Class 10 Mathematics Chapter 4 Notes: Quadratic Equations

Contents hide

1 Introduction: from polynomial to equation

2 Quadratic equation and its standard form

3 Turning a situation into a quadratic equation

4 Roots of a quadratic equation

5 Solving by factorisation

6 The quadratic formula

7 Discriminant and nature of roots

8 Pole in a circular park: Pythagoras meets the discriminant

What you will learn in this chapter

What a quadratic equation is, and how to write it in standard form $ax^2+bx+c=0$
How to turn a word problem (area, numbers, ages, speed, cost) into a quadratic equation
How to find the roots by factorisation (splitting the middle term)
The quadratic formula, and how the discriminant tells you the nature of the roots without solving

Quadratic equationStandard formRootFactorisationQuadratic formulaDiscriminantNature of roots

1 Introduction: from polynomial to equation

In Chapter 2 you met the quadratic polynomial $ax^2+bx+c$ (with $a\neq0$ ): a polynomial whose highest power of $x$ is 2 (its degree is 2). When we put this polynomial equal to zero, we get a quadratic equation. Such equations turn up naturally in real-life problems.

A charity trust wants to build a prayer hall with a carpet area (floor area) of 300 m², whose length is one metre more than twice its breadth. If the breadth is $x$ m, the length is $(2x+1)$ m, so

(2x+1)\cdot x = 300 \;\Rightarrow\; 2x^2+x-300=0

Figure 1: The prayer hall. Breadth x m and length (2x + 1) m give the equation 2x² + x − 300 = 0, which we solve in Section 5. — **Figure 1**: The prayer hall. Breadth x m and length (2x + 1) m give the equation 2x² + x − 300 = 0, which we solve in Section 5.

2 Quadratic equation and its standard form

Learn by heartDefinition 1: Quadratic equation

A quadratic equation in the variable $x$ is an equation of the form $ax^2+bx+c=0$ , where $a,b,c$ are real numbers and $a\neq0$ .

Learn by heartDefinition 2: Standard form

Any equation $p(x)=0$ , where $p(x)$ is a polynomial of degree 2, is a quadratic equation. When its terms are written in descending order of their degrees (the $x^2$ term first, then the $x$ term, then the number), we get the standard form $ax^2+bx+c=0$ , $a\neq0$ .

So $2x^2-3x+1=0$ , $4x-3x^2+2=0$ and $1-x^2+300=0$ are all quadratic equations. The second one in standard form is $-3x^2+4x+2=0$ , so there $a=-3$ , $b=4$ , $c=2$ .

Common Mistake

Do not decide by looks. Simplify both sides first (the book notes that often we need to simplify the given equation before deciding whether it is quadratic), then check whether an $x^2$ term is left and no higher power remains.

Equation (NCERT Example 2)	After simplifying	Quadratic?
$(x-2)^2+1=2x-3$	$x^2-6x+8=0$	Yes
$x(x+1)+8=(x+2)(x-2)$	$x^2+x+8=x^2-4$ , so $x+12=0$	No: the $x^2$ cancels, it only looked quadratic
$x(2x+3)=x^2+1$	$x^2+3x-1=0$	Yes
$(x+2)^3=x^3-4$	$x^3+6x^2+12x+8=x^3-4$ , so $6x^2+12x+12=0$ , i.e. $x^2+2x+2=0$	Yes: the $x^3$ cancels, it only looked cubic (degree 3)

3 Turning a situation into a quadratic equation

Step 1Let the unknown be

x

→

Step 2Write every other quantity in terms of

x

→

Step 3Use the given condition to form an equation

→

Step 4Simplify to

ax^2+bx+c=0

Solved Example 1: marbles (NCERT Example 1(i))

Q. John and Jivanti together have 45 marbles. Both lost 5 marbles each, and the product of the numbers they now have is 124. Represent this as a quadratic equation.

Solution: Let John have $x$ marbles. Then Jivanti has $45-x$ . After losing 5 each: John $x-5$ , Jivanti $40-x$ .

(x-5)(40-x)=124 \;\Rightarrow\; -x^2+45x-200=124 \;\Rightarrow\; x^2-45x+324=0

Solved Example 2: cost of toys (NCERT Example 1(ii))

Q. A cottage industry (a small business run from home) makes some toys in a day. The cost of each toy (in ₹) is 55 minus the number of toys made that day. On one day the total cost was ₹750. Form the equation.

