Examples: Quadratic Equations | Mathematics (Maths) for JEE Main & Advanced PDF Download

A. General polynomial

A function f defined by f(x) = a_nxⁿ + a_n-1 x^{n - 1} +.... + a_:x + a₀, where a₀, a₁ ,a₂,.... , a_n ∈R is called n degree polynomial while coefficient (a_n ≠ 0, n ∈ W) is real. if a₀, a₁ ,a₂ ,..... , a_n  ∈ C, then it is called complex cofficient polynomial.

B. Quadratic polynomial

A polynomial of degree two in one variable f(x) = y = ax² + bx + c, where a ≠ 0 & a, b, c ∈ R a → leading coefficient, c → absolute term / constant term

If a = 0 then y = bx + c → linear polynomial b ≠ 0

If a = 0, c = 0 then y = bx → odd linear polynomial

C. Quadratic equation 

1. The solution of the quadratic equation , ax² + bx + c = 0 is given by  

The expression b² - 4 ac = D is called the discriminant of the quadratic equation.

2. If α & β are the roots of the quadratic equation ax² + bx + c = 0 , then ;

(i) αβ = - b/a

(ii) α + β = c/a

D. nature of roots 

(1) Consider the quadratic equation ax² + bx + c = 0 where  then ;

(i) D > 0 ⇔ roots are real & distinct (unequal)

(ii) D = 0 ⇔ roots are real & coincident (equal)

(iii) D < 0 ⇔ roots are imaginary

(iv) If p + i q is one root of a quadratic equation, then the other must be the conjugate p - i q  &  vice versa.

(2) Consider the quadratic equation ax² + bx + c = 0 where  then ;

(i) If D > 0 & is a perfect square , then roots are rational & unequal .

(ii) If  is one root in this case, (where p is rational &  is a surd) then the other root must be the conjugate of it  & vice versa.

 E. Graph of Quadratic expression 

(i) The graph between  x , y  is  always a parabola.

(ii) If a > 0  then  the  shape  of  the parabola is concave upwards & if a < 0  then the shape of the parabola is concave  downwards  .

(iii) The co–ordinate of vertex are (-b/2a,-D/4a)

(iv) The parabola intersect the y–axis at point (0, c)

(v) The x-co-ordinate of point of intersection of parabola with x-axis are the real roots of the quadratic equation f(x) = 0. Hence the parabola may or may not intersect the x-axis at real points.

Consider  the  quadratic  expression ,  y = ax² + bx + c ,  a ≠ 0  &  a , b , c ∈ R  then ;
(i) "  x ∈ R ,  y > 0  only  if  a > 0  &  b² – 4ac < 0  (figure 3).
(ii) "  x ∈ R ,  y < 0  only  if  a < 0  &  b² – 4ac < 0  (figure 6).

Relation Between Roots & Coefficients

A quadratic  equation  whose  roots  are α & β is  (x – α)(x – β) = 0

i.e. x² – (α + β) x + αβ = 0 i.e.  x² - (sum of  roots) x + product of roo

 Ex.1 A quadratic polynomial p(x) has  as roots and it satisfies p(1) = 2. Find the quadratic polynomial.

 Sol. sum of the roots = 2, product of the roots = - 4

let p(x) = a(x² – 2x – 4)  ⇒ p(1) = 2  ⇒  2 = a(1² – 2 · 1 – 4)  ⇒  a = – 2/5

p (x) = – 2/5 (x² – 2x – 4)

Ex.2 The quadratic equation x² + mx + n = 0 has roots which are twice those of x² + px + m = 0 and m, n and p  Find the value of .

Sol. 

2(α + β) = – m ....(1) 4 αβ= n ....(2)

and α + β = – p ....(3) αβ = m ....(4)

(1) and (3) ⇒ 2p = m and (2) and (4) ⇒ 4m = n

 Ex.3 Find the range of the variable x satisfying the quadratic equation,

Sol. 

Roots of the equation   x² + (2 cosØ)x – sin²Ø = 0 are  x₁ = – cosØ + 1  or  x₂ = – (1 + cosØ)

hence  x ∈ [0, 2]  or  x ∈ [–2, 0] = x ∈ [– 2, 2]

Ex.4 If  α & β  are the roots of the equation  x² – ax + b = 0 and  vn = an + bn, show that v_{n + 1} = a v_n – b v_{n – 1}  and hence obtain the value of  α⁵ + β⁵

Sol. 

