quadratic equation

Best Approach


Quadratic Equation
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,Maths IIT-JEE ‘Best Approach’ (MC SIR) Quadratic Equation
KEY CONCEPTS (QUADRATIC EQUATIONS)
The general form of a quadratic equation in x is , ax2 + bx + c = 0 , where a , b , c  R & a  0.
RESULTS:
b b 2 4ac
1. The solution of the quadratic equation , ax² + bx + c = 0 is given by x =
2a
2
The expression b – 4ac = D is called the discriminant of the quadratic equation.
2. If  &  are the roots of the quadratic equation ax² + bx + c = 0, then;
(i) +  = – b/a (ii)  = c/a (iii) –  = D /a .
3. NATURE OF ROOTS:
(A) Consider the quadratic equation ax² + bx + c = 0 where a, b, c  R & a  0 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. (p , q  R & i = 1 ).
(B) Consider the quadratic equation ax2 + bx + c = 0 where a, b, c  Q & a  0 then;
(i) If D > 0 & is a perfect square , then roots are rational & unequal.
(ii) If  = p + q is one root in this case, (where p is rational & q is a surd)
then the other root must be the conjugate of it i.e.  = p  q & vice versa.
4. A quadratic equation whose roots are  &  is (x  )(x  ) = 0 i.e.
x2  (+ ) x +  = 0 i.e. x2  (sum of roots) x + product of roots = 0.
5. Remember that a quadratic equation cannot have three different roots & if it has, it becomes an identity.
6. Consider the quadratic expression , y = ax² + bx + c , a  0 & a , b , c  R then ;
(i) The graph between x , y is always a parabola . If a > 0 then the shape of the
parabola is concave upwards & if a < 0 then the shape of the parabola is concave downwards.
(ii)  x  R , y > 0 only if a > 0 & b²  4ac < 0 (figure 3) .
(iii)  x  R , y < 0 only if a < 0 & b²  4ac < 0 (figure 6) .
Carefully go through the 6 different shapes of the parabola given below.
Fig. 1 Fig. 2 Fig. 3
y y y

a>0
a>0 a>0 D<0
D>0 D=0



x1 O x2 x O x O x

Roots are real & distinct Roots are coincident Roots are complex conjugate

Fig. 4 Fig. 5 Fig. 6
y
y y


O x O x
a<0
D>0 a<0
x1 x2 a<0
D=0 D<0
O x


Roots are real & distinct Roots are coincident Roots are complex conjugate


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,Maths IIT-JEE ‘Best Approach’ (MC SIR) Quadratic Equation
7. SOLUTION OF QUADRATIC INEQUALITIES:
ax2 + bx + c > 0 (a  0).
(i) If D > 0, then the equation ax2 + bx + c = 0 has two different roots x1 < x2.
Then a > 0  x  (, x1)  (x2, )
a < 0  x  (x1, x2)
(ii) If D = 0, then roots are equal, i.e. x1 = x2.
In that case a > 0  x  (, x1)  (x1, )
a < 0  x 
P (x ) <
(iii) Inequalities of the form 0 can be quickly solved using the method of intervals.
Q (x ) >

8. MAXIMUM & MINIMUM VALUE of y = ax² + bx + c occurs at x =  (b/2a) according as ;
 4 ac  b 2   4 ac  b 2 
a < 0 or a > 0 . y   ,  if a > 0 & y     ,  if a < 0 .
 4a   4a 

9. COMMON ROOTS OF 2 QUADRATIC EQUATIONS [ONLY ONE COMMON ROOT] :
Let  be the common root of ax² + bx + c = 0 & ax2 + bx + c = 0 . Therefore