Solution: Let $x$ toys be made. Cost of each toy $=₹(55-x)$ , so total cost $=x(55-x)$ .

x(55-x)=750 \;\Rightarrow\; -x^2+55x-750=0 \;\Rightarrow\; x^2-55x+750=0

The same four steps handle these common types. Here is how the unknown and the condition look in each:

Type	Let $x$ be	Condition used	Equation
Area (plot of 528 m², length $=2\times$ breadth $+1$ )	breadth	length × breadth = area	$2x^2+x-528=0$
Consecutive integers (one after the other, like 17 and 18) with product 306	smaller integer	$x(x+1)=306$	$x^2+x-306=0$
Consecutive integers with sum of squares 365	smaller integer	$x^2+(x+1)^2=365$	$x^2+x-182=0$ (13 and 14)
Right triangle: altitude 7 cm less than the base, hypotenuse 13 cm	base in cm (altitude $x-7$ )	Pythagoras: $x^2+(x-7)^2=13^2$	$x^2-7x-60=0$ (base 12 cm, altitude 5 cm)
Ages: a mother is 26 years older than her son; 3 years from now the product of their ages will be 360	son’s present age (mother’s is $x+26$ )	after 3 years: $(x+3)(x+29)=360$	$x^2+32x-273=0$
Speed: a train covers 480 km; if it went 8 km/h slower, it would take 3 hours more	speed in km/h	$\dfrac{480}{x-8}-\dfrac{480}{x}=3$	$x^2-8x-1280=0$

Exam Tip

In speed problems use time = distance ÷ speed. The slower journey takes MORE time, so write (slower time) − (usual time) = extra hours. Then multiply both sides by $x(x-8)$ to clear the fractions.

4 Roots of a quadratic equation

Learn by heartDefinition 3: Root

A real number $\alpha$ (read “alpha”) is called a root of the quadratic equation $ax^2+bx+c=0$ , $a\neq0$ , if $a\alpha^2+b\alpha+c=0$ . We also say that $x=\alpha$ is a solution of the equation, or that $\alpha$ satisfies it.

For example, putting $x=1$ in $2x^2-3x+1$ gives $2-3+1=0$ , so 1 is a root of $2x^2-3x+1=0$ . It is also a zero of the polynomial $2x^2-3x+1$ , that is, a value of $x$ that makes the polynomial equal to 0.

Two facts about roots

Roots = zeroes: the zeroes of the polynomial $ax^2+bx+c$ and the roots of the equation $ax^2+bx+c=0$ are the same.
At most two roots: a quadratic polynomial has at most two zeroes (Chapter 2), so a quadratic equation has at most two roots.

5 Solving by factorisation

In Class IX you factorised quadratic polynomials by splitting the middle term. The same skill finds the roots: factorise $ax^2+bx+c$ into two linear factors (factors of degree 1, such as $2x-3$ ), then put each factor equal to zero.

Step 1Find two numbers whose product is

a\times c

and whose sum is

b

↓

Step 2Split

bx

into two terms using these numbers, and take out common factors in pairs

↓

Step 3Write the equation as (linear factor)(linear factor)

=0

↓

ResultEach factor

=0

gives one root

Solved Example 3 (NCERT Example 3)

Q. Find the roots of $2x^2-5x+3=0$ by factorisation.

Solution: $a\times c=6$ and $b=-5$ , so use $-2$ and $-3$ (product 6, sum $-5$ ):

2x^2-2x-3x+3 = 2x(x-1)-3(x-1) = (2x-3)(x-1)=0

So $2x-3=0$ or $x-1=0$ . The roots are $\dfrac32$ and $1$ .

Solved Example 4 (NCERT Example 4)

Q. Find the roots of $6x^2-x-2=0$ .

Solution: $a\times c=-12$ , $b=-1$ : use $3$ and $-4$ .

6x^2+3x-4x-2 = 3x(2x+1)-2(2x+1) = (3x-2)(2x+1)=0

The roots are $\dfrac23$ and $-\dfrac12$ .

Solved Example 5: a repeated root (NCERT Example 5)

Q. Find the roots of $3x^2-2\sqrt6\,x+2=0$ .

Solution: $a\times c=6$ , so split $-2\sqrt6\,x$ as $-\sqrt6\,x-\sqrt6\,x$ (product 6, sum $-2\sqrt6$ ). Since $\sqrt6=\sqrt3\cdot\sqrt2$ :

3x^2-\sqrt6\,x-\sqrt6\,x+2 = \sqrt3x(\sqrt3x-\sqrt2)-\sqrt2(\sqrt3x-\sqrt2) = (\sqrt3x-\sqrt2)(\sqrt3x-\sqrt2)=0

Both factors are the same, so $x=\sqrt{\dfrac23}$ comes twice, once for each repeated factor. The roots are $\sqrt{\dfrac23},\ \sqrt{\dfrac23}$ .

