quadratic eq

Contents

1. INTRODUCTION
2. SOLVING A QUADRATIC EQUATION
   1. METHOD OF COMPLETING SQUARES
   2. THE QUADRATIC FORMULA
   3. FACTORISATION
3. NATURE OF THE ROOTS OF A QUADRATIC EQUATION
4. GRAPHICAL REPRESENTATION OF A QUADRATIC EQUATION
5. HIGHER ORDER EQUATIONS
   1. SOLVING HIGHER ORDER EQUATIONS
   2. DESCARTES’ RULE OF SIGNS
6. FINDING THE COMMON ROOT

Quadratic and Higher Order Equations

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INTRODUCTION

A second degree polynomial is called a quadratic polynomial. An equation of the form ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0, is called a quadratic equation in variable x. The values of x for which the equation holds true are called the roots of the equation. Since a quadratic equation is of degree 2, it can have at most two roots.

Here are some examples of quadratic equations.

5x² = 0

3x² + 2x = 0

x² – 4x + 7 = 0

SOLVING A QUADRATIC EQUATION

Solving a quadratic equation is finding all its roots. This can be done in the following ways.

METHOD OF COMPLETING SQUARES

In this method, we express the given quadratic ax² + bx + c = 0 as (mx n)² = k², where m, n and k are numbers and then find the value of x by taking the square root.

The steps to be followed while using this method are as explained in the following example.

Consider the quadratic equation 5x² + 20x + 13 = 0. Here a = 5, b = 20 and c = 13.

Step 1: Express the equation as

Step 3: Express the equation as

Step 4: Take the square root and find the value of x.

Example 1:

Find the roots of the equation, x2 + 12x 20 = 0. Solution: x2 + 12x 20 = 0 x2 + 2 6x + 36 = 20 + 36 (x + 6)2 = 56 Hence, the roots of the equation are

Example 2:

Find the roots of the equation, 9x2 42x + 24 = 0. Solution: 9x2 42x + 24 = 0 (3x)2 2 3 7x + 49 = 49 – 24 (3x 7)2 = 25 3x = 7 5 3x = 2 or 12 Hence, the roots of the equation are 2/3 and 4.

Note: The method of completing squares can be used to find the minimum value of an algebraic expression without using concepts of higher mathematics.

Example 3:

If x and y are real numbers, then the minimum value of x2 + 4xy + 6y2 – 4y + 4 is [XAT 2010] (1) ‒4(2) 0 (3) 2(4) 4 (5) None of the above Solution: x2 + 4xy + 6y2 – 4y + 4 = x2 + 4xy + 4y2 + 2y2 – 4y + 4 = (x + 2y)2 + 2(y2 – 2y + 1 + 1) = (x + 2y)2 + 2(y – 1)2 + 2 This expression will take the minimum value when both the square terms are zero (y – 1)2 = 0 y = 1 and (x + 2y)2 = 0 x = –2 The minimum value of this expression is 2. Hence, option 3.

THE QUADRATIC FORMULA

The method of completing squares can be used to find a general formula for finding the roots of a quadratic equation.

The two roots of the quadratic equation, ax² + bx + c = 0 are given by:

Explanation

ax² + bx + c = 0

Dividing both the sides by a we get

Example 4:

Find the roots of the equation, x2 8x + 11 = 0. Solution: In the equation a = 1, b = –8, c = 11 Hence, the roots of the equation are: Hence, the two roots of the equation are

Example 5:

Find the roots of the equation, 7x2 + 12x 2 = 0. Solution: In the quadratic equation, 7x2 + 12x 2 = 0, we have a = 7, b = 12, c = 2. Hence, the roots of the equation are:

Example 6:

Find the value of x if, Solution: Multiplying both the sides by (x + 2)(x 1), (x + 1)(x 1) + (2x + 1)(x + 2) = 0 x2 – 1 + 2x2 + 4x + x + 2 = 0 3x2 + 5x + 1 = 0

Sum and Product of Roots of a Quadratic Equation

Consider the equation, ax² + bx + c = 0...(i)

This equation can be written as

Let and be the two roots of this equation.

