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QUADRATIC EQUATIONS 1.
Quadratic Polynomial A polynomial in the form p (x ) = ax 2 + bx + c , where a 0 and a, b, c are real numbers and x is a real variable is called a quadratic polynomial.
2.
Quadratic Equation An equation p (x ) = 0 where p (x ) is a quadratic polynomial is called a quadratic equation, i.e., ax 2 + bx + c = 0 , where a 0. e.g. 2 x 2 − 3 x + 7, 8 x 2 + x − 19
3.
Pure Quadratic Equation Equation of type ax 2 + c = 0 is known as pure quadratic equation.
4.
Roots of a Quadratic Equation The values of variable satisfying the given quadratic equation are called its roots. For Example: Determine whether the given values of x are the roots of the given equation or not: x 2 + 6 x + 5 = 0; x = −1, x = −5
Consider p(x ) = x 2 + 6 x + 5
…(i)
Substituting x = –1 in (i), we get
p( −1) = ( −1) 2 + 6 ( −1) + 5 =1–6+5 = 6−6 = 0. Hence x = −1 is the root of given equation. Again substituting x = –5 in (i), we get
p(−5) = ( −5) 2 + 6 ( −5) + 5 = 25 – 30 + 5 = 30 – 30 = 0 Hence x = −5 is the root of given equation. 5.
x = –1 and x = –5 are the roots of given equation. Methods to Find Roots of a Quadratic Equation (i)
Discriminant Method Let quadratic equation be ax 2 + bx + c = 0
Step 1: Find D = b 2 − 4ac ; D is called the discriminant of the equation. Step 2: (a) (b)
If D 0, roots are given by x =
−b+ D −b− D . , 2a 2a
If D = 0, equation has equal roots and root is given by x =
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−b . 2a
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(c) For Example:
Using quadratic formula solves the quadratic equation: p2 x 2 + (p2 − q 2 ) x − q 2 = 0
We have, p2 x 2 + (p2 − q 2 ) x − q 2 = 0
Comparing the equation with ax 2 + bx + c = 0, we have a = p 2 , b = p 2 − q 2 and c = −q 2
D = b 2 − 4ac = ( p 2 − q 2 ) 2 − 4 p 2 −q 2 = ( p 2 − q 2 ) 2 + 4 p 2 q 2 = p 4 + q 4 − 2 p 2 q 2 + 4 p 2 q 2 = ( p 2 + q 2 )2 0
[ square of any real number is non-negative] So, the given equation has real roots given by
and,
=
− b + D − ( p 2 − q 2 ) + ( p 2 + q 2 ) 2q 2 q 2 = = = 2a 2p 2 2p 2 p 2
=
− b − D − (p2 − q 2 ) − (p2 + q 2 ) = = −1 2a 2p 2
Hence the roots are
q2 p2
and -1
(ii) Factorisation method Step 1: Resolve the quadratic polynomial ax 2 + bx + c into the product of linear factors. i.e., ax 2 + bx + c = ( Ax + B)(Cx + D) where A, B, C and D are real numbers and A, C 0 Step 2: Put the linear factors equal to zero and get the values of x. (Ax + B)(Cx + D) = 0 Ax + B = 0, Cx + D = 0 −B −D ,x = are the two roots of quadratic equation. A C For Example: x=
Solve by factorisation: a 2 b 2 x 2 + b 2 x − a 2 x − 1 = 0 a 2b 2 x 2 + b 2 x − a 2 x − 1 = 0
b 2 x (a 2 x + 1) − 1(a 2 x + 1) = 0
(a 2 x + 1) b 2 x − 1 = 0
(
)
Either (a 2 x + 1) = 0
or
( b 2 x − 1) = 0
or
b2 x = 1
a 2 x = −1
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x=−
1 a2
Hence, x = −
x=
or
1 b2
1 1 are the required solutions , a2 b2
(iii) Completion of square method Let quadratic equation be ax 2 + bx + c = 0 b c Step 1: Make the coefficient of x 2 equal to 1, i.e., a x 2 + x + = 0 a a
x2 +
b c x+ =0 a a
Step 2: Add and subtract the square of half of coefficient of x, b b2 b2 c − + =0 i.e., x 2 + x + a 4a 2 4a 2 a Step 3: Express in the form of a perfect square added with a constant, 2 b b2 c i.e., x + − 2 + =0 2a a 4a 2 b b2 c b 2 − 4ac Step 4: x + = − = 2a 4a 2 a 4a 2
Step 5: x +
b b 2 − 4ac = 2a 4a 2
Step 6: x =
−b b 2 − 4ac 2a 4a 2
For Example: Solve the equation x 2 − ( 3 + 1) x + 3 = 0 by the method of completing the square. We have,
x 2 − ( 3 + 1) x + 3 = 0
x 2 − ( 3 + 1) x = − 3 2
3 + 1 on both sides we have Adding 2 2
3 + 1 x + 3 + 1 = − 3 + 3 + 1 x − 2 2 2 2
x − 3 + 1 = − 3 + 3 + 1 + 2 3 2 4
2
2
2
=
− 4 3 + 3 + 1+ 2 3 4
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3 + 1− 2 3 4
2
x − 3 + 1 = 3 − 1 2 2
x−
3 +1 3 −1 = 2 2
x= x−
or
2
3 −1 3 +1 2 3 + = = 3 2 2 2
3 +1 3 −1 =− 2 2
3 − 1 + 3 +1 = 2 = 1 x = − 2 2 2
x = 3, 1
Hence, the roots are 6.
