
If the sum of the roots of the equation $a{{x}^{2}}+bx+c=0$ is equal to sum of their squares, then
(A). ${{a}^{2}}+{{b}^{2}}={{c}^{2}}$
(B). ${{a}^{2}}+{{b}^{2}}=a+b$
(C). $ab+{{b}^{2}}=2ac$
(D). $ab-{{b}^{2}}=2ac$
Answer
597k+ views
Hint: Use the relation between the zeroes and the coefficients of a polynomial. Also, try to draw some results using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
Complete step-by-step solution:
For the standard quadratic equation $a{{x}^{2}}+bx+c=0$ , the roots are given by:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now we let the roots of the equation be l and m.
Also, the relation between the coefficients and the roots of a general quadratic equation comes out to be:
Sum of the roots of quadratic equation = $l+m=\dfrac{-\left( \text{coefficient of x} \right)}{\text{coefficient of }{{\text{x}}^{2}}}=\dfrac{-b}{a}$ .
Product of the roots of quadratic equation = \[lm=\dfrac{\text{constant term}}{\text{coefficient of }{{\text{x}}^{2}}}\text{=}\dfrac{\text{c}}{\text{a}}\] .
Now we will move to the actual part of the solution.
So, as it is given in the question that the sum of the roots of the equation $a{{x}^{2}}+bx+c=0$ is equal to the sum of their squares, we can say:
$l+m={{l}^{2}}+{{m}^{2}}$
Now we know ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, which implies ${{\left( a+b \right)}^{2}}-2ab={{a}^{2}}+{{b}^{2}}$ . So, using this our equation can be written as:
$l+m={{\left( l+m \right)}^{2}}-2lm$
Now, if we use the values from the above results we found, we get
$\dfrac{-b}{a}={{\left( \dfrac{-b}{a} \right)}^{2}}-2\times \dfrac{c}{a}$
$\Rightarrow \dfrac{-b}{a}=\dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{2c}{a}$
$\Rightarrow -ab={{b}^{2}}-2ac$
$\Rightarrow ab+{{b}^{2}}=2ac$
Therefore, the necessary condition for the sum of the roots of the equation $a{{x}^{2}}+bx+c=0$ is equal to the sum of their squares is $ab+{{b}^{2}}=2ac$ . Hence, the answer to the above question is option (c).
Note: Be careful about the signs while using the relation between the coefficients and roots of the polynomial. Also, you need to learn all the formulas related to algebra and exponents, as they are used very often.
Complete step-by-step solution:
For the standard quadratic equation $a{{x}^{2}}+bx+c=0$ , the roots are given by:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now we let the roots of the equation be l and m.
Also, the relation between the coefficients and the roots of a general quadratic equation comes out to be:
Sum of the roots of quadratic equation = $l+m=\dfrac{-\left( \text{coefficient of x} \right)}{\text{coefficient of }{{\text{x}}^{2}}}=\dfrac{-b}{a}$ .
Product of the roots of quadratic equation = \[lm=\dfrac{\text{constant term}}{\text{coefficient of }{{\text{x}}^{2}}}\text{=}\dfrac{\text{c}}{\text{a}}\] .
Now we will move to the actual part of the solution.
So, as it is given in the question that the sum of the roots of the equation $a{{x}^{2}}+bx+c=0$ is equal to the sum of their squares, we can say:
$l+m={{l}^{2}}+{{m}^{2}}$
Now we know ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$, which implies ${{\left( a+b \right)}^{2}}-2ab={{a}^{2}}+{{b}^{2}}$ . So, using this our equation can be written as:
$l+m={{\left( l+m \right)}^{2}}-2lm$
Now, if we use the values from the above results we found, we get
$\dfrac{-b}{a}={{\left( \dfrac{-b}{a} \right)}^{2}}-2\times \dfrac{c}{a}$
$\Rightarrow \dfrac{-b}{a}=\dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{2c}{a}$
$\Rightarrow -ab={{b}^{2}}-2ac$
$\Rightarrow ab+{{b}^{2}}=2ac$
Therefore, the necessary condition for the sum of the roots of the equation $a{{x}^{2}}+bx+c=0$ is equal to the sum of their squares is $ab+{{b}^{2}}=2ac$ . Hence, the answer to the above question is option (c).
Note: Be careful about the signs while using the relation between the coefficients and roots of the polynomial. Also, you need to learn all the formulas related to algebra and exponents, as they are used very often.
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