We know that the two roots of this equation are

and

Using (ii), (iii) and (iv), we can rewrite (i) as

x² ( + )x + () = 0...(v)

Thus,

x² (Sum of the roots)x + (Product of the roots) = 0

Equation (v) can be written as (x – )(x – ) = 0

Example 7:

Ujakar and Keshab attempted to solve a quadratic equation. Ujakar made a mistake in writing down the constant term. He ended up with the roots (4, 3). Keshab made a mistake in writing down the coefficient of x. He got the root as (3, 2). What will be the exact roots of the original quadratic equation? [CAT 2001] (1) (6, 1)(2) (–3, –4) (3) (4, 3)(4) (–4, –3) Solution: In Ujakar’s case, the constant term is wrong. Product of roots will be wrong but the sum of roots will be correct. Correct Sum of the roots = 4 + 3 = 7 In Keshab’s case, coefficient of x is wrong. Sum of roots will be wrong but the product of roots will be correct. Correct product of roots = 3 2 = 6 Correct quadratic equation is x2 – 7x + 6 = 0 (x – 6)(x – 1) = 0 x = 6 or x = 1 Hence, option 1.

Example 8:

If the roots of the equation x3 ax2 +bx – c = 0 are three consecutive integers, then what is the smallest possible value of b? [CAT 2008] (3) 0(4) 1 Solution: If p, q and r are the roots of a cubic equation ax3 + bx2 + cx + d = 0, then pq + pr + qr = c/a If a is 1, then pq + qr + pr = c Let the three roots of the given equation be (n – 1), n and (n + 1). (n – 1)n + n(n + 1) + (n - 1)(n + 1) = b n2 – n + n2 + n + n2 – 1 = b 3n2 – 1 = b For b to take minimum value, the LHS must take minimum value. This is possible for n = 0. From this, we can see that minimum value of b = – 1 Hence, option 2.

Example 9:

If one root of the equation ax2 + bx + c = 0 is double of the other, then 2b2 = [IIFT 2008] (1) (2) (3) (4) None of the above Solution: ax2 + bx + c = 0 Let the roots of the given equation are and 2. 2b2 = 9ac Hence, option 1.

FACTORISATION

Consider the equation, ax² + bx + c = 0

In this method we write b as a sum of two terms in such a way that the product of the two terms is equal to ac.

Example 10:

Find the roots of the equation 2x2 + x – 6 = 0. Solution: In the given equation a = 2, b = 1 and c = 6 Hence ac = 2 (6) = 12 The numbers whose sum is 1 (b) and product is 12 (ac) are 4 and 3. Hence the given equation can be split as 2x2 + 4x 3x 6 = 0 2x(x + 2) 3(x + 2) = 0 (x + 2)(2x 3) = 0 x + 2 = 0 or 2x – 3 = 0 x = 2 or x = 3/2 Hence, the roots of the equation are 2 and 3/2.

Example 11:

Find the roots of the equation, 3x2 23x + 44 = 0. Solution: In the given equation a = 3, b = 23 and c = 44 Hence ac = 3 44 = 132 = 12 11 Thus the factors of 132 whose sum is equal to 23 are 11 and 12. 3x2 – 11x – 12x + 44 = 0 x(3x – 11) – 4(3x – 11) = 0 (x – 4)(3x – 11) = 0 x – 4 = 0 or 3x – 11 = 0 x = 4 or 11/3 Hence, the roots of the equation are 4 and 11/3.

Example 12:

Solution: 3x2 + 4 = 7x 3x2 7x + 4 = 0 3x2 4x 3x + 4 = 0 x(3x 4) 1(3x 4) = 0 (x 1)(3x 4) = 0 x 1 = 0 or 3x 4 = 0 x = 1 or 4/3 Value of x can be 1 or 4/3.

Example 13:

There are as many mangoes in a crate as there are crates in a fruit shop. Pablo takes away one crate and then the shop has only 90 mangoes left. How many mangoes does each crate have? Solution: Let there be x mangoes in each crate. Then there are x crates in the shop. Total number of mangoes in the shop = x2 Pablo takes away one crate i.e. x mangoes. x2 – x = 90 x2 – x – 90 = 0 x2 – 10x + 9x – 90 = 0 (x – 10)(x + 9) = 0 x = 10 or x = –9 As the number of mangoes cannot be negative, x = 10. Each crate has 10 mangoes.

