get,

\(a\left( {{\beta ^2} – {\gamma ^2}} \right) + b(\beta – \gamma ) = 0\)

\( \Rightarrow (\beta – \gamma )[ a(\beta + \gamma ) + b] = 0\)

\( \Rightarrow a(\beta + \gamma ) + b = 0\) [\(\because \beta \) and \(\gamma \) are distinct \(\therefore \beta – \gamma \ne 0\)] \(….(v)\)

Subtracting \((v)\) from \((iv)\), we get,

\(a(\alpha – \gamma ) = 0\)

\(\therefore \alpha = \gamma \quad [\because a \ne 0]\)

But, this is not possible because \(\alpha \) and \(\gamma \) are distinct.

Thus, the assumption that a quadratic equation has three distinct real roots is wrong.

Hence, it is proved that a quadratic equation cannot have more than \(2\) roots.

Remark: It follows from the above theorem that if a quadratic equation is satisfied by more than two values of \(x\), it is satisfied by every value of \(x\), and it is an identity.

Quadratic Equations with Real Coefficients

Consider the quadratic equation:

\(a{x^2} + bx + c = 0\,\,\,…(i)\)

Where, \(a,b,c \in R\) and \(a \ne 0\)

Multiplying both sides of \((i)\) by \(a\), we get

\({a^2}{x^2} + abx + ac = 0\)

\( \Rightarrow {a^2}{x^2} + abx + \frac{{{b^2}}}{4} = \frac{{{b^2}}}{4} – ac\)

\( \Rightarrow {\left( {ax + \frac{b}{2}} \right)^2} = \frac{{{b^2} – 4ac}}{4}\)

\( \Rightarrow ax + \frac{b}{2} = \pm \frac{{\sqrt {{b^2} – 4ac} }}{2}\)

\( \Rightarrow ax = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{2}\)

\( \Rightarrow x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)

Thus, the quadratic equation \(a{x^2} + bx + c = 0\) has two roots, say \(\alpha \) and \(\beta \) given by

\(\alpha = \frac{{ – b + \sqrt {{b^2} – 4ac} }}{{2a}}\)

\(\beta = \frac{{ – b – \sqrt {{b^2} – 4ac} }}{{2a}}\)

Where \(a,b,c \in R\), and \(a \ne 0\).

Discriminant of a Quadratic Equation

We observed that the roots of the equation \(a{x^2} + bx + c = 0\) are given by

\(D = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)

where \(a,b,c \in R\), and \(a \ne 0\)

Thus, the nature of the roots depends on the quantity under the square root sign, i.e. \({b^2} – 4ac\). This quantity is called the discriminant of the equation and is usually denoted by \(D\).

Nature of Roots of a Quadratic Equation

Case 1: When \(a,b,c\) are real numbers, \(a \ne 0\).

Case 2: When \(a,b,c\) are rational numbers, \(a \ne 0\)

Graph of Quadratic Function

\(f(x) = a{x^2} + bx + c = a\left( {{x^2} + \frac{b}{a}x + \frac{c}{a}} \right)\)

\( = a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} – \frac{{{b^2} – 4ac}}{{4{a^2}}}} \right]\)

Let \({b^2} – 4ac = D\)

\(\therefore f(x) = a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} – \frac{D}{{4{a^2}}}} \right]\) i.e. \(y = a{\left( {x + \frac{b}{{2a}}} \right)^2} – \frac{D}{{4a}}\)

\(y + \frac{D}{{4a}} = a{\left( {x + \frac{b}{{2a}}} \right)^2}\) represents a parabola with vertex at \(\left( { – \frac{b}{{2a}}, – \frac{D}{{4a}}} \right)\) and faces upwards if \(a > 0\) and downwards if \(a < 0\).

Case 1: When \(D=0\), the equation \(a{x^2} + bx + c = 0\) has two equal real roots \(x = – \frac{b}{{2a}}, – \frac{b}{{2a}}\), and \(f(x) = a{\left( {x + \frac{b}{{2a}}} \right)^2}\) which is a parabola touching \(x-\)axis.

