Question

solue for xe by using quadratic formula : 9x2 -9(a+b)x + (2 a2+5ab +b2)=0​

Answer 1

$\huge \orange{AηsωeR} ✍$

$\tt \:  ⟼ {9x}^{2} - 9(a + b)x + ( {2a}^{2} + 5ab + {2b}^{2} ) = 0$

$\tt \:  ⟼ \: on \: comparing \: with \: A {x}^{2} + Bx + C = 0 \: we \: get$

$\begin{gathered}\begin{gathered}\bf \begin{cases} &\sf{A \: = 9} \\ &\sf{B = - 9(a +b) } \\ &\sf{ {2a}^{2} + 5ab + {2b}^{2} } \end{cases}\end{gathered}\end{gathered}$

$\tt \:  Discriminant, \: D = {B}^{2} - 4 \times A \times C$

$\tt \: D = {\bigg( - 9(a + b)}\bigg)^{2} - 4 \times 9 \times ( {2a}^{2} + 5ab + {2b}^{2} )$

$\tt \:  D = 81 ({a}^{2} + 2ab + {b}^{2} ) - 36( {2a}^{2} + 5ab + {2b}^{2} )$

$\tt \:  D = 81 {a}^{2} + {81b}^{2} + 162ab - 72 {a}^{2} - 180ab - 72 {b}^{2}$

$\tt \:  D = {9a}^{2} - 18ab + {9b}^{2}$

$\tt \:  D = 9( {a}^{2} - 2ab + {b}^{2} )$

$\tt \:  D = 9 {(a - b)}^{2}$

☆ This implies D > 0 and having real and distinct roots.

☆ Solution of equation is given by

$\tt \:  ⟼ \: x \: = \dfrac{ - \: B \: \pm \: \sqrt{D} }{2 \: A}$

$\tt \:  ⟼ \: \dfrac{9 \: (a \: + \: b) \: \pm \: \sqrt{9 \: {(a \: - \: b)}^{2} } }{2 \: \times \: 9}$

$\tt \:  ⟼ \: x \: = \: \dfrac{9a \: + \: 9b \: \pm(3a \: - \: 3b)}{18}$

$\tt \:  ⟼x = \dfrac{9a + 9b + 3a - 3b}{18} \: or \: \dfrac{9a + 9b - 3a + 3b}{18}$

$\tt \:  ⟼x = \dfrac{12a + 6b}{18} \: or \: \dfrac{6a + 12b}{18}$

$\tt \:  ⟼x = \dfrac{2a + b}{3} \: or \: \dfrac{a + 2b}{3}$