Example 14:

Find the value of x if Solution: This can be rewritten as x2 – 110 = x x2 – x – 110 = 0 (x – 11)(x + 10) = 0 x = 11 or x = –10 But x is positive. x = 11

REMEMBER:

• The Method of Factorisation is very commonly used if b can be expressed as a sum of two integers whose product is ac.

After sufficient practice in solving quadratic equations by factorisation, some quadratic equations can be solved just by observation.

Example 15:

Find the roots of the equation, x2 11x + 28 = 0 Solution: The sum of the roots of this quadratic equation is 11, and the product of the roots is 28. Hence, the roots are 4 and 7.

Example 16:

Find the roots of the equation, x2 + x 30 = 0 Solution: The sum of the roots of this quadratic equation is 1 and the product of the roots is 30. Hence, the roots are 6 and 5.

REMEMBER:

• The expressions for the sum and product of the roots of a quadratic equation can be used to find the values of symmetric expressions involving the roots. If and are the roots of a quadratic equation then (^{2 2}), (^{3 3}), (/ + /) etc. are symmetric expressions in and .

Example 17:

If and are the roots of the equation 3x2 7x + 11 = 0, then find the value of 3 + 3. Solution: In the quadratic equation 3x2 7x + 11 = 0, we have a = 3, b = 7, c = 11 Therefore, + = (b/a) = (7/3) and = (c/a) = (11/3) As we know, 3 + 3 = ( + )3 3( + ) Substituting the values that we have, we get 3 + 3 = (343 693)/27 = 350/27

Example 18:

If one root of the equation, 4x2 17x + k = 0 is 16 times the other, find k. Solution: In the quadratic equation, 4x2 17x + k = 0 a = 4, b = 17, c = k Let and be the roots of the equation, and = 16 Now, + = (b/a) = (17/4) + 16 = 17/4 17 = 17/4 = (1/4) Now, = (c/a) = (k/4) 162 = (k/4) 16(1/4)2 = k/4 k = 4

Example 19:

Arjun and Abhi are together told to find the roots of the equation x2 + mx + n = 0. They are given two clues. If m is replaced by some other number, the roots of the new equation are 3 and –6. If n is replaced by some other number, the roots of the resulting equation are –2 and –5. Find the roots of the equation. Solution: Let and be the roots of the equation x2 + mx + n = 0 = n and + = –m When n is replaced, only the value of gets affected, where as + is correct. + = –2 – 5 = –7...(i) When m is replaced, only the value of + gets affected, where as is correct. = –18...(ii) The actual equation is x2 + 7x – 18 = 0 and its roots are –9 and 2.

Example 20:

The roots of the equation x2 + 10x + 2 = 0 are and . Find the equation with roots /2 and /2. Solution: x2 + 10x + 2 = 0 + = –10 and = 2 The equation with roots /2 and /2 is

Example 21:

One root of the equation ax2 + 2px + b2 = 0 is five times the other. Find the relation between p, a and b. Solution: Let and 5 be the roots of the equation ax2 + 2px + b2 = 0 and 5p2 = 9ab2

Example 22:

The sum of two numbers is 2/3 and the sum of their cubes is 7/54. Find the quadratic equation, the roots of which are reciprocals of these two numbers. Solution: Let the two numbers be u and v. Then, u + v = 2/3 and u3 + v3 = 7/54 (u + v)3 = u3 + v3 + 3uv(u + v) uv = 1/12 The required equation is x2 – 8x + 12 = 0

Example 23:

Solution: x2 + 8x + 20 = (x + 4)2 + 4 This value will be minimum when (x + 4)2 = 0 Thus the minimum value of this expression will be 4. = 0.25

Example 24:

Solution: x(x + 5) = 36 x2 + 5x – 36 = 0 (x + 9)(x – 4) = 0 x = –9 or x = 4 As logarithm of a negative number is not defined,

NATURE OF THE ROOTS OF A QUADRATIC EQUATION

The factor b² – 4ac in the quadratic formula is called the discriminant and is denoted by ∆. The nature of the roots of the equation depends on ∆.

The equation is a perfect square and has equal roots.

If ∆ > 0 and is a perfect square, roots are real and rational.

