Question

If $p\left( x \right)=a{{x}^{2}}+bx+c$ find:
$\left( i \right){{\alpha }^{2}}+{{\beta }^{2}}$

Answer 1

Hint: To solve the given question, we will use the given quadratic equation $a{{x}^{2}}+bx+c$ to find the relation between its roots that are $\alpha $ and \[\beta \]. Then, we will convert given expression ${{\alpha }^{2}}+{{\beta }^{2}}$ in the form of perfect square ${{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta $. Then, we will substitute the corresponding values in the obtained expression and will use appropriate mathematical operations to simplify the expression and will get the required answer.

Complete step by step answer:
Since, we have the quadratic equation as:
$\Rightarrow a{{x}^{2}}+bx+c$
The properties of the quadratic equation:
1. The sum of roots of quadratic equation is equal to the negative ratio of coefficient of $x$ to coefficient of ${{x}^{2}}$ as:
$\Rightarrow \alpha +\beta =-\dfrac{b}{a}$
2. The product of roots of quadratic equation is equal to the ratio of constant to coefficient of ${{x}^{2}}$ as:
$\Rightarrow \alpha \beta =\dfrac{c}{a}$
Now, we will have the given question as:
 $\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}$
Now, we will convert it into a sum of roots and product of roots to make the process easier. So, we will add and subtract $2\alpha \beta $ in the above expression as:
$\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -2\alpha \beta $
As we know, the formula $\left( a+b \right)={{a}^{2}}+{{b}^{2}}+2ab$. So, we will use it in the above step and will convert the expression as:
$\Rightarrow {{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta $
Now, we will substitute $-\dfrac{b}{a}$ for \[\left( \alpha +\beta \right)\] and $\dfrac{c}{a}$ for \[\alpha \beta \] in the above expression as:
$\Rightarrow {{\left( -\dfrac{b}{a} \right)}^{2}}-2\times \dfrac{c}{a}$
Here, we will do the square of $-\dfrac{b}{a}$ and will get $\dfrac{{{b}^{2}}}{{{a}^{2}}}$ as:
$\Rightarrow \dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{2c}{a}$
Now, we will use the method of subtraction of fraction as:
$\Rightarrow \dfrac{{{b}^{2}}-2ac}{{{a}^{2}}}$
Hence, this is the required solution of the given question while using a given quadratic equation.

Note: As we know that the quadratic equation is a equation expressed as $p\left( x \right)=a{{x}^{2}}+bx+c$, where $x$ is variable and $a,b,c$ are constants. Since, the degree of the quadratic equation is \[2\]. So, it has two roots named $\alpha $ and $\beta $. We can use discriminant formula $\left( {{b}^{2}}-4ac \right)$ to check the types of roots and can find the value of roots as:
$\Rightarrow \alpha =\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$
$\Rightarrow \beta =\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$