Case 2: When \(D<0\) the equation \(a{x^2} + bx + c = 0\) has no real roots, and \(f(x) = a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} – } \right.\left. {\frac{D}{{4{a^2}}}} \right]\), which represents a parabola with vertex at \(\left( { – \frac{b}{{2a}}, – \frac{D}{{4a}}} \right)\).

Case 3: When \(D>0\), the equation has two distinct real roots, say \(\alpha ,\beta (\alpha < \beta )\). Then \(f(x) = a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} – \frac{\Delta }{{4{a^2}}}} \right]\) represents a parabola with vertex at \(\left( { – \frac{b}{{2a}}, – \frac{D}{{4a}}} \right)\).

Sign of the Expression \(a{x^2} + bx + c\)

The sign of \(a{x^2} + bx + c\) for various values of \(D\), and \(a\) are tabulated below.

Maxima and Minima

From the graphs, we see that maximum or minimum occurs at \(x = – \frac{b}{{2a}}\), and its corresponding value is \( – \frac{D}{{4a}}\). When \(a>0\), we have a minima, and when \(a<0\), we have maxima.

Quadratic Inequalities

A quadratic inequality is an equation of second degree that uses an inequality sign instead of an equal sign.
Example: \({x^2} – 6x – 16 \le 0,2{x^2} – 11x + 12 > 0\)

Steps to Solve Quadratic Inequalities

Step 1: Consider the quadratic inequality \({x^2} + bx + c > 0\), i.e. \(f(x) > 0\), where \(f(x) = {x^2} + bx + c > 0\).

Step 2: Make linear factors in L.H.S. using the method of factorisation.

Step 3: Let the roots of the equation \({x^2} + bx + c = 0\) be \(\alpha \) and \(\beta \). So, write the  given inequation as \((x – \alpha )(x – \beta ) > 0\), where \(\alpha < \beta \).

Step 4: Thus, the critical points corresponding to the given inequation are \(\alpha \) and \(\beta \). Plot these critical points on a real number line

Step 5: Divide the number line into three sub intervals \(( – \infty ,\alpha ),(\alpha ,\beta )\) and \((\beta ,\infty )\) such that \(f(x) > 0\) on \((\beta ,\infty )\) or \(( – \infty ,\alpha )\), and \(f(x) < 0\) on \((\alpha ,\beta )\).

Note: The values of \(f(x)\) for the right most interval on the above number line is positive, and then for the remaining intervals, the sign of \(f(x)\) changes alternatively.

Step 6: The required solution for \(f(x) > 0\) is \(( – \infty ,\alpha ) \cup (\beta ,\infty )\).

Note:

• If we have inequality sign as \(f(x) \ge 0\) then the solution for this inequality will be \(( – \infty ,\alpha ] \cup [\beta ,\infty )\).
• Similarly, the solutions for \(f(x) < 0\) is \((\alpha ,\beta )\), and the solution for \(f(x) \le 0\) is \([\alpha ,\beta ]\).

Relationship Between Roots and Coefficients of a Quadratic Equation

If \(\alpha \) and \(\beta \) are roots of a quadratic equation, \(a{x^2} + bx + c = 0\), where \(a \ne 0\) then, we have

\(\alpha + \beta = – \frac{b}{a},\quad \alpha \beta = \frac{c}{a}\)

Sum of roots \( = – \frac{{{\rm{ coefficient}}\,{\rm{of }}\,{x}}}{{{\rm{ coefficient}}\,{\rm{of }}\,{x^2}}}\)

Product of roots \( = \frac{{{\rm{ constant}}\,{\rm{term }}}}{{{\rm{ coefficient}}\,{\rm{of }}\,{x^2}}}\)

Formation of Quadratic Equation From Roots

Let \(\alpha \) and \(\beta \) be the two roots of a quadratic equation \(a{x^2} + bx + c = 0\).

