Question

# If m and n are the two roots of the equation ${{x}^{2}}-5x+3=0$, find the values of:(i) ${{\left( m+n \right)}^{2}}+{{\left( m-n \right)}^{2}}$(ii) ${{\left( m+n \right)}^{3}}+4mn$

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.

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.