If ∆ > 0 but is not a perfect square, roots are real and irrational.

If ∆ < 0, the roots are complex in nature.

Complex roots and irrational roots always occur as conjugate pairs.

If is one root of a quadratic equation, the other root is .

Example 25:

Find the nature of the roots of the following quadratic equations. (1) x2 + x 6 = 0 (2) 3x2 8x + 1 = 0 (3) 4x2 12x + 9 = 0 (4) x2 4x + 13 = 0 Solution: 1) ∆ = b2 4ac = 1 + 24 = 25 Since 25 > 0 the roots are real. Further, 25 is a perfect square and hence the roots are rational. 2) ∆ = b2 4ac = 64 12 = 52 Since 52 > 0, the roots are real and since 52 is not a perfect square, the roots are irrational and are conjugates of each other. 3) ∆ = b2 4ac = 144 144 = 0 Since ∆ = 0, the roots are real and equal. 4) ∆ = b2 4ac = 16 52 = 36 Since ∆ < 0, the roots are imaginary and are conjugates of each other.

Example 26:

If the equation x3 – ax2 + bx – a = 0 has three real roots, then it must be the case that, [CAT 2000] (1) b = 1 (2) b ≠ 1 (3) a = 1 (4) a ≠ 1 Solution: x3 – ax2 + bx – a = x3 + bx – ax2 – a = x(x2 + b) – a(x2 + 1) x(x2 + b) – a(x2 + 1) = 0 If b = 1 then, x(x2 + 1) – a(x2 + 1) = (x – a)(x2 + 1) = 0 (x – a) = 0 or (x2 + 1) = 0 But then (x2 + 1) = 0 does not have real roots. b ≠ 1 Hence, option 2.

Example 27:

[XAT 2008] (1) Both 0 (2) Imaginary (3) Real and both positive (4) Real and of opposite sign (5) Real and both negative Solution: We have, Roots are real and after checking for the arbitrary values of p say 0.5, we get both the roots will have negative roots. Hence, option 5.

Example 28:

If the roots of the equation 2ax2 + 2ax + 1 = 0 are real and distinct, then the number of possible integer value(s) a can take between 1 and 4 (inclusive) is: [JMET 2010] (1) 1 (2) 4 (3) 3 (4) 2 Solution: Roots of equation 2ax2 + 2ax + 1 = 0 are real and distinct ∆ > 0 4a2 8a > 0 2a(a 2) > 0 a > 0 and a > 2 or a < 0 and a < 2 But a can take values between 1 and 4 (inclusive) only. a > 2 a = 3, 4 So, a can take 2 values. Hence, option 4.

GRAPHICAL REPRESENTATION OF A QUADRATIC EQUATION

The quadratic function y = ax² + bx + c is a parabola whose axis is parallel to the y-axis.

If a > 0, then the parabola opens upwards. If a < 0, then the parabola opens downwards.

Nature of Roots

When ∆ > 0, the parabola intersects the x-axis in two points and hence the roots are real and distinct in nature. This condition is shown by the curve 1.

When ∆ = 0, the parabola intersects the x-axis at only one point and hence we get only one root which is given by b/2a. This condition is shown by the curve 2.

When ∆ < 0, then the parabola does not intersect the x-axis and hence we do not have any real roots. The roots are complex in nature. This condition is shown by the curve 3.

HIGHER ORDER EQUATIONS

An expression of the form a₁xⁿ + a₂xⁿ–1 + a₃xⁿ-2 + … + a_n-1x² + a_nx + a_n+1, where a₁, a₂, a₃, ..., a_n+1 are real numbers is called an n^th degree polynomial in variable x. An equation of the form a₁xⁿ + a₂xⁿ-1 + a₃xⁿ-2 + … + a_n-1x² + a_nx + a_n+1 = 0 is called an n^th degree equation in variable x. An n^th degree equation has n roots.

