Question

If alpha and beta are the roots of equation x² - 5x + 6 = 0, then find the value of alpha - beta.​

Answer 1

Answer:

1

Step-by-step explanation:

In an equation written in form of x^2 - Sx + P = 0, S is the sum of roots and  P is the product of roots.   If α and β are its roots:

​⇒ α + β = 5     &  αβ = 6

  square on both sides of α+β=5

​⇒ α² + β² + 2αβ = 25

​⇒ α² + β² = 25 - 2(6) = 13

  Add - 2αβ to both sides:

⇒ α² + β²  - 2αβ = 13 - 2αβ

⇒  (α - β)² = 13 - 2(6) = 1

⇒ α - β = √1 = 1

Answer 2

$\underline{\underline{\sf{{\maltese\:\:Given}}}}$

$\sf\alpha + \beta$are the roots of the equation x² - 5x + 6 = 0.

$\underline{\underline{\sf{\maltese\:\:To\: Find}}}$

$\sf \: \alpha - \beta = \: ?$

$\underline{\underline{\sf{\maltese\:Calculations \:}}}$

$\bf\red{STEP \:1:-}$

Compare x² + 5x - 6 = 0 with ax² + bx + c = 0 to get

a = 1, b = 5 and c = 6.

$\bf\red{STEP \:2:-}$

Sum of roots $\sf \: \alpha + \beta = \frac{ - b}{a} = \frac{ - ( - 5)}{1}$

or,

$\sf\alpha + \beta = 5$

$\bf\red{STEP \:3:-}$

Product of roots $\sf \: \alpha \beta = \frac{c}{a} = \frac{6}{1}$

or,

$\sf \: \alpha \beta = 6$

$\bf\red{STEP \:4:-}$

$\alpha - \beta = ± \sqrt{ {( \alpha + \beta) }^{2} - 4 \alpha \beta }$

substituting the values,

$\longrightarrow \alpha - \beta = ± \sqrt{ {5}^{2} - 4 \times 6 }$

$\longrightarrow \alpha - \beta = ± \sqrt{25 - 24}$

$\longrightarrow \alpha - \beta = ± \sqrt{1}$

$\purple{\alpha - \beta = 1}$

and

$\pink{\alpha - \beta = - 1}$

HENCE, the value of $\alpha - \beta$ is -1 and +1