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(i) \[{{\left( m+n \right)}^{2}}+{{\left( m-n \right)}^{2}}\]

(ii) \[{{\left( m+n \right)}^{3}}+4mn\]

Answer
Verified

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 donâ€™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 donâ€™t solve it as a quadratic equation to get the value.