Question

If $\alpha \,$and $\beta$ are the zeros of the polynomial$2{{x}^{2}}-4x+5$, then find the value of
A) ${{\alpha }^{2}}+{{\beta }^{2}}$
B) $\dfrac{1}{\alpha }+\dfrac{1}{\beta }$
C) ${{\left( \alpha -\beta \right)}^{2}}$
D) $\dfrac{1}{{{\alpha }^{2}}}+\dfrac{1}{{{\beta }^{2}}}$
E) ${{\alpha }^{3}}+{{\beta }^{3}}$

Answer 1

In the given question, we have been given a quadratic equation. Quadratic equation is the equation in which the highest degree is equal to zero. The sum and the product of the roots of the quadratic equation can be calculated by using a formula i.e. ,
Sum of the roots = $\dfrac{-b}{a}$
Product of the roots = $\dfrac{c}{a}$,
Where $a$ is the coefficient of ${{x}^{2}}$, $b$ is the coefficient of $x$ and $c$ is the constant term of a given quadratic equation i.e. $a{{x}^{2}}+bx+c=0$.

Complete step-by-step answer:
We have the given polynomial,
$\Rightarrow 2{{x}^{2}}-4x+5$
Standard form of quadratic equation i.e. $a{{x}^{2}}+bx+c$, where $\alpha$ and $\beta$ are the roots of the equation
Thus,
Sum of the roots = $\alpha +\beta =\dfrac{-b}{a}$
Product of the roots = $\alpha \times \beta =\dfrac{c}{a}$
Compare it with the given polynomial i.e.$2{{x}^{2}}-4x+5$, we get
$\Rightarrow a=2$
$\Rightarrow b=-4$
$\Rightarrow c=5$
Thus,
Sum of the roots = $\alpha +\beta =\dfrac{-b}{a}=-\dfrac{-4}{2}=\dfrac{4}{2}=2$
Product of the roots = $\alpha \times \beta =\dfrac{c}{a}=\dfrac{5}{2}$
Now,
a) ${{\alpha }^{2}}+{{\beta }^{2}}$
Using the identity of ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
We have now,
$\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta$
Substituting the values and solving, we will get
$\Rightarrow {{2}^{2}}-2\times \left( \dfrac{5}{2} \right)=4-5=-1$
Thus,
$\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=-1$

b) $\dfrac{1}{\alpha }+\dfrac{1}{\beta }$
Taking the LCM of the denominators, we will get
$\Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\beta }=\dfrac{\alpha +\beta }{\alpha \beta }$
Substituting the values and solving, we will get
$\Rightarrow \dfrac{2}{\dfrac{5}{2}}=\dfrac{2}{5}\times 2=\dfrac{4}{5}$
Thus,
$\Rightarrow \dfrac{1}{\alpha }+\dfrac{1}{\beta }=\dfrac{4}{5}$

c) ${{\left( \alpha -\beta \right)}^{2}}$
Rewrite the given question in the form of sum of the roots and the products of the roots.
We will get,
$\Rightarrow {{\left( \alpha -\beta \right)}^{2}}={{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta$
Substituting the values and solving, we will get
$\Rightarrow {{2}^{2}}-4\times \dfrac{5}{2}=4-10=-6$
Thus,
$\Rightarrow {{\left( \alpha -\beta \right)}^{2}}=-6$

d) $\dfrac{1}{{{\alpha }^{2}}}+\dfrac{1}{{{\beta }^{2}}}$
Taking the LCM of the denominators, we will get
$\Rightarrow \dfrac{1}{{{\alpha }^{2}}}+\dfrac{1}{{{\beta }^{2}}}=\dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{{{\left( \alpha \beta \right)}^{2}}}$
Substituting the values and solving them, we will get,
$\Rightarrow\dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{{{\left( \alpha \beta \right)}^{2}}}=\dfrac{{{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta }{{{\left( \dfrac{5}{2} \right)}^{2}}}=\dfrac{{{2}^{2}}-2\times \dfrac{5}{2}}{\dfrac{25}{4}}=\dfrac{4-5}{\dfrac{25}{4}}=\dfrac{-1}{\dfrac{25}{4}}=\dfrac{-1}{25}\times 4=\dfrac{-4}{25}$
Thus,
$\Rightarrow \dfrac{1}{{{\alpha }^{2}}}+\dfrac{1}{{{\beta }^{2}}}=-\dfrac{4}{25}$

e) ${{\alpha }^{3}}+{{\beta }^{3}}$
Rewrite the given question in the form of sum of the roots and the products of the roots.
We will get,
Using the identity of ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$
$\Rightarrow {{\alpha }^{3}}+{{\beta }^{3}}={{\left( \alpha +\beta \right)}^{3}}-3\alpha \beta \left( \alpha +\beta \right)$
Substituting the values and solving them, we will get,
$\Rightarrow {{2}^{3}}-3\times \dfrac{5}{2}\left( 2 \right)=8-15=-7$
Thus,
$\Rightarrow {{\alpha }^{3}}+{{\beta }^{3}}=-7$

Note: An equation is said to be quadratic if the highest value of degree is equal to 2. Roots of the quadratic equation are the values of x, when substituted in the equation satisfies the equation. A root of the polynomial is a value where the polynomial is equal to zero. To solve these types of questions, we need to write the given polynomial in the standard form i.e. decreasing order. .