\(\therefore \,\alpha + \beta = – \frac{b}{a}\,\,…(i)\,\,\,\,\alpha \beta = \frac{c}{a}\,\,…(ii)\)

Now, \(a{x^2} + bx + c = 0\).

\( \Rightarrow {x^2} + \frac{b}{a}x + \frac{c}{a} = 0\)

Using \(i\), and \(ii\), we have

\({x^2} – (\alpha + \beta )x + \alpha \beta = 0\)

Hence, if \(\alpha \) and \(\beta \) be the roots of the quadratic equation, then the quadratic equation is given by

\({x^2} – (\alpha + \beta )x + \alpha \beta = 0\)

Solved Examples – Introduction to Quadratic Equations and Finding the Roots

Q.1. Discuss the sign of the following quadratic function:
\({x^2} + x + 1\)
Ans: Given: \({x^2} + x + 1\)
Comparing it with \(a{x^2} + bx + c\), we have
\(a = 1 > 0,h = 1,c = 1\)
Substituting the known values we get,
\(D = {b^2} – 4ac\)
\( = {(1)^2} – 4 \times 1 \times 1 = – 3 < 0\)
Thus, \(a > 0,D < 0\)
Hence, \({x^2} + x + 1 > 0\) for all real \(x\). It carries a positive sign.

Q.2. Solve the following equation
\(\sqrt 3 {x^2} – \sqrt 2 x + 3\sqrt 3 = 0\)
Ans: Given: \(\sqrt 3 {x^2} – \sqrt 2 x + 3\sqrt 3 = 0\)
Comparing it with \(a{x^2} + bx + c = 0\), we get
\(a = \sqrt 3 ,b = – \sqrt 2 ,c = 3\sqrt 3 \)
Discriminant \( = {b^2} – 4ac = {( – \sqrt 2 )^2} – 4 \times \sqrt 3 \times 3\sqrt 3 \)
\({\rm{ = 2 – 36 = – 34}}\)
\(\therefore x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)
\( = \frac{{ – ( – \sqrt 2 ) \pm \sqrt { – 34} }}{{2 \times \sqrt 3 }}\)
\(x = \frac{{\sqrt 2 \pm \sqrt {34} i}}{{2\sqrt 3 }}\)

Q.3. Find the roots of the equation:
\(\frac{1}{{x + 1}} + \frac{2}{{x + 2}} = \frac{4}{{x + 4}}\)
Ans: Given: \(\frac{1}{{x + 1}} + \frac{2}{{x + 2}} = \frac{4}{{x + 4}}\)
\( \Rightarrow \frac{{1(x + 2) + 2(x + 1)}}{{(x + 1)(x + 2)}} = \frac{4}{{x + 4}}\)
\( \Rightarrow (3x + 4)(x + 4) = 4(x + 1)(x + 2)\)
\( \Rightarrow 3{x^2} + 12x + 4x + 16 = 4{x^2} + 12x + 8\)
\( \Rightarrow – {x^2} + 4x + 8 = 0\)
\( \Rightarrow {x^2} – 4x – 8 = 0\)
Comparing it with \(a{x^2} + bx + c = 0\), we get
\(a = 1,b = – 4,c = – 8\)
\(\therefore x = \frac{{ – ( – 4) \pm \sqrt {{{( – 4)}^2} – 4(1)( – 8)} }}{{2(1)}}\)
\( = \frac{{4 \pm \sqrt {16 + 32} }}{2}\)
\( = \frac{{4 \pm \sqrt {48} }}{2}\)
\( = \frac{{4 \pm 4\sqrt 3 }}{2}\)
\(x = 2 \pm 2\sqrt 3 \)
Hence, the roots of the given equation are \(2 + 2\sqrt 3 \) and \(2 – 2\sqrt 3 \).