If the roots of an n^th order equation are ₁, ₂, ₃, …, _n-1, _n then the equation can be rewritten as follows.

a₁(x ₁)(x ₂)(x – ₃)…(x _n-1)(x _n) = 0

Explanation:

The root of an equation is the value of x that satisfies the equation. Since ₁ is a root of the equation the equation should be satisfied when the value of x is substituted as ₁ in the equation. Hence (x – ₁) has to be one of the factors of the LHS of the equation. Similarly using the same logic for the other roots the equation can be rewritten as shown

Relation between Roots and Coefficients

Let S₁ = Sum of roots taken one at a time

i.e. ₁ + ₂ + ₃ +…

S₂ = Sum of roots taken two at a time

i.e. ₁₂ + ₂₃ + ₃₄ +…

S₃ = Sum of roots taken three at a time

i.e. ₁₂₃ + ₂₃₄ +…

In the equation ax² + bx + c = 0, we have

S₁ = b/a and S₂ = c/a

In the equation ax³ + bx² + cx + d = 0

S₁ = b/a and S₂ = c/a and S₃ = d/a

In equation ax⁴ + bx³ + cx² + dx + e = 0

S₁ = b/a and S₂ = c/a and S₃ = d/a and S₄ = e/a

This can be generalised for any equation of the form

a₁xⁿ + a₂xⁿ 1 + a₃xⁿ 2 + … + a_n 1x² + a_nx + a_n + 1 = 0

Thus,

S₁ = –a₂/a₁

S2 = a₃/a₁

S3 = –a₄/a₁ and so on till

S_n = (–1)ⁿ (a_n+1/a₁)

Example 29:

If a, b, c, d are the roots of the equation x4 – 3x3 + 2x2 – 5x + 1 = 0, Solution: For the equation, x4 – 3x3 + 2x2 – 5x + 1 = 0, S1 = 3, S2 = 2, S3 = 5 and S4 = 1 bcd + acd + abd + abc = S3 = 5 abcd = S4 = 1

Example 30:

If a, b and c are roots of x3 – 6x2 + 11x – 6 = 0 and the roots of the equation x3 – px2 + qx – r = 0 are a + b, b + c and c + a, then r equals: [JMET 2010] (1) 40(2) 50 (3) 60(4) 70 Solution: Here x3 – 6x2 + 11x – 6 = 0 (x 1)(x2 – 5x + 6) = 0 (x 1)(x 3) (x 2) = 0 x = 1, 2, 3 Roots of the equations x3 – px2 + qx r = 0 are a + b, b + c and c + a Its roots will be 1 + 2 = 3, 2 + 3 = 5 and 1 + 3 = 4 So the equation will be (x 3)(x 4 )(x 5) = 0 r = 3 4 5 = 60 Hence, option 3.

SOLVING HIGHER ORDER EQUATIONS

In this section we would restrict ourselves to the equations solvable by using simple tips and tricks.

1. EQUATIONS THAT CAN BE CONVERTED TO A QUADRATIC EQUATION

Please refer to the following example.

Example 31:

Find the roots of the equation, x4 5x2 + 4 = 0. Solution: We have to find out roots of the equation, x4 5x2 + 4 = 0 Substituting x2 = y in this equation, y2 5y + 4 = 0, which is nothing but a simple quadratic equation (y – 1)(y – 4) = 0 y = 1 or y = 4 x2 = 4 or x2 = 1 x = 2 or x = 1 Roots of the equation x4 5x2 + 4 = 0 are: 2, 2, 1 and 1.

2. USING SYNTHETIC DIVISION FOR FACTORISATION

Please refer to the following example.

Example 32:

Find the roots of 5x3 + 4x2 5x 4 = 0. Solution: On observation we can see that if we substitute x = 1 the equation is satisfied. Hence one of the roots is 1 and consequently one of the factors of the LHS is (x 1). Using synthetic division we can factorise the equation. 5x3 + 4x2 5x 4 = (x 1)(5x2 + 9x + 4) (x 1)(5x2 + 9x + 4) = 0 (x 1)(5x + 4)(x + 1) = 0 x = 1 or x = –4/5 or x = –1 Alternatively, If 1, 2 and 3 are the roots of the equation, we have, 1 + 2 + 3 = 4/5 12 + 23 + 31 = 5/5 = –1 123 = 4/5 We have three equations and three variables using which the value of roots can be found out.

3. USING ALGEBRAIC FORMULAE

Equations may be simplified using algebraic formulae.

