Answer
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Hint: It is said that the two roots are m and n. Thus the product of m and n will be 3 and their sum will be (+5). Thus substitute these values in the expression to be verified and get the values.
Complete step-by-step answer:
We have been given the quadratic equation \[{{x}^{2}}-5x+3=0.......(1)\]
Thus the quadratic equation is similar to the general equation \[a{{x}^{2}}+bx+c=0\]. Thus comparing both the equation (1) and the general equation, we get the values as a = 1, b = 5, c = 3.
It is said that m and n are roots of the equation. Thus we can write the equation as \[{{x}^{2}}-(m+n)x+mn=0\].
We get the values as m + n = 5 and mn = 3.
Now let us substitute these values in the given expressions.
(i) \[{{\left( m+n \right)}^{2}}+{{\left( m-n \right)}^{2}}\]
We can write \[{{\left( m+n \right)}^{2}}\] as \[{{\left( m+n \right)}^{2}}-4mn\].
Thus \[{{\left( m+n \right)}^{2}}+{{\left( m-n \right)}^{2}}={{\left( m+n \right)}^{2}}+{{\left(
m+n \right)}^{2}}-4mn=2{{\left( m+n \right)}^{2}}-4mn\]
Put \[{{\left( m+n \right)}^{2}}=5\] and mn = 3.
\[\begin{align}
& \therefore {{\left( m+n \right)}^{2}}+{{\left( m-n \right)}^{2}}=2{{\left( m+n
\right)}^{2}}-4mn \\
& =2\times {{5}^{2}}-4\times 3=2\times 25-(4\times 3) \\
& =50-12=38. \\
\end{align}\]
Similarly, let us find the value of \[{{\left( m+n \right)}^{3}}+4mn\].
\[{{\left( m+n \right)}^{3}}+4mn={{5}^{3}}+4\times 3=125+12=137.\]
Thus we got, \[{{\left( m+n \right)}^{2}}+{{\left( m-n \right)}^{2}}=38.\]
\[{{\left( m+n \right)}^{3}}+4mn=137.\]
Note: We might try to get the roots of the quadratic equation first and then take the value of roots as m and n, which is wrong. We have been given the roots as m and n already. So don’t solve it as a quadratic equation to get the value.
Complete step-by-step answer:
We have been given the quadratic equation \[{{x}^{2}}-5x+3=0.......(1)\]
Thus the quadratic equation is similar to the general equation \[a{{x}^{2}}+bx+c=0\]. Thus comparing both the equation (1) and the general equation, we get the values as a = 1, b = 5, c = 3.
It is said that m and n are roots of the equation. Thus we can write the equation as \[{{x}^{2}}-(m+n)x+mn=0\].
We get the values as m + n = 5 and mn = 3.
Now let us substitute these values in the given expressions.
(i) \[{{\left( m+n \right)}^{2}}+{{\left( m-n \right)}^{2}}\]
We can write \[{{\left( m+n \right)}^{2}}\] as \[{{\left( m+n \right)}^{2}}-4mn\].
Thus \[{{\left( m+n \right)}^{2}}+{{\left( m-n \right)}^{2}}={{\left( m+n \right)}^{2}}+{{\left(
m+n \right)}^{2}}-4mn=2{{\left( m+n \right)}^{2}}-4mn\]
Put \[{{\left( m+n \right)}^{2}}=5\] and mn = 3.
\[\begin{align}
& \therefore {{\left( m+n \right)}^{2}}+{{\left( m-n \right)}^{2}}=2{{\left( m+n
\right)}^{2}}-4mn \\
& =2\times {{5}^{2}}-4\times 3=2\times 25-(4\times 3) \\
& =50-12=38. \\
\end{align}\]
Similarly, let us find the value of \[{{\left( m+n \right)}^{3}}+4mn\].
\[{{\left( m+n \right)}^{3}}+4mn={{5}^{3}}+4\times 3=125+12=137.\]
Thus we got, \[{{\left( m+n \right)}^{2}}+{{\left( m-n \right)}^{2}}=38.\]
\[{{\left( m+n \right)}^{3}}+4mn=137.\]
Note: We might try to get the roots of the quadratic equation first and then take the value of roots as m and n, which is wrong. We have been given the roots as m and n already. So don’t solve it as a quadratic equation to get the value.
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