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Algebra
Quadratic Equations: 1
Quadratic Equations:
Quadratic Expression or Quadratic Function:
The general form of a quadratic function is f(x) = ax2 + bx + c, where a, b and c are real numbers and a ≠ 0. The degree of the expression is two. i.e. the highest exponent of the variable is two. Here a, b are the coefficients of x2 , x and c is the constant term.
Let us say f(x) = x2 + 1. Now if we plot the curve of f(x), we get a parabola:
Input = x 
2 
1 
0 
1 
2 
Output = f(x) 
5 
2 
1 
2 
5 
Here the vertex of the parabola lies at the point (0,1). The point of minima is x = 0 and the minimum value of f(x) = 1.
Quadratic Equation:
If f(x) = ax2 + bx + c = 0 , then it is called a quadratic equation. This means that the value of the quadratic expression is zero. Here we find the values of x for which the quadratic assumes a value equal to zero. In other words we can say that the values of x for which the graph of f(x) cuts the X axis.
Roots of a Quadratic Equation:
Let f(x) = ax2 + bx + c = 0
The roots are those values of x for which the value of the quadratic expression becomes zero or the values of x where the curve of f(x) cuts the X axis.
Let’s find the value of the roots.
Nature of roots:
1. Real and Unequal Roots:
4. Rational Roots:
If the quantity under the square root, i.e. the discriminant is a perfect square, then we can say that the roots are rational. So, if we want the roots to be rational then b2 4ac must be a perfect square.
Sum and Product of the Roots of a Quadratic Equation:
Remember: If a > the minimum value of f(x) occurs at x = b/2a
If a < 0, then thGraph of a Quadratic Expression:
If we look at the curve of ax2 + bx + c, then definitely it is a parabola. Now if a > 0 , i.e. if the coefficient of x2 is positive, then it is open upwards and if a < 0, then the parabola is open downwards.
a). If the roots are real and unequal, then the curve of f(x) intersects the X axis at two points which are the roots say α and β. Part of the curve lies above the X axis and a part of it lies below the X axis. The curve looks like:
So, we can say that when a > 0, and D > 0 then, f(x) > 0 when x < α and x > β
f(x) = 0 when x = α and β
f(x) < 0 when α < x < β
b). If the roots are real and equal
If D = 0, then the graph of f(x) touches the X axis at one point and at x = b/2a which is the point of minima or maxima depending on the sign of a. So, the graph of f(x) looks like:
So, if a >0 and D =0 , then f(x) > 0 for all values of x except at x = b/2a where f(x) = 0
a < 0 and D = 0, then f(x) < 0 for all values of x except at x = b/2a where f(x) = 0
c). If roots are imaginary, then there exist no real roots. So, the entire curve is entirely above the X axis or below it depending on the sign of a. The minimum or the maximum value occurs at x = b/2a .
So, if a > 0 and D < 0, then f(x) > 0 for all real values of x
If a < 0 and D < 0, then f(x) < 0 for all real values of x
Let us say we have a quadratic : x2 – 2x 3
This can be factorized as x2 – 3x + x – 3 or x (x3) + 1(x3)
Or x2 – 2x – 3 = (x3)(x+1)
Here x = 3 and 1 are the roots of this quadratic and we have written the above quadratic in terms of its roots.
So if ax2 + bx + c has two roots α and B, then we can write the quadratic in terms of its roots as:
ax2 + bx + c = a (x α)(xβ)
Forming a New Quadratic Equation by changing the roots of a given quadratic equation:
1. A quadratic equation whose roots are ‘n’ more than the roots of a quadratic ax2 + bx + c is :
a(xn)2 + b(xn) + c = 0
2. A quadratic equation whose roots are ‘n’ less than the roots of a quadratic ax2 + bx + c is :
a(x+n)2 + b(x+n) + c = 0
3. A quadratic equation whose roots are ‘n’ times the roots of a quadratic ax2 + bx + c is :
a(x/n)2 + b(x/n) + c = 0
4. A quadratic equation whose roots are 1/n times the roots of a quadratic ax2 + bx + c is :
a(nx)2 + b(nx) + c = 0
5. A quadratic equation whose roots are the reciprocals of the roots of a quadratic ax2 + bx + c is :
a(1/x)2 + b(1/x) + c =0
6. A quadratic equation whose roots are negative of the roots of a quadratic ax2 + bx + c is :
a(x)2 + b(x) + c = 0
7. A quadratic equation whose roots are the squares of the roots of a quadratic ax2 + bx + c is :
a(√x)2 + b(√x) + c = 0
8. A quadratic equation whose roots are the cubes of the roots of a quadratic ax2 + bx + c is :
a(x1/3)2 + b(x1/3) + c = 0
Example 1: Find the maximum value of 4  5x  4x2.
Solution: Here f(x) = 4 – 5x – 4x2 , we know that the maxima occurs at x = b/2a, here b = 5, a = 4
So maxima occurs at x = b/2a =  (5/8) = 5/8
Maximum value of f(x) = 4 – 5(5/8) – 4(5/8)2 = 89/16
Example 2: If ax2 + bx  15 = 0 does not have 2 distinct real roots, then find the maximum value of 3a + b?
Solution: Let f(x) = ax2 + bx – 15
Since f(x) = 0 does not have 2 distinct real roots, hence either
f(x) 0 for every x R or f(x) 0 for every x R.
