Question

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

Answer 1

$\large \green {\underline{answer}} \\ \\ \large \implies{ \boxed{a \: = \: 12 \: \: \: and \: \: \: a \: = \: 14}} \\ \\ \large \implies \orange{ \underline{step \: \: by \: \: step \: \: explanation}} \\ \\ \large \implies \orange{ \underline{find \: \: value \: \: of \: \: a}} \\ \\ \large \implies \orange{ \underline{given}} \\ \\ \large \implies{equation \: \: has \: \: real \: \: roots} \\ \\ \large \implies \green{ \boxed{d \: = \: {b}^{2} - 4ac }} \\ \\ \large \implies{(a - 12) {x}^{2} + 2(a - 12)x + 2 = 0 } \\ \\ \large \implies{d \: = \: 4(a - 12) {}^{2} - 4(a - 12)(2) = 0 } \\ \\ \large \implies{4( {a}^{2} + 144 - 24a) - 8(a - 12) = 0 } \\ \\ \large \implies{4a {}^{2} + 576 - 96a - 8a + 96 = 0 } \\ \\ \large \implies{4a {}^{2} - 104a + 672 = 0 } \\ \\ \large \implies{ {a}^{2} - 26a + 168 = 0 } \\ \\ \large \implies{ {a}^{2} - 14a - 12a + 168 = 0 } \\ \\ \large \implies{a(a - 14) - 12(a - 14) = 0} \\ \\ \large \implies{(a - 12)(a - 14) = 0} \\ \\ \large \implies \therefore \green{ \boxed{a \: = 12 \: \: and \: \: a \: = 14}} \\ \\ \large \implies{ \underline{related \: \: formula}} \\ \\ \large \implies{d > 0 \: \: roots \: \: are \: \: real \: \: and \: \: unequal} \\ \\ \large \implies{d \: = 0 \: roots \: \: are \: \: real \: \: and \: \: equal} \\ \\ \large \implies{d < 0 \: \: roots \: \: are \: \: imaginary}$

Answer 2

$\tt\pink{Given,}$

→$\tt{Equation\:have\: real \:roots.}$

→$\tt{D=b^2-4ac}$

$\tt\pink{To\:find:-}$

→$\tt{The\:value\:of\:a}$

$\tt\pink{Solution,}$

→$\tt{(a-12)x^2+2(a-12)x+2=0}$

→$\tt{D=4(a-12)^2-4(a-12)2}$

→$\tt{4(a^2+144-24a)-8(a-12)=0}$

→$\tt{4a^2+576-96a-8a+96=0}$

→$\tt{4a^2-104a+672=0}$

→$\tt{a^2-26a+168=0}$

→$\tt{a^2-14a-12a+168=0}$

→$\tt{a(a-14)-12(a-14)=0}$

→$\tt{a(a-14)-12(a-14)=0}$

→$\tt{(a-12)(a-14)}$

$\tt\pink{We\:get,}$

$\fbox{a=12}$

$\fbox{a=14}$