English, asked by nagendra123net, 3 months ago

Solve the quadratic equation (x - 1)² -5(x-1)+6=0​

Answers

Answered by Anonymous
5

Given equation:-

  • (x - 1)² - 5(x - 1) + 6 = 0

To:-

  • Solve and find the zeroes of the equation.

Solution:-

(x - 1)² - 5(x - 1) + 6 = 0

By using the identity:-

(a - b)² = a² + b² - 2ab

= {(x)² + (1)² - 2 × x × 1} - 5x + 5 + 6 = 0

= x² + 1 - 2x - 5x + 5 + 6 = 0

Taking like terms together,

x² - 2x - 5x + 1 + 5 + 6 = 0

=> x² - 7x + 12 = 0

By splitting the middle term,

= x² - 4x - 3x + 12 = 0

=> x(x - 4) - 3(x - 4) = 0

=> (x - 4) (x - 3) = 0

Either,

x - 4 = 0

=> x = 4

Or,

x - 3 = 0

=> x = 3

Therefore, the two zeroes of the given quadratic equation are 4 and 3.

________________________________

Verification:-

Sum of zeroes = -(Coefficient of x)/Coefficient of x²

= 4 + 3 = -(-7)/1

=> 7 = 7

Product of zeroes = Constant Term/Coefficient of x²

= 4 × 3 = 12/1

=> 12 = 12

Hence, Verified!!!

________________________________

How to solve?

→ Firstly we removed the first bracket by applying the identity and then the second bracket. Then we brought the like terms together. After adding the like terms we got a quadratic equation which is in the form ax² + bx + c. So by splitting the middle term we got the two zeroes of the quadratic equation.

________________________________

Answered by niha123448
0

Explanation:

Given equation:-

(x - 1)² - 5(x - 1) + 6 = 0

To:-

Solve and find the zeroes of the equation.

Solution:-

(x - 1)² - 5(x - 1) + 6 = 0

By using the identity:-

(a - b)² = a² + b² - 2ab

= {(x)² + (1)² - 2 × x × 1} - 5x + 5 + 6 = 0

= x² + 1 - 2x - 5x + 5 + 6 = 0

Taking like terms together,

x² - 2x - 5x + 1 + 5 + 6 = 0

=> x² - 7x + 12 = 0

By splitting the middle term,

= x² - 4x - 3x + 12 = 0

=> x(x - 4) - 3(x - 4) = 0

=> (x - 4) (x - 3) = 0

Either,

x - 4 = 0

=> x = 4

Or,

x - 3 = 0

=> x = 3

Therefore, the two zeroes of the given quadratic equation are 4 and 3.

________________________________

Verification:-

Sum of zeroes = -(Coefficient of x)/Coefficient of x²

= 4 + 3 = -(-7)/1

=> 7 = 7

Product of zeroes = Constant Term/Coefficient of x²

= 4 × 3 = 12/1

=> 12 = 12

Hence, Verified!!!

________________________________

How to solve?

→ Firstly we removed the first bracket by applying the identity and then the second bracket. Then we brought the like terms together. After adding the like terms we got a quadratic equation which is in the form ax² + bx + c. So by splitting the middle term we got the two zeroes of the quadratic equation.

________________________________

hope this helps you!!

thank you ⭐

Similar questions