Question

please solve this question...........​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

$\mathfrak{ \large \red{ \underline { \underline{question}}}} \\ \implies{ \large{show \: that \: the \: roots \: of \: equation \: are \: real \: or \: unequal}} \\ { \large{(a - b) {x}^{2} + 5(a + b)x - 2(a - b) = 0}} \\ \\ \mathfrak{ \large \blue{ \underline{solution : }}} \\ \implies{ \large{compare \: give \: equation \: with \: the \: general \: from \: of \: quadatric \: a {x}^{2} + bx + c = 0 }} \\ \implies{ \large{a = ( a - b) \: and \: b = 5(a + b) \: and \: c = - 2(a - b)}} \\ { \large \green{ \underline{find \: discriminant}}} \\ \implies{ \large{d = {b}^{2} - 4ac}} \\ \implies{ \large{(5(a + b)) {}^{2} - 4(a - b)( - 2(a - b)) }} \\ \implies{ \large{25(a + b) {}^{2} + 8a(a - b) {}^{2} }} \\ \implies{ \large{17(a + b) {}^{2} + (8(a + b) {}^{2} + 8(a - b) {}^{2} )}} \\ \implies{ \large{17(a + b) {}^{2} + 16( {a}^{2} + {b}^{2} ) }} \\ \\ { \large \orange{ \underline{proved}}} \\ \implies{ \large{which \: is \: always \: greater \: than \: zero}} \\ \implies{ \large{equation \: has \: real \: and \: unequal \: root}}$