Now since f(0) = 15
f(x) 0 for every x R.
In particular f (3)0.
9a + 3b 15 0.
3a + b 5.
Hence the maximum value of 3a + b = 5.
Location of Roots/ Interval in which Roots Lie
In some problems we want the roots of the equation ax2 + bx + c = 0 to lie in a given Interval. For this we impose conditions on a, b and c.
Let f(x) = ax2 + bx + c.
Case 1: If both the roots are positive i.e. they lie in (0, ),
Then the sum of the roots as well as the product of the roots must be positive
α+ β = b/a >0 And αβ = c/a > 0 with b24ac≥ 0
Similarly, if both the roots are negative i.e. they lie in (– , 0) then the sum of
the roots will be negative and the product of the roots must be positive
i.e. α+ β = –b/a < 0 and αβ = c/a >0 with b2 – 4ac ≥ 0.
Case 2 : Both the roots are greater than a number ‘k’
In this case we observe that three conditions hold:
(i) D ≥ 0 (ii) af (k) >0 (iii) k < (b/2a) i.e. k should be less than the point of maxima or minima.
Both the roots are greater so the equation has real roots or equal roots. Hence D ≥ 0. Further we see that if a >0, then f(k) >0 and if a < 0 then f(k) is also <0. So the product of ‘a’ and f(k) is positive. As k is less than both the roots so k must be less than –b/2a.
Case3: If both the roots are less than ‘k’
In this case we see the following conditions hold:
(i) D ≥ 0 (ii) af (k) > 0 (iii) k > (b/2a)
Again the roots are real or equal so D ≥0. As we see that if a >0, then f(k) >0 and if a < 0 then f(k) is also <0. so the product a and f(k) is positive. As k is greater than both the roots so k > (b/2a).
Case 4: If ‘k’ lies between the two roots
In this case we have the following conditions:
i) D > 0 (ii) af (k) < 0
First of all k lies between the two roots. So D > 0 for roots to be real. Also when a < 0 then we see f (k) >o and when a is > 0 then f(k) is < 0. So product of ‘a’ and ‘f(k)’ is negative.
Case 5: if both the roots lie in the interval (k1, k2)
Here the conditions that hold are:

D ≥ 0 (ii) af(k1) >0 (iii) af(k2) >0 (iv) k1 < (b/2a) < k2 .

Here we are assuming that k1< k2.
Case6: Exactly one root lies in the interval (k1, k2)
In this case we have the following conditions:
(i)D > 0 (ii) f(k1). f(k2) < 0
As the equation has two roots. Discriminant must be positive .Also we observe that f(k1) and f(k2) are opposite in signs.
So their product is negative.
Case 7: If k1and k2 lie between the two roots
The conditions that apply in this case are:
(i) D >0 (ii) a(f(k1) < 0 (iii) a f(k2) < 0
Example 3: For all ‘x’, x2+ 2ax + (103a) >0, then the interval in which ‘a’ lies is :
A. a < 5 B. 5 < a < 2 C. a > 5 D. 2 < a <5
Solution: Here a is positive and the expression is positive for all x.
So D < 0,
4a24(103a) < 0 or a2+3a10< 0
a2+5a2a10 < 0
(a+5)(a2) < 0
or we can say 5 < a < 2
Hence option (b) is correct.
Example 4: Let a,b,c be real. If ax2+bx+c=0 has two real roots α and β, where α < 1 and β > 1, then show that
1+ c/a+b/a < 0
Solution: Here the equation ax2+bx+c=0 has two real roots. So, discriminant is positive.
a>0 a< 0
? 1 1 β ? 1 1 β
We see that when a is positive then f(1) and f(1) both are negative
And when a is negative then f(1) and f(1) both are positive.
af(1) < 0 and af(1) < 0
f(1)= a+b+c and f(1)= ab+c
af (1) = a (a+b+c) < 0 and af (1)= a(ab+c) < 0
Divide both side of inequation by a2
1+ (b/a) + (c/a) < 0 and 1(b/a) +c/a< 0
Combining both these inequations we get
Or 1+b/a+c/a< 0
Example 5: Find two consecutive even natural numbers, sum of whose squares is 884
A. 20, 18 B. 22, 24 C. 16, 18 D. 20, 22
Solution: Let the two consecutive numbers be 2m and 2m +2,
Given that (2m)2 + (2m+2)2 = 884
4m2 + 4m2 + 4 + 8m = 884
Or 8m2 + 8m = 880 or m2 + m = 110
Or solving we get m = 10 and 11 , but m cannot be negative as m is a natural number
So, the numbers are 20 and 22
Option (D) is correct
Example 6: An aeroplane takes 2 hour less for the journey of 2400 Km, if its speed is increased by 100 Km/hr, from its usual speed . Find its usual speed.
A. 300 Km/hr
B. 400 Km/hr
C. 500 Km/hr
D. 600 Km/hr
Solution: Let the usual speed of the aeroplane be x km/hr
Time taken initially = 2400/x {Time = Distance/speed}
New Speed = x+100
New time taken = 2400/(x+100)
But now the time taken is 2 hours less
So 2400/x  2400/(x+100) = 2
2400*100x(x+100)=2
Or 240000 = 2x2 + 200x
Or x2 + 100x = 120000
Or x2 + 400x – 300x 120000 = 0