Ex.5 One root of mx² - 10x + 3 = 0 is two third of the other root. Find the sum of the roots.

Sol. 

Ex.6  If x₁ ∈ N and x₁ satisfies the equation. If x² + ax + b + 1 = 0, where a, b ≠ 1 are integers has a root in natural numbers then prove that a² + b² is a composite.

Sol. 

Let a and b be the two roots of the equation where a ∈ N. Then

α + b = –a ...(1)

α . b = b + 1 ...(2)

b = –α – α is an integer. Also, since b + 1 ≠ 0, b ≠ 0

From eq. (1) & eq. (2), we get a² + b² = (α + α)² + (αb – 1)² = α² + b²+ α²b² + 1 = (1 + α²) (1 + b²)

Now, as a ∈ N and b is a nonzero integer, 1 + α² > 1 and 1 + b² > 1.
Hence a² + b² is composite number

Note : If b = –1, then a2 + b2 can not be a composite number.

Consider a = –6, b = –1

x² – 6x + (–1) + 1 = 0, its are 6 and 0.

a² + b² = 36 + 1 = 37, a prime number.

 Ex.7 Find a quadratic equation whose roots x₁and x₂ satisfy the condition

 (Assume that x₁, x₂ are real)

Sol. 

We have  (x₁ + x₂)² = 5 + 4 = 9 ⇒  x¹ + x² = ± 3 (if x₁ x₂ = 2)

(x₁ + x₂)² = 5 + 2(–10/3) = –5/3 (if x₁x₂ = –10/3)

Ex.8 Form a quadratic equation with rational coefficients if one of its root is cot²18°.

Sol. 

Ex.9 Let a & c be prime numbers and b an integer. Given that the quadratic equation a x² + b x + c = 0 has rational roots, show that one of the root is independent of the co-efficients. Find the two roots.

Sol. 

 Ex.10 Find all integers values of a such that the quadratic expressions (x + a) (x + 1991) + 1 can be factored as (x + b) (x + c), where b and c are integers. 

Sol. 

If the difference between two perfect square is 4, then one of them is 4 and the other is zero.

Therefore, 1991 - a = ± 2,          (b - c)² = 0

⇒ a = 1991 + 2 = 1993 and b = c or a = 1991 - 2 = 1989 and b = c

But b + c = 2b = 1991 + a = 1991 + 1993 or 1991 + 1989 ⇒ b = c = 1992 or 1990

So, the only 2 values of a are 1993 and 1989.

Ex.11 Find a, if ax² - 4x + 9 = 0 has integral roots.

Sol. 

This equation has integeral roots if b is an integer and 16b² - 36b is a perfect square

For any other factorization of 81, b will not be an integer.

Ex.12 

Sol. 

This is true only if n is an even integer.

Ex.13 Find all values of the parameter a for which the quadratic equation (a+1) x² + 2(a + 1) x + a - 2 = 0

(a) has two distinct roots, 

(b) has no roots, 

(c) has two equal roots.

Sol. By the hypothesis this equation is quadratic, and therefore  and the discriminant of this equation   then this equation has two distinct roots. For   then this equation has no roots. This equation can not have two equal roots since D = 0 only for a = -1, and this contradicts the hypothesis.

Ex.14 If the equation ax² + 2bx + c = 0 has real roots, a, b, c being real numbers and if m and n are real numbers such that m² > n > 0 then prove that the equation ax² + 2mbx + nc = 0 has real roots.

Sol. 

Hence roots of equation ax² + 2mbx + nc = 0 are real.

Ex.15 Show that the expression x² + 2(a + b + c) x + 3(bc + ca + ab) will be a perfect square if a = b = c.

Sol. Given quadratic expression will be a perfect square if the discriminant of its corresponding equation is zero.

Ex.16 If c < 0 and ax² + bx + c = 0 does not have any real roots then prove that

(i) a - b + c < 0 

(ii) 9a + 3b + c < 0.

Sol. 

G. Equation v/s Identity 

A quadratic equation is satisfied by exactly two values of `x' which may be real or imaginary. The equation,      a x² + b x + c = 0 is :

If a quadratic equation is satisfied by three distinct values of `x', then it is an identity.

(x + 1)² = x² + 2x + 1 is an identity in x.

