Verified Answer :

mariamikayla

\(a+b=31=> \boxed{a=b-31} \\ab=234 \\\\ \\\\ b*(b-31)=234 \\\\ b^2-31b-234=0 \\ a=1 \\ b=-31 \\ c=-234 \\\\ \Delta= (-31)^2-4*1*(-234)= 961-936=25 \\\\ x_1;x_2=\frac{-(-31)+/-\sqrt{25}}{2*1} =\frac{31+/-5}{2} \\\\ x_1=\frac{31+5}{2}=\frac{36}{2}\to\boxed{18} \\\\ x_2=\frac{31-5}{2}=\frac{26}{2}\to\boxed{13} \\\\ (x-18)(x-13)=0 \\\\ 1)x-18=0 \ => \boxed{x=18} \\\\ 2)x-13=0 \ => \boxed{x=13}\)

calculista

Let

x-------> the first number

y------> the second number


we know that

\(x+y=31\)

\(x=31-y\)

equation \(1\)


\(x*y=234\)

equation \(2\)


substitute equation 1 in equation 2

\((31-y)*y=234\\ 31y-y^{2} =234\\ -y^{2} +31y-234=0\)



using a graph tool-----> to resolve the second order equation

see the attached figure


the solution is

\(13 and 18\)

View image calculista

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