Example 33:

Find the values of x and y if x3 + y3 = 169 and x + y = 1. Solution: x3 + y3 = 169 ...(i) x + y = 1...(ii) We know that x3 + y3 = (x + y)(x2 xy + y2) Hence the first equation can be written as: (x + y)(x2 xy + y2) = 169 (x2 xy + y2) = 169 ...(iii) Squaring (ii), x2 + 2xy + y2 = 1...(iv) Subtracting equation (i) from equation (ii) we get, 3xy = 168 x = 56/y From (ii), 56/y + y = 1 56 y + y2 = 0 y = 8 or 7 x = 7 or 8

Example 34:

(x2 – 5x + 6) completely divides x3 + px + q. Find the quadratic equation having roots p and q. Solution: x2 – 5x + 6 = (x – 2)(x – 3) x = 2 and x = 3 satisfy the equation x3 + px + q = 0 8 + 2p + q = 0...(i) 27 + 3p + q = 0 ...(ii) Solving (i) and (ii), p = –19 and q = 30 The quadratic equation having roots p and q is x2 – (p + q)x + pq = 0 x2 – 11x – 570 = 0 Alternatively, x3 + px + q = (x – 2)(x – 3)(x – k) x3 + px + q = x3 + (k – 5)x2 + (6 – 5k)x + 6k k – 5 = 0, 6 – 5k = p and 6k = q p = –19 and q = 30 The quadratic equation having roots p and q is x2 – (p + q)x + pq = 0 x2 – 11x – 570 = 0

DESCARTES’ RULE OF SIGNS

If the coefficients of all the powers of x of an equation are positive, the equation does not have any positive root.

For example, the equation x⁴ + 4x³ + 5x² + 9x + 1 = 0 will have no positive root.

If the coefficients of all even powers of x in an equation are of one sign and the coefficients of all the odd powers of x are of the opposite sign, the equation does not have any negative root.

For example, the equation x⁵ – 2x⁴ + 9x³ 5x² + 7x 20 = 0 will not have any negative root.

Example 35:

If a > 3, how many positive roots does the following equation have? x9 + (2a + 7)x7 + (2a – 5)x5 + (2a + 3)x3 + (2a – 1)x + 2a = 0. Solution: As a > 3, (2a + 7), (2a – 5), (2a + 3), (2a – 1) and 2a are all positive. Thus, by Descartes’ rule of signs, the equation has no positive roots.

REMEMBER:

• Consider a quadratic equation ax² + bx + c = 0

The roots of this equation are given by

If and are the roots of the quadratic equation,

• Sum of the roots = + = –b/a

• Product of the roots = = c/a

• The equation can be expressed as:

x² (Sum of the roots)x + (Product of the roots) = 0

• The discriminant ∆ of a quadratic equation is given by

∆ = b² – 4ac

If ∆ = 0, the equation has real roots.

If ∆ > 0 and is a perfect square, roots are real and rational.

If ∆ > 0 but is not a perfect square, roots are real and irrational.

If ∆ < 0, the roots are complex in nature.

FINDING THE COMMON ROOT

Let f(x) = 0 and g(x) = 0 be two quadratic equations in x then their common roots will also be the roots of the equation f(x) g(x) = 0.

The roots of f(x) g(x) = 0 may or may not be the roots of f(x) = 0 and g(x) = 0. Therefore, to find the common roots of f(x) = 0 and g(x) = 0 follow the following procedure.

Step 1:

Find f(x) g(x)

Step 2:

Find the root(s) of f(x) g(x) = 0

Step 3:

Substitute the roots of f(x) g(x) = 0 in f(x) and g(x). If the roots satisfy both the equations then they are the common roots.

Example 36:

The number of roots common between the two equations x3 + 3x2 + 4x + 5 = 0 and x3 + 2x2 + 7x + 3 = 0 is: [CAT 2003 Re-Test] (1) 0 (2) 1 (3) 2 (4) 3 Solution: x3 + 3x2 + 4x + 5 = 0... (i) x3 + 2x2 + 7x + 3 = 0... (ii) Equating equations (i) and (ii), we get, x2 – 3x + 2 = 0 (x – 1)(x – 2) = 0 x = 1 or x = 2 Since both x = 1 and x = 2, do not satisfy either (i) or (ii), there exists no common roots for equations (i) and (ii). Hence, option 1.