Here highest power of x in the given relation is 2 and this relation is satisfied by three different values x = 0, x = 1 and x = -1 and hence it is an identity because a polynomial equation of n^th degree cannot have more than n distinct roots.

Ex.17 If a + b+ c = 0 , a n² + b n + c = 0  and   a + b n + c n² = 0  then prove that a  =  b  =  c  =  0 .

Sol. 

Note that  a x² + b x + c = 0  is satisfied by  x = 1 ;   x = n & x = 1/n  where   n  ≠ 1/n

⇒ Q.E.  has 3 distinct real roots which implies that it must be an identity.

Ex.18 If tan α, tan β are the roots of x² - px + q = 0 and cot α, cot β are the roots of x² - rx + s = 0 then find the value of rs in terms of p and q.

Sol. 

hence roots of 2^nd are reciprocal of (1)

H. Solution of Quadratic Inequalities 

The values of ‘x’ satisfying the inequality, ax² + bx + c > 0 (a ≠ 0) are :

(i) If D > 0, i.e. the equation ax² + bx + c = 0 has two different roots α < β.
Then  

(ii) If D = 0, i.e. roots are equal, i.e. α = β.

then

(iii) If D < 0, i.e. the equation ax² + bx + c = 0 has no real roots.

Then 

(iv) Inequalities  of  the  form P(x)/Q(x) = 0 can be solved using the method of intervals.

Ex.19 Find the solution set of k so that y = kx is secant to the curve y = x² + k.

Sol. put y = kx in  y = x² + k  ⇒ kx = x² + k = 0  ⇒  x² – kx + k = 0 for line to be secant,  D > 0

⇒ k² – 4k > 0 k(k – 4) > 0

hence k > 4  or   k < 0  

⇒  k ∈ (– ∝, 0) U (4, ∝)

Ex.20 Find out the values of 'a'  for which any solution of the inequality,  is also a solution of the inequality,  x² + (5 – 2 a) x ≤ 10 a .

Sol. 

Ex. 21 Find the values of 'p' for which the inequality,   is valid for all real x.

Sol. 

(2 – t) x² + 2 (1 + t) x – 2 (1 + t) > 0 when   t = 2 ,   6 x – 6 > 0     which is not true x ∈ R.

Let   t ≠ 2   ;  t < 2  ......(1)    and  4 (1 + t)² + 8 (1 + t) (2 – t) < 0  (for given inequality to be valid)

or (t – 5) (t + 1) > 0 

= t > 5  or   t < – 1 ......(2)

From  (1) and (2) 

I. Range of Quadratic Expression f(x) = ax² + bx + c 

(i) Range when x ε R : 

Maximum & Minimum Value of y = ax² + bx + c occurs at x = - (b/2a) according as a < 0 or a > 0 respectively

(ii) 

Ex.22 Find the minimum value of f(x) = x² - 5x + 6.

Sol. 

Ex.23 Let P(x) = ax² + bx + 8 is a quadratic polynomial. If the minimum value of P(x) is 6 when x = 2, find the values of a and b.

Sol. 

P(x) = ax² + bx + 8 ....(1)

P(2) = 4a + 2b + 8 = 6 ....(2)

-b/2a = 2; 4a = -b

from (2), we get – b + 2b = – 2 ⇒ b = –2 

Therefore, 4a = – (– 2)   = a = 1/2

Ex.24 If min (x² + (a – b)x + (1 – a – b)) > max (–x² + (a + b) x – (1 + a + b)), prove that a² + b² < 4.

Sol. 

Let f(x) = x² + (a – b)x + (1 – a – b)

Let g(x) = –x² + (a + b)x – (1 + a + b)

= 4(1 – a – b) – (a – b)² > (a + b)² > (a + b)² – 4(1 + a + b)  or 8 > 2(a² + b²) = a² + b² < 4

Ex.25 The coefficient of the quadratic equation ax² + (a + d)x + (a + 2d) = 0 are consecutive terms of a positively valued, increasing arithmetic sequence. Determine the least integral value of d/a such that the equation has real solutions.

