Question

Answer ,the questions in the pic nd one more question ,

find the real roots if exist
Q 3x²-bx+2=0

correct answers will b marked brainliest and rewarded 15 points..​

Answer 1

Answer:

$x=\sqrt{\cfrac{2}{3}}$

Answers for questions is pic:

$\text{(i) } n = -6 \text{ or } n = 2$

$\text{(ii) } n = \cfrac{2}{3} \text{ or } n = -\cfrac{1}{3}$

$\text{(iii) } n = \cfrac{2b}{a} \text{ or } n = \cfrac{b}{a}$

$\text{(iv) } n = - 4a \text{ or } n = \cfrac{3b}{a}$

Step-by-step explanation:

For question, if real roots exist:

Given equation is $3x^2 - bx + 2 = 0$

Using discriminant formula:

$D = b^2 - 4ac$

$0 \leq (-b)^2 - 4(3)(2)$

$b^2 \geq 24$

$b \geq 2\sqrt{6}$

Now, given equation becomes, $3x^2 - 2\sqrt{6}x + 2 = 0$

Using quadratic formula,

$x = \cfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\therefore x = \cfrac{-(-2\sqrt{3}) \pm \sqrt{24 - 24}}{2(3)}$

$\Rightarrow x = \cfrac{2\sqrt{6}}{2 \times 3} = \cfrac{\sqrt{6}}{3} = \sqrt{\cfrac{2}{3}}$

For questions in pic:

(i)

$\cfrac{1}{n-1} - \cfrac{1}{n + 5} = \cfrac{6}{7}$

$\Rightarrow \cfrac{n + 5 - n + 1}{(n - 1) (n + 5)} = \cfrac{6}{7}$

$\Rightarrow \cfrac{6}{(n-1)(n+5)} = \cfrac{6}{7}$

$\Rightarrow (n-1)(n+5) = 7$

$\Rightarrow n^2 + 4n - 5 - 7 = 0$

$\Rightarrow n^2 + 4n - 12 = 0$

$\Rightarrow n^2 + 6n - 2n - 12 = 0$

$\Rightarrow n(n + 6) -2(n + 6) = 0$

$\Rightarrow (n+6)(n-2) - 0$

$\therefore n = -6 \text{ or }n = 2$ Ans.

(ii)

$9n^2 - 3n - 2 = 0$

$\Rightarrow 9n^2 - 6n + 3n - 2 = 0\\\Rightarrow 3n(3n-2) + 1 (3n - 2) = 0 \\\Rightarrow (3n - 2)(3n + 1) = 0 \\\Rightarrow n = \cfrac{2}{3} \text{ or } n = -\cfrac{1}{3}$

(iii)

$a^2n^2 - 3abn + 2b^2 = 0$

$\Rightarrow a^2n^2 -2abn -abn + 2b^2 = 0$

$\Rightarrow an(an - 2b) - b(an - 2b) = 0$

$\Rightarrow (an-2b)(an - b) = 0$

$\therefore n = \cfrac{2b}{a} \text{ or } n = \cfrac{b}{a}$

(iv)

$an^2 + (4a^2 - 3b)n - 12ab = 0$

$\Rightarrow an^2 + 4a^2n - 3bn - 12ab = 0$

$\Rightarrow an(n + 4a) - 3b (n + 4a) = 0$

$\Rightarrow (n + 4a)(an - 3b) = 0$

$\therefore n = - 4a \text{ or } n = \cfrac{3b}{a}$

Answer 2

Answer:

Answer is in the attachment