# Examples: Quadratic Equations - Mathematics (Maths) Class 11 - Commerce

A. General polynomial

A function f defined by f(x) = anxn + an-1 xn - 1 +.... + a:x + a0, where a0, a1 ,a2,.... , an ∈R is called n degree polynomial while coefficient (an ≠ 0, n ∈ W) is real. if a0, a1 ,a2 ,..... , an  ∈ C, then it is called complex cofficient 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

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.

(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(1– 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   x+ (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 vn + 1 = a vn – b vn – 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)= α2 + b2+ α2b2 + 1 = (1 + α2) (1 + b2)

Now, as a ∈ N and b is a nonzero 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 ax+ 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 nth 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 2nd are reciprocal of (1)

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 a+ 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 - 19x- 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 2ax2 - 4ax + 2a - 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 kx - k2 + 1 > 0  is valid for all real  x  which do not exceed unity in the absolute value .

Sol.

The document Examples: Quadratic Equations | Mathematics (Maths) Class 11 - Commerce is a part of the Commerce Course Mathematics (Maths) Class 11.
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## FAQs on Examples: Quadratic Equations - Mathematics (Maths) Class 11 - Commerce

 1. What is a quadratic equation?
Ans. A quadratic equation is a polynomial equation of second degree that can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable.
 2. How do you solve a quadratic equation?
Ans. There are different methods to solve a quadratic equation, but the most common one is the quadratic formula, which is x = (-b ± sqrt(b^2 - 4ac)) / 2a. Another method is factoring the equation, and a third method is completing the square.
 3. What are the roots of a quadratic equation?
Ans. The roots of a quadratic equation are the values of x that make the equation true, or in other words, the solutions to the equation. If the discriminant (b^2 - 4ac) is positive, the equation has two real roots; if it is zero, the equation has one real root; and if it is negative, the equation has two complex roots.
 4. What is the discriminant of a quadratic equation?
Ans. The discriminant of a quadratic equation is the expression b^2 - 4ac that appears inside the square root in the quadratic formula. It determines the nature of the roots of the equation: if it is positive, the equation has two real roots; if it is zero, the equation has one real root; and if it is negative, the equation has two complex roots.
 5. What are some real-life applications of quadratic equations?
Ans. Quadratic equations can be used to model many physical phenomena, such as the trajectory of a projectile, the shape of a parabolic reflector, the spread of a disease, or the growth of a population. They are also used in engineering, finance, and computer graphics, among other fields.

## Mathematics (Maths) Class 11

157 videos|210 docs|132 tests

## Mathematics (Maths) Class 11

157 videos|210 docs|132 tests

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