Sol. 

ax² + (a + d)x + (a + 2d) = 0

a,   a + d,  a + 2d    are in   ­ A.P.   (d > 0) (note that positively valued terms   ⇒  a > 0) for real roots  

D ≥ 0   ⇒  (a + d)² – 4a(a + 2d) ≥ 0 ⇒ a² + d² + 2ad – 4a² – 8ad ≥ 0

⇒ (d – 3a)² – 12a² ≥ 0

⇒ (d – 3a – 12 a)(d – 3a + 12 a) ≥ 0

Ex.26 The set of real parameter 'a' for which the equation x⁴ – 2ax² + x + a² – a = 0 has all real solutions, is given by [m/n,∝)where m and n are relatively prime positive integers, find the value of (m + n).

Sol. 

We have  a² – (2x² + 1)a + x⁴ + x = 0

– ve sign   2a = 2x² – 2x + 2 = a = x² – x + 1  If x² – x + 1 – a

for  x to be real   a ≥ 3/4  and  a ≥ – 1/4 Þ a ≥ 3/4  =  3 + 4 = 7

 Ex.27   

Sol. 

J. Resolution of a Second Degree Expression in X and Y 

The condition that a quadratic function f (x , y) = ax² + 2 hxy + by² + 2 gx + 2 fy + c may be resolved into two

Ex.28 If x is real and 4y² + 4xy + x + 6 = 0, then find the complete set of values of x for which y is real.

Sol. 

Ex.29 Find the greatest and the least real values of x & y satisfying the relation, x² + y²  =  6 x - 8 y.

Sol. writing as a quadratic in y , y² + 8 y + x² - 6 x = 0

Ex.30 If x, y ans z are three real numbers such that x+y+z = 4 and x²+ y²+z² = 6, then show that each

Sol. 

K. Theory of Equations 

Note : 

Ex.31 If x = 1 and x = 2 are solutions of the equation x³ + ax² + bx + c = 0 and a + b = 1, then find the value of b.

Sol.

Ex.32 A polynomial in  x  of  degree greater than 3 leaves the remainder 2, 1 and  - 1 when divided by (x - 1); (x + 2) & (x + 1) respectively . Find the remainder, if the polynomial is divided by, (x² - 1) (x + 2) . 

Sol. 

Ex.33 Find the roots of the equation x⁴ + x³ - 19x² - 49x - 30, given that the roots are all rational numbers.

Sol. Since all the roots are rational because, they are the divisors of -30.

The divisors of -30 are ± 1, ± 2, ± 3, ± 5, ± 6, ± 10, ± 15, ± 30.

By remainder theorem, we find that -1, -2, - 3, and 5 are the roots.

Hence the roots are -1, -2, -3 and + 5.

Ex.34 Let u, v be two real numbers such that u, v and uv are roots of a cubic polynomial with rational coefficients. Prove or disprove uv is rational.

Sol. Let x³ + ax² + bx + c = 0 be the cubic polynomial of which u, v and uv are the roots and a, b, c are all rationals.

u + v + uv = -a         

⇒        u + v = -a - uv,          

uv + uv² + u² v = b       

and         u²v² = - c

⇒ b = uv + uv² + u²v = uv (1 + v + u)       

= uv (1 - a - uv)                   

= (1 - a) uv - u²v² = (1 - a) uv + c

i.e., uv =  and since a, b, c are rational, uv is rational.

Ex.35  Prove that the equation x⁴ + 2x³ + 5 + ax³ + a = 0 has at the most two real roots for all values of a ∈ R – {–5}

Sol. The given expression is (x³ + 1)² + a (x³ + 1) + 4 = 0

If discriminant of the above equation is less than zero i.e. D < 0

Then we have six complex roots and no real roots.

we will get two real roots and other roots will be complex except when t = 1 is one of the roots

⇒ f(1) = 0 ⇒ a = -5.

Ex.36

Sol. 

Ex.37 If p(x) is a polynomial with integer coefficients and a, b, c are three distinct integers, then show that it is impossible to have p(a) = b, p(b) = c and p(c) = a.

Sol. Suppose a, b, c are distinct integers such that p(a) = b, p(b) = c and p(c) = a. Then

p(a) - p(b) = b - c, p(b) - p(c) = c - a, p(c) - p(a) = a - b.

But for any two integers m ≠ n, m - n divides p(m) - p(n). Thus we get,

a - b | b - c, b - c | c - a, c - a|a - b.

Therefore a = b = c, a contradiction, Hence there are no integers a, b, and c such that p(a) = b, p(b) = c and p(c) = a.

Ex.38 Find all cubic polynomials p(x) such that (x - 1)² is a factor of p(x) + 2 and (x + 1)² is a factor of p(x) - 2.

