Question

MCQ QUESTIONS
Consider a quadratic equation X^2+bx+c. If both b and care modified so that c increases increased by 44%
and one of the roots increases by what is the percentage
change in b?

Question 8

Cannot be determined
20% decrease
20% increase
No change
​

Answer 1

Complete question:

Consider a quadratic equation x^2 + bx + c = 0. If both b and c are modified such that c increases by 44% and one of the roots increases by 20%, what is the percentage change in b?

Solution:

The given quadratic equation is

x^2 + bx + c = 0

Given:

• Both b and c are modified; c is increased by 44%
• One of the root is increased by 20%

To find:

• The percentage of change in b = ?

We consider, the changed value of b is b'.

Let's find the changed value of c first:

The changed value of c is

= c * (1 + 44/100)

= 144c/100

We find roots of the given quadratic equation:

Solving by quadratic formula, we can conclude that the two roots of x^2 - bx + c = 0 are

x = {- b + √(b^2 - 4c)}/2, {- b - √(b^2 - 4c)}/2

We take the root, x = {- b + √(b^2 - 4c)}/2 to solve the problem.

Changed value of the root by change of b and c:

Changed value of the root by the change of b and c is

= [- b' + √{b'^2 - 4 (144c/100)}]/2 ..... (1)

Changed value of the root by increase of 20%:

Changed value of {- b + √(b^2 - 4c)}/2 by 20% is

= {- b + √(b^2 - 4c)}/2 * (1 + 20/100)

= {- b + √(b^2 - 4c)}/2 * 120/100

Rearranging:

= [- 120b/100 + √{(120/100b)^2 - 4 (144c/100)}]/2

..... (2)

By the given condition, (1) and (2) values must be equal by a change of b.

Then,

b' = 120b/100

= b + 20b/100

= b (1 + 20/100)

Clearly, b is increasing here by 20%.

Option (C) 20% increase of b.

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