a ² + b+ c = 0 ; a² + b + c = 0. By Cramer’s Rule 2  1
 
bc  bc a c  ac ab  a b
ca ca bcbc
Therefore,  =  .
aba b a cac
So the condition for a common root is (ca  ca)² = (ab ab)(bc bc).
10. The condition that a quadratic function f (x , y) = ax² + 2 hxy + by² + 2 gx + 2 fy + c may be resolved
into two linear factors is that ;
a h g
abc + 2 fgh  af2  bg2  ch2 = 0 OR h b f = 0.
g f c
11. THEORY OF EQUATIONS :
If 1, 2, 3, ......n are the roots of the equation;
f(x) = a0xn + a1xn-1 + a2xn-2 + .... + an-1x + an = 0 where a0, a1, .... an are all real & a0  0 then,
a1 a a a
 1 =  ,  1 2 = + 2 , 1 2 3 =  3 , ....., 1 2 3 ........n = (1)n n
a0 a0 a0 a0
Note : (i) If  is a root of the equation f(x) = 0, then the polynomial f(x) is exactly divisible by (x ) or
(x ) is a factor of f(x) and conversely .
(ii) Every equation of nth degree (n 1) has exactly n roots & if the equation has more than n roots,
it is an identity.
(iii) If the coefficients of the equation f(x) = 0 are all real and  + i is its root, then   i is also
a root. i.e. imaginary roots occur in conjugate pairs.
(iv) If the coefficients in the equation are all rational &  +  is one of its roots, then   is also
a root where ,   Q &  is not a perfect square.
(v) If there be any two real numbers 'a' & 'b' such that f(a) & f(b) are of opposite signs, then
f(x) = 0 must have atleast one real root between 'a' and 'b' .
(vi) Every equation f(x) = 0 of degree odd has atleast one real root of a sign opposite to that of its
last term.



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, Maths IIT-JEE ‘Best Approach’ (MC SIR) Quadratic Equation
12. LOCATION OF ROOTS:
Let f (x) = ax2 + bx + c, where a > 0 & a, b, c  R.
(i) Conditions for both the roots of f (x) = 0 to be greater than a specified number ‘d’ are
b2  4ac  0; f (d) > 0 & ( b/2a) > d.
(ii) Conditions for both roots of f (x) = 0 to lie on either side of the number ‘d’ (in other words
the number ‘d’ lies between the roots of f (x) = 0) is f (d) < 0.
(iii) Conditions for exactly one root of f (x) = 0 to lie in the interval (d , e) i.e. d < x < e are
b2  4ac > 0 & f (d) . f (e) < 0.
(iv) Conditions that both roots of f (x) = 0 to be confined between the numbers p & q are
(p < q). b2  4ac  0; f (p) > 0; f (q) > 0 & p < ( b/2a) < q.

13. LOGARITHMIC INEQUALITIES
(i) For a > 1 the inequality 0 < x < y & loga x < loga y are equivalent.
(ii) For 0 < a < 1 the inequality 0 < x < y & loga x > loga y are equivalent.
(iii) If a > 1 then loga x < p  0 < x < ap
(iv) If a > 1 then logax > p  x > ap
(v) If 0 < a < 1 then loga x < p  x > ap
(vi) If 0 < a < 1 then logax > p  0 < x < ap

SOLVED EXAMPLES

1. Solve the following equations :
2x2 – (3 + 7i)x + (9i – 3) = 0
Sol. We have 2x2 – (3 + 7i)x + (9i – 3) = 0.
Here a = 2, b = –(3 + 7i), c = 9i – 3

b  b 2  4ac 3  7i  [(3  7i)]2  4(2)(9i  3) 3  7i  9  49i 2  42i  72i  24
 x  
2a 2 2 4

3  7i  16  30i
or x
4
Let square root of –16 – 30i be a + ib.
 (a + ib)2 = – 16 – 30i
 (a2 – b2) + (2ab)i = – 16 – 30i  a2 – b2 = –16 and 2ab = –30.
Now, (a2 + b2)2 = (a2 – b2)2 + (2ab)2.
 (a2 + b2)2 = (– 16)2 + (30)2 = 1156
 a2 + b2 = 34 ( a2 + b2  0)
Also, a2 – b2 = –16
Adding, we get 2a2 = 18, a2 = 9, a =  3
30
a=3 b  5  a + ib = 3 – 5i
2(3)

30
a = –3  b  5  a + ib = –(3 – 5i)
2(3)

 a + ib =  (3 – 5i)
 Square root of –16 – 30i =  (3 + 5i)



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