How to Construct a Quadratic Equation Based on Given Conditions in the CAT?
Tuesday, August 18th, 2020
Quadratic Equations:
As we all know a quadratic equation is a second-degree polynomial in x equated to zero and also if the co-efficient of x2 is zero, it becomes a linear equation.
It would be a waste of time to explain how to find roots to a CAT aspirant so skipping the very normal basics let’s understand the various types of question on quadratic that appears in CAT exam.
Constructing AÂ Quadratic Equation
Very often we are required to obtain a quadratic equation when some conditions are given. The following conditions occur frequently.
- The roots of the quadratic equation are given.
- The sum of the roots and the products of the roots of the quadratic equation are given.
- The relation between the roots of the equation to be obtained and the roots of another equation is given.
a. If the roots of the quadratic equation are given as α and β, the equation can be written as:
(x- α)(x- β) = 0 i.e., x2 – x(α+β) + αβ = 0
b. If p is the sum of the roots of the quadratic equation and q is the product of the roots of the quadratic equation, then the equation can be written as x2 – px + q = 0
c. For this condition, we shall consider 5 possible relations.
If we are given a quadratic equation, we can obtain a new quadratic equation by changing the roots of this equation in the manner specified to us.
i) A quadratic equation whose roots are the reciprocals  of the roots of the given equation ax2 + bx + c=0 i.e. the roots of the required equation are 1/α and 1/β.
This can be obtain by substituting 1/x in place of x in the given equation giving us cx2 +bx + a =0, i.e. we get the equation required by interchanging the co-efficient of x2 and the constant term.
ii) A quadratic equation (say in y) whose roots are k more than the roots of the equation ax2 + bx + c=0, i.e. the roots are (α + k) and (β + k) or x+k=y
The required equation can be obtained by substituting (y-k) in place of x in the given equation.
iii) A quadratic equation (say in y) whose roots are k less than the roots of the equation ax2 + bx + c=0, i.e. the roots are (α – k) and (β – k) or x-k=y
The required equation can be obtained by substituting (y+k) in place of x in the given equation.
iv) A quadratic equation (say in y) whose roots are k times the roots of the equation ax2 + bx + c=0, i.e. the roots are kα and kβ or kx =y
The required equation can be obtained by substituting y/k in place of x in the given equation
v) A quadratic equation (say in y) whose roots are 1/k times the roots of the equation ax2 + bx + c=0, i.e. the roots are α/k and β/k or x/k =y
The required equation can be obtained by substituting ky in place of x in the given equation.
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Examples:
1. Construct a quadratic equation from x2 + 7x + 16 = 0
i) 4 more than the current roots
ii) Reciprocal of the current roots
Sol:
i. For 4 more, as discussed in point ii. Replace x by (x-4) in the above equation. On solving we will get x2 + 15x + 16= 0
ii. For reciprocal of current roots, replace x by 1/x then on solving we will get the new equation as: 16x2+ 7x + 1 = 0
2. Equation x2 + 5x – 7 = 0 has roots a and b. Equation 2x2 + px + q = 0 has roots a + 1 and b + 1. Find p + q.
Sol:Â
There are two ways of solving it.
First way is to find out the roots then add 1 to them and then the form new equation. However, in this case the roots would be irrational so it is going to be an arduous task.
The another way is to form a new equation by replacing x to (x-1) {as roots are incremented by 1}.
On replacing x by (x-1), the new equation would be: (x-1)2 + 5(x-1) – 7 = 0
On solving this you will get; x2 + 3x – 11= 0; on multiplying by 2 we get: 2x2 +6x – 22 = 0
On comparing , we get p =6 and q = -22.
Hence p+q= -16
Common Roots:
Condition for one common root:
Let α be the common root of the given equations a1x2 +b1x + c1=0 and
a2x2 + b2x + c2=0. Then the common root is given by:
α = (c1a2-c2a1)/(a1b2-a2b1) or α = (b1c2-b2c1)/(c1a2-c2a1)
E.g:
a)Â Find the condition that the equations x2 + px + q = 0 and
x2 + qx + p = 0 (p≠q) have one root in common.
Sol: If one of the equations a1x2 +b1x + c1=0 and a2x2 + b2x + c2=0
have one root in common then (c1a2 – c2a1)2 = (b1c2 – b2c1)(a1b2-a2b1)
Here a1 = a2 =1
b1 =p= c2
c1 =q= b2
Therefore (q-p)2 = (p.p –q.q) (q-p)
⇒ (q-p)2 = (p2 – q2)(q-p)
⇒ q-p = (p-q)(p+q) ⇒ p+q = -1
b) If the equations x2 + 7ax + 10 = 0 and x2 + 3bx -85 = 0, where a and b are integers, have a common root, then the value of b would be?
Sol: Given the both equations have exactly one root in common, as the product of the roots of the two equations are different. This root must be rational.
The roots are rational, their sum in an integer and their product is rational therefore the roots are integers.
For equation (1) the roots are factors of 10 (±1, ±2, ±5, ±10) ad for (2) they are factors of 85 (±1, ±5, ±17, ±85) only if the roots are ±2, ±5
Suppose 5 is the common root; the other root of x2 + 3bx -85 = 0 is -17 then -3b = -17 +5
⇒ b =4
Suppose -5 is the common root; then the other root of x2 + 3bx -85 = 0 is -17 then -3b = 17 -5
⇒ b =4
Hence the value of b can be 4.
c) If a, b, c belong to R and equations ax2 + bx + c=0  and  x2 + 2x + 9 =0 have a common root, show that a:b:c = 1:2:9
Sol:  The roots of the equation x2 + 2x + 9 =0 are imaginary as the discriminant D= 4-36= -32 , which is negative.
As the imaginary roots of a quadratic equation are conjugate to each other, so both the roots of the equations will be common.
So, a/1 = b/2 =c/9
or, a:b:c = 1:2:9
The questions in CAT based on forming the equations by changing the roots of the current equation and on common are up to a limited extent especially the former one type. However, continue to practice the common roots questions by solving more questions.
Quadratic Equations based on x + 1/x type and other algebraic identities
Word Problems that can be solved using Quadratic Equation Techniques
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