Question

Please don't spam
Please give correct answer​

Answer 1

Answer:

Question :-

$\dfrac{5}{6}$ and $\dfrac{4}{3}$ are the two roots of the quadratic equation $x + \dfrac{1}{x} = \dfrac{13}{6}$ - Whether True or False.

Solution :-

$\because x + \dfrac{1}{x} =\: \dfrac{13}{6}$

$\Rightarrow$ $\dfrac{x^2 + 1}{x} = \dfrac{13}{6}$

$\Rightarrow$ 6x² + 6 = 13x

$\Rightarrow$6x² - 13x + 6 = 0 [a quadratic equation of the form ax² + bx + c = 0 (a ≠ 0) ]

Now, 6x² - 9x - 4x + 6 = 0

$\Rightarrow$ 3x(2x - 3) - 2(2x - 3) = 0

$\Rightarrow$ (2x - 3) (3x - 2) = 0

$\therefore$ 2x - 3 = 0

$\Rightarrow$ 2x = 3

$\Rightarrow$ $x = \dfrac{3}{2}$

$\Rightarrow$ 3x - 2 = 0

$\Rightarrow$ 3x = 2

$\Rightarrow$ $x = \dfrac{2}{3}$

Hence, the roots of the equation are : $\dfrac{3}{2}$ , $\dfrac{2}{3}$

Thus, the statement is false.

Answer 2

GIVEN :–

• Quadratic equation x + 1/x = 13/6 have two roots ⅚ & 4/3 .

To Check :–

• Is the given condition is true or false ?

SOLUTION :–

$\\ \implies \bf x + \dfrac{1}{x} = \dfrac{13}{6}$

$\\ \implies \bf \dfrac{ {x}^{2} + 1}{x} = \dfrac{13}{6}$

$\\ \implies \bf6({x}^{2} + 1) =13x$

$\\ \implies \bf6{x}^{2}+6 =13x$

$\\ \implies \bf6{x}^{2} - 13x+6 = 0$

• We know that –

$\\ \implies \bf Sum \: \: of \: \: roots = - \dfrac{b}{a}$

$\\ \implies \bf \dfrac{5}{6} + \dfrac{4}{3} = - \dfrac{( - 13)}{6}$

$\\ \implies \bf \dfrac{5 + 8}{6}=\dfrac{13}{6}$

$\\ \implies \bf \dfrac{13}{6}=\dfrac{13}{6} (\checkmark)$

• We also know that –

$\\ \implies \bf Product \: \: of \: \: roots = \dfrac{c}{a}$

$\\ \implies \bf \dfrac{5}{6} \times \dfrac{4}{3} = \dfrac{6}{6}$

$\\ \implies \bf \dfrac{20}{18}=\dfrac{6}{6}$

$\\ \implies \bf \dfrac{10}{9}=1 (\sf X)$

• Hence , Given statement is false.