
If a and b are the roots of the equation$3{{m}^{2}}=6m+5$, find the value of $\dfrac{a}{b}+\dfrac{b}{a}$
Answer
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Hint:If a and b are the roots of a quadratic equation $p{{x}^{2}}+qx+r=0$ , the sum of the roots is $a+b=\dfrac{-q}{p}$ and the product of the roots is $ab=\dfrac{r}{p}$.Using this concept try to solve the given question.
Complete step-by-step answer:
The given quadratic equation $3{{m}^{2}}=6m+5$ can be rewritten as $3{{m}^{2}}-6m-5=0$ is in the form $p{{x}^{2}}+qx+r=0$ where $p=3,q=-6$ and $r=-5$.
Let consider $a$ and $b$ are the roots of a quadratic equation.
We know that, the Sum of the roots = $\dfrac{-q}{p}$
$a+b=\dfrac{-q}{p}$
Now put the values of p and q in the above equation, we get
$a+b=\dfrac{-(-6)}{3}=\dfrac{6}{3}=2$
And the product of the roots =$\dfrac{r}{p}$
$ab=\dfrac{r}{p}$
Now put the values of p and r in the above equation, we get
$ab=\dfrac{-5}{3}$
Now to find the value of ${{a}^{2}}+{{b}^{2}}$ as follows
We have
${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Rearranging the terms, we get
${{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab$
Now put the values of $(a+b)$and$ab$, we get
${{a}^{2}}+{{b}^{2}}={{(2)}^{2}}-2\left( \dfrac{-5}{3} \right)$
Rearranging the terms, we get
${{a}^{2}}+{{b}^{2}}=4+\dfrac{10}{3}$
Taking the LCM on the right side, we get
${{a}^{2}}+{{b}^{2}}=\dfrac{22}{3}$
Therefore the values of ${{a}^{2}}+{{b}^{2}}$ is $\dfrac{22}{3}$ .
Let us consider
$\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{{{a}^{2}}+{{b}^{2}}}{ab}$
Now put the values of ${{a}^{2}}+{{b}^{2}}$ and $ab$ in the above equation, we get
$\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{\left( \dfrac{22}{3} \right)}{\left( \dfrac{-5}{3} \right)}$
Cancelling the terms, we get
$\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{22}{-5}$
Hence, the value of the given expression is $\dfrac{-22}{5}$ .
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, $a{{x}^{2}}+bx+c=0$.Students should remember formula of sum and product of roots of quadratic equation for solving these types of questions.
Complete step-by-step answer:
The given quadratic equation $3{{m}^{2}}=6m+5$ can be rewritten as $3{{m}^{2}}-6m-5=0$ is in the form $p{{x}^{2}}+qx+r=0$ where $p=3,q=-6$ and $r=-5$.
Let consider $a$ and $b$ are the roots of a quadratic equation.
We know that, the Sum of the roots = $\dfrac{-q}{p}$
$a+b=\dfrac{-q}{p}$
Now put the values of p and q in the above equation, we get
$a+b=\dfrac{-(-6)}{3}=\dfrac{6}{3}=2$
And the product of the roots =$\dfrac{r}{p}$
$ab=\dfrac{r}{p}$
Now put the values of p and r in the above equation, we get
$ab=\dfrac{-5}{3}$
Now to find the value of ${{a}^{2}}+{{b}^{2}}$ as follows
We have
${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Rearranging the terms, we get
${{a}^{2}}+{{b}^{2}}={{(a+b)}^{2}}-2ab$
Now put the values of $(a+b)$and$ab$, we get
${{a}^{2}}+{{b}^{2}}={{(2)}^{2}}-2\left( \dfrac{-5}{3} \right)$
Rearranging the terms, we get
${{a}^{2}}+{{b}^{2}}=4+\dfrac{10}{3}$
Taking the LCM on the right side, we get
${{a}^{2}}+{{b}^{2}}=\dfrac{22}{3}$
Therefore the values of ${{a}^{2}}+{{b}^{2}}$ is $\dfrac{22}{3}$ .
Let us consider
$\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{{{a}^{2}}+{{b}^{2}}}{ab}$
Now put the values of ${{a}^{2}}+{{b}^{2}}$ and $ab$ in the above equation, we get
$\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{\left( \dfrac{22}{3} \right)}{\left( \dfrac{-5}{3} \right)}$
Cancelling the terms, we get
$\dfrac{a}{b}+\dfrac{b}{a}=\dfrac{22}{-5}$
Hence, the value of the given expression is $\dfrac{-22}{5}$ .
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, $a{{x}^{2}}+bx+c=0$.Students should remember formula of sum and product of roots of quadratic equation for solving these types of questions.
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