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Mathematics For JEE

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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.

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  Doc: Quadratic Equations JEE Notes | EduRev

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 Doc: Quadratic Equations JEE Notes | EduRev 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.

Doc: Quadratic Equations JEE Notes | EduRev

(2) Consider the quadratic equation ax² + bx + c = 0 where Doc: Quadratic Equations JEE Notes | EduRev then ;

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

(ii) If Doc: Quadratic Equations JEE Notes | EduRev is one root in this case, (where p is rational & Doc: Quadratic Equations JEE Notes | EduRev is a surd) then the other root must be the conjugate of it Doc: Quadratic Equations JEE Notes | EduRev & vice versa.

Doc: Quadratic Equations JEE Notes | EduRev
Doc: Quadratic Equations JEE Notes | EduRev
Doc: Quadratic Equations JEE Notes | EduRev  

 E. Graph of Quadratic expression 

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

(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.

Doc: Quadratic Equations JEE Notes | EduRev 

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

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 Doc: Quadratic Equations JEE Notes | EduRev 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 Doc: Quadratic Equations JEE Notes | EduRev Find the value of Doc: Quadratic Equations JEE Notes | EduRev.

Sol. 

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

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

Doc: Quadratic Equations JEE Notes | EduRev

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

Doc: Quadratic Equations JEE Notes | EduRev

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. Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

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

 

Sol. 

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

 

Doc: Quadratic Equations JEE Notes | EduRev

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 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

Doc: Quadratic Equations JEE Notes | EduRev (Assume that x1, x2 are real)

Sol. 

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

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. 

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

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. 

Doc: Quadratic Equations JEE Notes | EduRev

 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. 

Doc: Quadratic Equations JEE Notes | EduRev

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. Doc: Quadratic Equations JEE Notes | EduRev

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

Doc: Quadratic Equations JEE Notes | EduRev

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

Ex.12 Doc: Quadratic Equations JEE Notes | EduRev

Sol. 

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

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 Doc: Quadratic Equations JEE Notes | EduRev and the discriminant of this equation  Doc: Quadratic Equations JEE Notes | EduRev then this equation has two distinct roots. For  Doc: Quadratic Equations JEE Notes | EduRev 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. 

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

  Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

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.

Doc: Quadratic Equations JEE Notes | EduRev

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. 

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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 :

Doc: Quadratic Equations JEE Notes | EduRev

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. 

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

hence roots of 2nd are reciprocal of (1)

Doc: Quadratic Equations JEE Notes | EduRev

 Doc: Quadratic Equations JEE Notes | EduRev 

 Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

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  Doc: Quadratic Equations JEE Notes | EduRev

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

then Doc: Quadratic Equations JEE Notes | EduRev

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

Then 

Doc: Quadratic Equations JEE Notes | EduRev

(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,  Doc: Quadratic Equations JEE Notes | EduRevis also a solution of the inequality,  x2 + (5 – 2 a) x ≤ 10 a .

Sol. 

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

 

Ex. 21 Find the values of 'p' for which the inequality,  Doc: Quadratic Equations JEE Notes | EduRev 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) 

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

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

(i) Range when x ε R : 

Doc: Quadratic Equations JEE Notes | EduRev

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

(ii) 

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

Sol. 

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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. 

Doc: Quadratic Equations JEE Notes | EduRev

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)

Doc: Quadratic Equations JEE Notes | EduRev

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

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

= 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

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

 

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

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

– ve sign   2a = 2x2 – 2x + 2 = a = x2 – x + 1  If x2 – x + 1 – a Doc: Quadratic Equations JEE Notes | EduRev

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

 Ex.27   Doc: Quadratic Equations JEE Notes | EduRev

 

Sol. 

Doc: Quadratic Equations JEE Notes | EduRev

 

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 Doc: Quadratic Equations JEE Notes | EduRev

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. 

Doc: Quadratic Equations JEE Notes | EduRev

 

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

Doc: Quadratic Equations JEE Notes | EduRev

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 eachDoc: Quadratic Equations JEE Notes | EduRev

Sol. 

Doc: Quadratic Equations JEE Notes | EduRev

K. Theory of Equations 

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Note : 

Doc: Quadratic Equations JEE Notes | EduRev

 

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.

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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. 

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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 = Doc: Quadratic Equations JEE Notes | EduRev 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.

Doc: Quadratic Equations JEE Notes | EduRev

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

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Sol. 

Doc: Quadratic Equations JEE Notes | EduRev

 

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. 

Doc: Quadratic Equations JEE Notes | EduRev
+ 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 : 

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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 

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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.

Doc: Quadratic Equations JEE Notes | EduRev

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

Sol. 

Doc: Quadratic Equations JEE Notes | EduRev

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. 

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Ex.42

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

Doc: Quadratic Equations JEE Notes | EduRev

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 = Doc: Quadratic Equations JEE Notes | EduRev.

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.

Doc: Quadratic Equations JEE Notes | EduRev

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)

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

Ex.45

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

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

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M. Location of Roots 

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Doc: Quadratic Equations JEE Notes | EduRev

 

Ex.46

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Sol. 

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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. 

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Ex.48 

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Sol.

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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. 

  Doc: Quadratic Equations JEE Notes | EduRev

Doc: Quadratic Equations JEE Notes | EduRev

 

Ex.50 The coefficients of the equation ax2 + bx + c = 0 where Doc: Quadratic Equations JEE Notes | EduRev satisfy the inequality

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

 

Sol. 

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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. 

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Ex.52

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Doc: Quadratic Equations JEE Notes | EduRev

 

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. 

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Doc: Quadratic Equations JEE Notes | EduRev

 

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. 

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