Question

(3n-1)^2=12n+8 whats the value of n ?

Answer 1

Step-by-step explanation:

(3n-1)^2=12n+8

(a-b)^2=a^2 -2ab +b^2

(3n)^2 -2×3n×1 + (1)^2 =12n+8

9n^2 -6n + 1 =12 n+8

9n^2 = 12n +6n +8 -1

9n^2 = 18n +7

9n^2 -18n -7 = 0

9n^2 - 21n +3n -7 =0

3n (3n-7) +1(3n-7)=0

(3n+1) + (3n-7)=0

3n+1=0 3n-7 =0

n=-1/3 n =7/3

Answer 2

(3n - 1)² = 12n + 8

We will apply this identity for the given equation ⇒ (a - b)² = a² - 2ab + b²

⇒ (3n - 1)² = 12n + 8

⇒ (3n)² - 2(3n)(1) + 1² = 12n + 8

⇒ 9n² - 6n + 1 = 12n + 8

Now we will subtract 1 from both sides of the equation.

⇒ 9n² - 6n + 1 - 1 = 12n + 8 - 1

⇒ 9n² - 6n = 12n + 7

Now we will add 6n to both sides of the equation.

⇒ 9n² - 6n + 6n = 12n + 7 + 6n

⇒ 9n² = 18n + 7

Subtract 18n+7 from both sides.

⇒ 9n² - (18n + 7) = 18n + 7 - (18n + 7)

⇒ 9n² - 18n - 7 = 18n + 7 - 18n - 7

⇒ 9n² - 18n - 7 = 18n - 18n + 7 - 7

⇒ 9n² - 18n - 7 = 0

Factor left side of equation.

⇒ (3n + 1) (3n - 7) = 0

Set factors equal to 0.

⇒ 3n + 1 = 0 or 3n - 7 = 0

⇒ 3n = -1 or 3n = 7

⇒ n = ⁻¹/₃ or n = ⁷/₃

∴ The value of 'n' is $\bf \frac{-1}{3}$ or $\bf \frac{7}{3}$.