Solved Example 6: the prayer hall (NCERT Example 6)

Q. Find the dimensions of the prayer hall from Section 1.

Solution: The breadth $x$ satisfies $2x^2+x-300=0$ . Here $a\times c=-600$ , $b=1$ : use $25$ and $-24$ .

2x^2-24x+25x-300 = 2x(x-12)+25(x-12) = (x-12)(2x+25)=0

So $x=12$ or $x=-12.5$ . A breadth cannot be negative, so the breadth is 12 m and the length is $2(12)+1=$ 25 m.

Exam Tip

Both roots may be algebraically correct, but only one may fit the situation. Length, age, speed and number of items can never be negative, and a number of items must be a whole number. Write one line saying which root you reject and why, as the book does for the hall: “Since $x$ is the breadth of the hall, it cannot be negative.”

6 The quadratic formula

Factorisation is quick when the roots are neat numbers. When they are not (for example $x^2-2x-4=0$ , where no two integers have product $-4$ and sum $-2$ ), use the formula. It works for every quadratic equation whose roots are real numbers (Section 7 shows how to check this first).

Formula: Quadratic formula

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

for

ax^2+bx+c=0

a\neq0

, provided

b^2-4ac\ge0

. The sign

\pm

gives the two roots.

Solved Example 7: when factorisation fails

Q. Solve $x^2-2x-4=0$ .

Solution: $a=1$ , $b=-2$ , $c=-4$ . So $b^2-4ac=4+16=20$ and $\sqrt{20}=2\sqrt5$ .

x=\dfrac{2\pm2\sqrt5}{2}=1\pm\sqrt5

The roots are $1+\sqrt5$ and $1-\sqrt5$ .

Common Mistake

Taking the sign of $b$ wrongly. In $x^2-2x-4=0$ , $b=-2$ , so $-b=+2$ and $b^2=(-2)^2=+4$ . Write $a$ , $b$ , $c$ with their signs before substituting.

Note: completing the square

Older books and guides have a section on solving by completing the square (rewriting $ax^2+bx+c$ using a perfect square such as $(x+k)^2$ ), which is how the formula is derived. The current NCERT book does not teach this method: it states the quadratic formula without deriving it. You must know and use the formula; you do not need to derive it or solve by completing the square.

7 Discriminant and nature of roots

Learn by heartDefinition 4: Discriminant

For $ax^2+bx+c=0$ , the quantity $b^2-4ac$ is called the discriminant, because it decides whether the equation has real roots or not. In these notes we write it as $D=b^2-4ac$ .

Value of $D=b^2-4ac$	Nature of roots	The roots
$D\gt0$	two distinct real roots	$\dfrac{-b+\sqrt{D}}{2a}$ and $\dfrac{-b-\sqrt{D}}{2a}$
$D=0$	two equal real roots (coincident roots)	both equal to $-\dfrac{b}{2a}$
$D\lt0$	no real roots	no real number has a negative square

Graphs of y = x squared minus 4x plus 3, plus 4 and plus 6, crossing the x-axis twice, touching it once, and missing it — **Graph 1**: A picture of the three cases, to help you remember them. $y=x^2-4x+3$ ( $D=4$ ) cuts the x-axis at 1 and 3; $y=x^2-4x+4$ ( $D=0$ ) just touches it at 2; $y=x^2-4x+6$ ( $D=-8$ ) never meets it. The roots of $ax^2+bx+c=0$ are the values of $x$ where the graph of $y=ax^2+bx+c$ meets the x-axis.

Solved Example 8 (NCERT Example 7)

Q. Find the discriminant of $2x^2-4x+3=0$ and hence the nature of its roots.

Solution: $a=2$ , $b=-4$ , $c=3$ . $D=(-4)^2-4(2)(3)=16-24=-8\lt0$ . So the equation has no real roots.

Solved Example 9 (NCERT Example 9)

Q. Find the discriminant of $3x^2-2x+\dfrac13=0$ , the nature of its roots, and the roots if they are real.

Solution: $D=(-2)^2-4(3)\left(\dfrac13\right)=4-4=0$ , so there are two equal real roots, each $-\dfrac{b}{2a}=\dfrac{2}{6}$ , i.e. $\dfrac13,\ \dfrac13$ .

