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Question: If a and b are the roots of the equation\(3{{m}^{2}}=6m+5\), find the value of \(\dfrac{a}{b}+\dfrac...

If a and b are the roots of the equation3m2=6m+53{{m}^{2}}=6m+5, find the value of ab+ba\dfrac{a}{b}+\dfrac{b}{a}

Explanation

Solution

Hint:If a and b are the roots of a quadratic equation px2+qx+r=0p{{x}^{2}}+qx+r=0 , the sum of the roots is a+b=−qpa+b=\dfrac{-q}{p} and the product of the roots is ab=rpab=\dfrac{r}{p}.Using this concept try to solve the given question.

Complete step-by-step answer:
The given quadratic equation 3m2=6m+53{{m}^{2}}=6m+5 can be rewritten as 3m2−6m−5=03{{m}^{2}}-6m-5=0 is in the form px2+qx+r=0p{{x}^{2}}+qx+r=0 where p=3,q=−6p=3,q=-6 and r=−5r=-5.
Let consider aa and bb are the roots of a quadratic equation.
We know that, the Sum of the roots = −qp\dfrac{-q}{p}
a+b=−qpa+b=\dfrac{-q}{p}
Now put the values of p and q in the above equation, we get
a+b=−(−6)3=63=2a+b=\dfrac{-(-6)}{3}=\dfrac{6}{3}=2
And the product of the roots =rp\dfrac{r}{p}
ab=rpab=\dfrac{r}{p}
Now put the values of p and r in the above equation, we get
ab=−53ab=\dfrac{-5}{3}
Now to find the value of a2+b2{{a}^{2}}+{{b}^{2}} as follows
We have
(a+b)2=a2+2ab+b2{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}
Rearranging the terms, we get
a2+b2=(a+b)2−2ab{{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab
Now put the values of (a+b)(a+b)andabab, we get
a2+b2=(2)2−2(−53){{a}^{2}}+{{b}^{2}}={{(2)}^{2}}-2\left( \dfrac{-5}{3} \right)
Rearranging the terms, we get
a2+b2=4+103{{a}^{2}}+{{b}^{2}}=4+\dfrac{10}{3}
Taking the LCM on the right side, we get
a2+b2=223{{a}^{2}}+{{b}^{2}}=\dfrac{22}{3}
Therefore the values of a2+b2{{a}^{2}}+{{b}^{2}} is 223\dfrac{22}{3} .
Let us consider
ab+ba=a2+b2ab\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{{{a}^{2}}+{{b}^{2}}}{ab}
Now put the values of a2+b2{{a}^{2}}+{{b}^{2}} and abab in the above equation, we get
ab+ba=(223)(−53)\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{\left( \dfrac{22}{3} \right)}{\left( \dfrac{-5}{3} \right)}
Cancelling the terms, we get
ab+ba=22−5\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{22}{-5}
Hence, the value of the given expression is −225\dfrac{-22}{5} .

Note: The roots of a function are the x-intercepts. By definition, the y-coordinate of points lying on the x-axis is zero. Therefore, to find the roots of a quadratic function, we set f (x) = 0, and solve the equation, ax2+bx+c=0a{{x}^{2}}+bx+c=0.Students should remember formula of sum and product of roots of quadratic equation for solving these types of questions.