Q.4. Solve for \(x\): \({x^2} – 5x + 6 \ge 0\).
Ans: Step 1: Factorise the L.H.S., \((x – 3)(x – 2) \ge 0\)
Step 2: Determine the critical values of \(x\).
Here, the critical values are \(x=3,2\).
Step 3: Plotting these critical values on the real number line, we have

Step 4: Let \(f(x) = (x – 3)(x – 2)\), we need to check for the sign of \(f(x)\) in the intervals, \(( – \infty ,2),(2,3)\) and \((3,\infty )\).
Thus, we have

Step 5: We observe that \(f(x) \ge 0\) for \(x \in ( – \infty ,2] \cup [3,\infty )\).
Hence, the required solution is \(( – \infty ,2] \cup [3,\infty )\)

Q.5. Determine a positive real value of \(k\) such that both the equations \({x^2} + kx + 64 = 0\) and \({x^2} – 8x + k = 0\) may have real roots.
Ans: The given equations are \({x^2} + kx + 64 = 0\) and \({x^2} – 8x + k = 0\)
As both the equations have real roots, the discriminant of each \(D \ge 0\).
Thus, we have
\({k^2} – 4 \times 1 \times 64 \ge 0\) and \({( – 8)^2} – 4 \times 1 \times k \ge 0\)
\( \Rightarrow {k^2} – 256 \ge 0\) and \(64 – 4k \ge 0\)
\( \Rightarrow (k + 16)(k – 16) \ge 0\) and \(16 – k \ge 0\)
\( \Rightarrow k \ge – 16\) or \(k \ge 16\) and \(k \le 16\)
But \(k\) is a positive real number. So, we have
\(k \ge 16\) and \(k \le 16 \Rightarrow k = 16\)
Hence, the required value of \(k\) is \(16\).

Summary

A quadratic equation is defined as an equation in one variable in which the maximum power of the variable is \(2\). The roots of the quadratic equation \(a{x^2} + bx + c = 0,a \ne 0\) are \(\frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\), where \(a,b,c \in R\). If a quadratic equation with real coefficients has discriminant \(D>0\), then the roots are real and equal; if \(D=0\), then the roots are real and repeated; if \(D<0\), then the roots are complex. Also, the sign of the quadratic function changes with the values of \(D\) and \(a\). If \(\alpha \) and \(\beta \) be the roots of the quadratic equation, then the quadratic equation is given by \({x^2} – (\alpha + \beta )x + \alpha \beta = 0\). Similarly, for \(a{x^2} + bx + c = 0\), where \(a \ne 0\), we know that \(\alpha + \beta = – \frac{b}{a},\alpha \beta = \frac{c}{a}\).

Frequently Asked Questions (FAQs)

Q.1. How do you find the roots of a quadratic equation?
Ans: The roots of the quadratic equation \(a{x^2} + bx + c = 0\), where \(a \ne 0\), are given by \(x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}\)

Q.2. How do you introduce quadratic equations?
Ans: A quadratic equation is defined as an equation in one variable in which the maximum power of the variable is \(2\). Examples of quadratic equations are \({x^2} – 7x + 12 = 0\), and \({x^2} + 5x – 1050 = 0\).

Q.3. What are quadratic roots?
Ans: The values of a variable that satisfy the given quadratic equation are the roots of that quadratic equation or quadratic roots.

Q.4. How do you write a quadratic equation with given roots and points?
Ans: If \(\alpha \) and \(\beta \) be the roots of the quadratic equation, then the quadratic equation is given by \({x^2} – (\alpha + \beta )x + \alpha \beta = 0\)
If the roots of a quadratic equation are given, then we can determine the quadratic equation, using the formula, \({x^2} – ({\rm{Sum}}\,{\rm{of}}\,{\rm{roots}}) + ({\rm{Product}}\,{\rm{of}}\,{\rm{roots}}) = 0\)

Q.5. How do you find if a quadratic equation has real roots?
Ans: The discriminant of a quadratic equation \(a{x^2} + bx + c = 0\) is given by \(D = {b^2} – 4ac\)
If \(D>0\) then the quadratic equation has real and unequal roots, and if \(D=0\), then the quadratic equation has real and equal roots.
Thus the roots of a quadratic equation are real for \(D \ge 0\).

NCERT Solutions for Class 10 Maths Chapter 4

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