Question

the sum of two number is 9 and the sum of their square is 41 .taking one number as x ,form an equation in x and solve it to find the numbers.​

Answer 1

Answer:

x=-4 or x=-5

Step-by-step explanation:

given the first no. is x

∴second no. is 9-x

B.T.P.

$x^{2}$+$(9-x)^{2}$=41

⇒$x^{2}$+81+$x^{2}$+2×9×x=41

⇒2$x^{2}$+18x+40=0

⇒$x^{2}$+9x+20=0

⇒$x^{2}$+4x+5x+20=0

⇒x(x+4)+5(x+4)=0

⇒(x+4)(x+5)=0

∴either (x+4)=0 or (x+5)=0

         ⇒x=-4       ⇒x=-5    

Answer 2

Answer:

if y=4,then x=5.if y=5 then x=4