Question

find two numbers whose sum is 27 and product is 182

Answer 1

Hi dear,
here is ur answer.........
let the two numbers X and Y
so, according to the question
X+Y=27
X=27-Y.............................................1.
XY=182
(27-Y)(Y)=182
27Y-Y^2=182
Y^2-27Y+182=0
Y^2-13Y-14Y+182=0
(Y-13)(Y-14)=0
so,
if Y=13 then X=14
if X=13 then Y =14
hope it helps uh...............

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Assume the number be m and 27 - m

★Situation :-

m(27 - m) = 182

27m - m² = 182

m² - 27m + 182 = 0

${\boxed{\sf\:{Factorise\;it\;we\;get}}}$

m² - 13m - 14m + 182 = 0

m(m - 13) - 14(m - 13) = 0

(m - 13) = 0

m = 13

(m - 14) = 0

m = 14

${\boxed{\sf\:{Therefore\;we\;get\;the\;two\;numbers\;13\;and\;14}}}$

$\boxed{\begin{minipage}{13 cm} Additional Information \\ \\ \ A\; Quadratic\; Equation\;has\;three\;equal\;roots \\ \\ 1)Real\;and\;Distinct \\ \\ 2)Real\;and\;Coincident \\ \\ 3) Imaginary \\ \\ Note:-Third\; Imaginary\;is\;not\;taken\;in\;class\;10th \\ \\ If\;p(x)\;is\;a\; quadratic\; polynomial\;then\;p(x)=0\;is\;called\; Quadratic\; Polynomial \\ \\ General\;Formula=ax^2+bx+c=0 \\ \\ A polynomial\;whose\;degree\;will\;be\;2\;is\; considered\;as\; Quadratic\; Polynomial \\ \\ Rules\;for\; solving\; Quadratic\; Equations:- \\ \\ Put\;all\;the\;terms\;into\;RHS\;and\;make\it\;zero \\ \\ Substitute\;all\; factors\;equal\;to\;Zero\;Get\;a\;equal\; solution \end{minipage}}$