Question

The equation (x^2 − 4ax + b )(x + a ) = 0 has three solutions and the sum of all solutions is 60.
Find a.

Answer 1

⠀⠀ || ✪ ϙᴜᴇsᴛɪᴏɴ ✪ ||

The equation (x^2 − 4ax + b )(x + a ) = 0 has three solutions and the sum of all solutions is 60. Find a.

⠀⠀ || ✪.ANSWER.✪ ||

Given:-

• equation, (x²-4ax+b)(x+a)=0
• Sum of all zeroes = 60

Find:-

• Value of a

⠀ || ✪.Explanation.✪ ||

Equation,

➠ (x²-4ax+b)(x+a) = 0

➠ (x²-4ax+b) = 0 Or, (x+a) = 0

➠ (x²-4ax+b) = 0 Or, x = -a

Now, calculate zeroes of (x²-4ax+b) = 0

Using Dharacharya Formula,

★ x = [ -B±√{B²-4AC}]/2a

where,

• B = -4a
• A = 1
• C = b

keep value in above equation,

➠ x = [ -(-4a)±√{(-4a)²-4.1.b}]/2.1

➠x = [ 4a ± √{ 16a² - 4b }]/2

➠x = [4a ± 2√(4a²-b)]/2

➠x = [ 2a ± √(4a² - b )]

First, take +ve sign,

➠ x = [ 2a + √(4a²-b)]

Now, Take -ve sign,

➠ x = [ 20 - √(4a²-b)]

So, A/c to question,

Sum of all zeroes = 60

➠ [ 2a + √(4a²-b)] + [ 2a - √(4a²-b)] + (-a) = 60

➠ 4a - a = 60

➠ 3a = 60

➠ a = 60/3

➠ a = 20

Thus:-

• Value of a = 20