Solved Example 10: finding k for equal roots

Q. Find $k$ so that $kx^2-6x+1=0$ has two equal roots.

Solution: Equal roots means $D=0$ : $(-6)^2-4(k)(1)=0$ , so $36-4k=0$ and $k=9$ . (If an answer ever came out as $k=0$ , reject it: then the $x^2$ term vanishes and the equation is no longer quadratic.)

Exam Tip

“Find the nature of the roots” needs only the sign of $D$ , not the roots. “Is it possible to …?” questions work the same way: form the equation, find $D$ . If $D\lt0$ the situation is not possible. If $D\ge0$ the equation has real roots; then solve it and keep only a root that fits the situation (a positive length, age or count), as the book does in NCERT Example 8, where $x=-12$ is ignored and $x=5$ is kept. NCERT asks this about a mango grove (a garden of mango trees), the ages of two friends, and a park. For a park of perimeter 80 m and area 400 m², length + breadth $=40$ m, so $x(40-x)=400$ gives $x^2-40x+400=0$ with $D=0$ , so it is possible: a 20 m by 20 m square.

8 Pole in a circular park: Pythagoras meets the discriminant

Solved Example 11 (NCERT Example 8)

Q. A pole has to be erected (set up) at a point on the boundary of a circular park of diameter 13 m so that the difference of its distances from two diametrically opposite gates A and B (gates at the two ends of a diameter) is 7 m. Is it possible? If yes, how far from each gate?

Solution: Let P be the pole and $BP=x$ m. Then $AP=(x+7)$ m. Since AB is a diameter, $\angle APB=90^\circ$ (angle in a semicircle: a diameter always makes a right angle at any point on the circle). By Pythagoras theorem:

(x+7)^2+x^2=13^2 \;\Rightarrow\; 2x^2+14x-120=0 \;\Rightarrow\; x^2+7x-60=0

$D=7^2-4(1)(-60)=49+240=289\gt0$ , so real roots exist and the pole can be erected. By the formula:

x=\dfrac{-7\pm\sqrt{289}}{2}=\dfrac{-7\pm17}{2}

, so

x=5

x=-12

A distance cannot be negative, so $x=5$ . The pole is 5 m from gate B and 12 m from gate A.

Figure 2: The circular park. Because AB is a diameter, the angle at P is 90°, so AP² + BP² = AB². — **Figure 2**: The circular park. Because AB is a diameter, the angle at P is 90°, so AP² + BP² = AB².

Did you know?

Many people believe the Babylonians were the first to solve quadratic equations. The Greek mathematician Euclid found such answers using geometry. In India, Brahmagupta (C.E. 598-665; C.E. means Common Era, the same years as A.D.) gave a clear formula for $ax^2+bx=c$ , and Sridharacharya (C.E. 1025) derived what we now call the quadratic formula, as quoted by Bhaskara II. The Arab mathematician Al-Khwarizmi (about C.E. 800) also studied these equations. In Europe, Abraham bar Hiyya Ha-Nasi gave complete solutions in his book Liber embadorum (C.E. 1145).

Quick Revision: read this the night before the exam

Quadratic equation: $ax^2+bx+c=0$ , $a,b,c$ real, $a\neq0$ ; simplify first, then decide
Root $\alpha$ : $a\alpha^2+b\alpha+c=0$ ; roots of the equation = zeroes of the polynomial; at most two roots
Factorisation: split $bx$ using two numbers with product $ac$ and sum $b$ ; each factor $=0$
Quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ , provided $b^2-4ac\ge0$
$D=b^2-4ac$ : $D\gt0$ two distinct real roots; $D=0$ two equal roots $-\dfrac{b}{2a}$ ; $D\lt0$ no real roots
Equal roots / find $k$ : put $D=0$ ; reject any $k$ that makes $a=0$
In word problems reject a negative length, age, speed or count, and say why

Practice Questions

1 markFind the discriminant of $x^2-5=0$ .
2 marksIs $(x+1)^2=x^2+5$ a quadratic equation? Give a reason.
3 marksSolve $3x^2-5x+2=0$ by factorisation.
3 marksFind the values of $k$ for which $x^2+kx+9=0$ has two equal roots, and find the roots in each case.

More for this chapter

Important Questions MCQ / Objective NCERT Solutions Short Notes