Sol. 

+ 2, we get a + b + c + d + 2 = 0.

Hence d = -a - b - c - 2 and

p(x) + 2  = a(x³ - 1) + b(x² - 1) + c(x - 1)    = (x - 1) {a(x² + x +1) + b(x + 1) + c}.

Since (x - 1)² divides p(x) + 2, we conclude that (x - 1) divides a(x² + x + 1) + b(x + 1) c. This implies that 3a + 2b + c = 0. Similarly, using the information that (x + 1)² divides p(x) - 2, we get two more relations : -a + b - c + d - 2 = 0; 3a - 2b + c = 0. Solving these for a, b, c, d, we obtain b = d = 0, and a = 1, c = -3. Thus there is only one polynomial satisfying the given condition : p(x) = x³ - 3x.

L. Common Roots of Quadratic Equations 

Atleast one Common Root : 

Note : If f(x) = 0 & g(x) = 0 are two polynomial equation having some common roots(s) then those common root(s) is/are also the root(s) of h(x) = a f(x) + bg (x) = 0.

Ex.39 

Sol. Given equations are : x² + 2x + 9 = 0 .... (i)            and ax² + bx + c = 0 .... (ii)

Clearly roots of equation (i) are imgainary since equation (i) and (ii) have a common root, therefore common root must be imaginary and hence both roots will be common.

Ex.40 Determine a such that x² - 11x + a and x² - 14x + 2a may have a common factor.

Sol. 

Solving (i) and (ii) by cross multiplication method, we get a = 24.

Ex.41 If the quadratic equations, x² + bx + c = 0 and bx² + cx + 1 = 0 have a common root then prove that either b + c + 1 = 0 or b² + c² + 1 = b c + b + c .

Sol. 

Ex.42

Sol. x² - (a + b) x + a b = 0            or         (x - a)  (x - b)  =  0          ⇒        x  =  a  or  b

Ex.43 If x² - ax + b = 0 and x² - px + q = 0 have a root in common and the second equation has equal roots show that b + q = .

Sol. Given equation are x² - ax + b = 0 ans x² - px + q = 0. Let a be the common root. Then roots of equation (2) will be a and a. Let b be the other root of equation (1). Thus roots of equation (1) are a, b and those of equation (2) are α, α.

from (7) and (8), L.H.S. = R.H.S.

Ex.44If each pair of the following three equations x² + ax + b = 0, x² + cx + d = 0, x² + ex + f = 0 has exactly one root in common, then show that (a + c + e)² = 4 (ac + ce + ea - b - d - f)

Sol. x² +ax + b = 0             ...(1)

x² + cx + d = 0                   ...(2)

x² + ex + f = 0                    ...(3)

From (7) & (8), (a + c + e)² = 4 (ac + ce + ea - b - d - f)

Ex.45

Sol. Since cubic is divisible by both x² + ax + b and x² + bx + a and

product of the roots be 1 · a · b = - 72 ...(1) and a + b + 1 = 0 ...(2) (from x² + ax + b = 0 put x = 1)

M. Location of Roots 

Ex.46

Sol. 

Ex.47 Find the set of values of  'p' for which the quadratic equation, (p - 5) x² - 2 px - 4 p = 0 has atleast one positive root.

Sol. 

Ex.48 

Sol.

Ex.49 Find all the values of `a' for which both the roots of the equation. (a - 2)x² + 2ax + (a + 3) = 0 lies in the interval (-2, 1).

Sol. 

Ex.50 The coefficients of the equation ax² + bx + c = 0 where  satisfy the inequality

(a + b + c)(4a - 2b + c) < 0. Prove that this equation has 2 distinct real solutions.

Sol. 

Ex.51 Find the value of k for which one root of the equation of x² - (k + 1)x + k² + k-8=0 exceed 2 and other is smaller than 2.

Sol. 

Ex.52

Ex.53 (a) For what values of `a' exactly one root of the equation 2^ax² - 4^ax + 2^a - 1 = 0, lies between a and 2.

(b) Find all values of a for which the equation 2x² - 2(2a + 1) x + a(a + 1) = 0 has two roots, one of which is greater than a and the other is smaller than a.

Sol. 

Ex.54Find all values of  k  for which the inequality, 2 x² - 4 k² x - k² + 1 > 0  is valid for all real  x  which do not exceed unity in the absolute value .

Sol.