3 and 1.
Nature of Roots Let quadratic equation be ax 2 + bx + c = 0 , a 0, and D = b 2 − 4ac , (i) If D 0, then equation has real and unequal roots. For Example: Find the values of k for which the given equation has real and unequal roots: kx 2 − 6 x − 2 = 0
We have kx 2 − 6 x − 2 = 0
Here,
a = k, b = −6 and c = –2
D = b 2 − 4ac = ( −6) 2 − 4 k −2 = 36 + 8k
The given equation will have real and unequal roots, if − 36 9 D 0 36 + 8k 0 8k −36 k k − 8 2 (ii) If D = 0, then equation has equal and real roots. For Example: Find the values of k for which the following equation has equal roots: (k − 12 ) x 2 + 2(k − 12 ) x + 2 = 0 We have, (k − 12 ) x 2 + 2 (k − 12 ) x + 2 = 0
Here, a = k − 12, b = 2 (k − 12) and c = 2
D = b 2 − 4ac = 4 (k − 12 ) 2 − 4 (k − 12 ) 2
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D = 4 (k − 12) {(k − 12) − 2}
D = 4 (k − 12) (k − 14)
The given equation will have equal roots, if D=0
4 (k − 12) (k − 14) = 0 k – 12 = 0 or k – 14 = 0
k = 12 or, k = 14 (iii) If D 0, then equation has no real roots. (iv) If a, b, c are rational, D 0 and perfect square, then roots are real, unequal and rational. (v) If a, b, c are rational, D > 0 and not a perfect square, then roots are real, unequal and conjugate irrational, i.e., if one root is p + q then other is p − q . 7.
Condition for real roots D 0 i.e., b 2 − 4ac 0.
8.
Results to remember b b (i) ax − b 0 x if a 0 and x if a 0 a a (ii)
x 2 − a 2 0 x −a or x a
(iii) x 2 − a 2 = 0 x = −a or x = a (iv) x 2 − a 2 0 − a x a (v) ( x − a)( x − b) 0, a b x a or x b (vi) ( x − a)( x − b) 0, a b a x b For Example: Find the values of k for which the equation x 2 + 5kx + 16 = 0 has no real roots. The given equation is x 2 + 5kx + 16 = 0 Comparing the given equation with ax 2 + bx + c = 0, we have a = 1, b = 5k, c = 16 D = b 2 − 4ac = (5k ) 2 − 4 1 16 = 25k 2 − 64
The given equation will have no real roots if D < 0
25k 2 − 64 0
64 25 k 2 − 0 25
9.
64 0 25 8 8 − k 5 5 k2 −
[If ab < 0 and a > 0, then b < 0] [If x 2 − a 2 0, then –a < x < a]
Applications of Quadratic Equations Step 1:
Change the word problems into the form of quadratic equations according to the given conditions.
Step 2:
Solve the quadratic equations.
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https://www.entrancei.com/ For example: Some students planned a picnic. The budget for food was Rs. 480. But eight of these failed to go and thus the cost of food for each member increased by Rs. 10. How many students attended the picnic? Let the x students planned for picnic, then share of each student = Rs. Share of each student when 8 students failed to go = Rs.
480 x −8
According to question 480 480 − = 10 x −8 x 480 x − 480( x − 8) = 10 x ( x − 8) 480 x − 480 x + 480 8 = 10 x ( x − 8)
3840 2
x − 8x
=
10 1
10 x 2 − 80 x = 3840
10 x 2 − 80 x − 3840 = 0
x 2 − 8 x − 384 = 0
x 2 + 16 x − 24 x − 384 = 0
x(x + 16) – 24(x + 16) = 0
(x + 16)(x – 24) x = 24, – 16
(Not possible)
24 students planned for picnic Students went for the picnic = 24 – 8 = 16.
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480 x
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