Question

If the roots of the equation ${{x}^{2}}+ax+b=0$ are $c$ and $d$, then roots of the equation ${{x}^{2}}+\left( 2c+a \right)x+{{c}^{2}}+ac+b=0$ are

Answer 1

Hint: To solve this question we need to have knowledge about solving quadratic equations. To solve this question we need to represent the quadratic equation as the product and sum of the roots, where roots are the number which actually make the quadratic equation result to zero.

Complete step by step answer:
The question ask us to find the value of find the roots of quadratic equation ${{x}^{2}}+\left( 2c+a \right)x+{{c}^{2}}+ac+b=0$ when the roots of the other quadratic equation ${{x}^{2}}+ax+b=0$is given as $c$ and $d$. To solve this question, our first step will be to consider the two roots of the quadratic equation ${{x}^{2}}+\left( 2c+a \right)x+{{c}^{2}}+ac+b=0$ be $p$ and $q$.
If $c$ is one of the root of the equation ${{x}^{2}}+ax+b=0$, then on substituting the $x$ with $c$ which is the root we will get the following equation:
$\Rightarrow {{c}^{2}}+ac+b=0$
On analysing the second equation ${{x}^{2}}+\left( 2c+a \right)x+{{c}^{2}}+ac+b=0$ whose roots are required to be found, we see that the constant part of the equation is same as the above equation which is ${{c}^{2}}+ac+b=0$. On removing the constant part of the equation we get:
$\Rightarrow {{x}^{2}}+\left( 2c+a \right)x+{{c}^{2}}+ac+b=0$
$\Rightarrow {{x}^{2}}+\left( 2c+a \right)x+0=0$
$\Rightarrow {{x}^{2}}+\left( 2c+a \right)x=0$
Now the quadratic equation which we get has no constant term. If we take $x$ as common, we get:
 $\Rightarrow x\left( x+\left( 2c+a \right) \right)=0$
On calculating the two roots we get is $x=0$ and $x=-(2c+a)$ .
We know that the sum of roots is equal to the negative of the coefficient of $x$ in the quadratic equation. So on applying the same in the second equation we get:
$\Rightarrow p+q=-\left( 2c+a \right)$
Since one of the roots is zero so,
$\Rightarrow 0+q=-\left( 2c+a \right)$
$\Rightarrow q=-\left( 2c+a \right)$
$\Rightarrow q=-2c-a$
So the second root is $-\left( 2c+a \right)$. We need to write the root in terms of $c$ and $d$.For this we will see the equation 1 which is ${{x}^{2}}+ax+b=0$. Applying the sum of root rule in this equation we get:
$\Rightarrow c+d=-a$
On this on the above expression we get:
$\Rightarrow q=-2c+c+d$
$\Rightarrow q=d-c$
So the roots of the equation are $p=0$ and $q=(d-c)$ .
$\therefore $ The roots of the equation ${{x}^{2}}+\left( 2c+a \right)x+{{c}^{2}}+ac+b=0$ are $0$ and $(d-c)$ .

Note: Always remember that for a quadratic equation having roots $a$ and $b$ can be written as ${{x}^{2}}-\left( a+b \right)x+ab=0$. Sometimes a function $f\left( x \right)$ is represented as $f\left( c \right)=0$, which means at $x=c$ the function $f\left( x \right)$ becomes zero.