# Find the zeroes of the quadratic polynomial ${x^2} + x - 2$, and verify the relationship between the zeroes and the coefficients?

Answer

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Hint – While analysing the given question, write down the equation and try to solve this quadratic equation by any method possible and find the zeros. After finding the zeros we will show the relationship between the zeroes and the coefficients by using the two basic formulas which are sum of zeroes and product of zeroes. For instance, we will use $\alpha {\text{ and }}\beta$ as the zeroes of the polynomial equation.

“Complete step-by-step answer:”

We have, ${x^2} + x - 2 = 0$

Here, we will solve this quadratic equation by using factorisation method

$

{x^{^2}} + 2x - x - 2 = 0 \\

x\left( {x + 2} \right) - 1\left( {x + 2} \right) = 0 \\

\left( {x - 1} \right)\left( {x + 2} \right) = 0 \\

\\

$

It means zeroes of the given quadratic equation are

$x = 1, - 2$ … (1)

From this ${x^2} + x - 2 = 0$ quadratic equation we will find the coefficient of ${x^2}$ is $1$ and the coefficient of $x$ is $1$ whereas the constant is $ - 2$.

Now, the relation between zeroes and quadratic polynomials can be verified by sum of zeroes and product of zeroes.

Let $\alpha {\text{ and }}\beta $ be two zeroes which are equal to $ - 2{\text{ and 1}}$ respectively from (1)

Now, if we put values of $\alpha = 1{\text{ and }}\beta {\text{ = - 2}}$ then,

$\alpha + \beta = 1 - 2 = - 1$ … (2)

As we know,

${\text{Sum of zeroes = }}\dfrac{{{\text{ - coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$

$\therefore {\text{ }}\alpha {\text{ + }}\beta {\text{ = }}\dfrac{{ - 1}}{1} = - 1$ … (3)

Hence, from (2) and (3) we can say that ${\text{L}}{\text{.H}}{\text{.S = R}}{\text{.H}}{\text{.S}}$

Similarly, if we put values of $\alpha = 1{\text{ and }}\beta {\text{ = - 2}}$ in,

$\alpha \times \beta = 1 \times - 2 = - 2$ … (4)

As we know,

${\text{product of zeroes = }}\dfrac{{{\text{constant }}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$

$\therefore {\text{ }}\alpha \times \beta {\text{ = }}\dfrac{{ - 2}}{1} = - 2$ … (5)

Hence, from (4) and (5) we can say that ${\text{L}}{\text{.H}}{\text{.S = R}}{\text{.H}}{\text{.S}}$

Hence, the relationship between zeroes and coefficients of polynomials is verified.

Note – The most crucial concept in these type of questions is that we should solve the given quadratic polynomial at very first point by any of the following methods:

$\left( 1 \right)$ Factorisation

$\left( 2 \right)$ Completing the square

$\left( 3 \right)$ Quadratic formula

$\left( 4 \right)$ Graphing

Do note that after finding zeros we can simply put the values of zeroes, coefficients and constants in those two formulas mentioned above, where the coefficients are defined as a constant by which an algebraic term is multiplied.

To get the required answer do remember those things. This will also help us to verify the relationship.

“Complete step-by-step answer:”

We have, ${x^2} + x - 2 = 0$

Here, we will solve this quadratic equation by using factorisation method

$

{x^{^2}} + 2x - x - 2 = 0 \\

x\left( {x + 2} \right) - 1\left( {x + 2} \right) = 0 \\

\left( {x - 1} \right)\left( {x + 2} \right) = 0 \\

\\

$

It means zeroes of the given quadratic equation are

$x = 1, - 2$ … (1)

From this ${x^2} + x - 2 = 0$ quadratic equation we will find the coefficient of ${x^2}$ is $1$ and the coefficient of $x$ is $1$ whereas the constant is $ - 2$.

Now, the relation between zeroes and quadratic polynomials can be verified by sum of zeroes and product of zeroes.

Let $\alpha {\text{ and }}\beta $ be two zeroes which are equal to $ - 2{\text{ and 1}}$ respectively from (1)

Now, if we put values of $\alpha = 1{\text{ and }}\beta {\text{ = - 2}}$ then,

$\alpha + \beta = 1 - 2 = - 1$ … (2)

As we know,

${\text{Sum of zeroes = }}\dfrac{{{\text{ - coefficient of x}}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$

$\therefore {\text{ }}\alpha {\text{ + }}\beta {\text{ = }}\dfrac{{ - 1}}{1} = - 1$ … (3)

Hence, from (2) and (3) we can say that ${\text{L}}{\text{.H}}{\text{.S = R}}{\text{.H}}{\text{.S}}$

Similarly, if we put values of $\alpha = 1{\text{ and }}\beta {\text{ = - 2}}$ in,

$\alpha \times \beta = 1 \times - 2 = - 2$ … (4)

As we know,

${\text{product of zeroes = }}\dfrac{{{\text{constant }}}}{{{\text{coefficient of }}{{\text{x}}^2}}}$

$\therefore {\text{ }}\alpha \times \beta {\text{ = }}\dfrac{{ - 2}}{1} = - 2$ … (5)

Hence, from (4) and (5) we can say that ${\text{L}}{\text{.H}}{\text{.S = R}}{\text{.H}}{\text{.S}}$

Hence, the relationship between zeroes and coefficients of polynomials is verified.

Note – The most crucial concept in these type of questions is that we should solve the given quadratic polynomial at very first point by any of the following methods:

$\left( 1 \right)$ Factorisation

$\left( 2 \right)$ Completing the square

$\left( 3 \right)$ Quadratic formula

$\left( 4 \right)$ Graphing

Do note that after finding zeros we can simply put the values of zeroes, coefficients and constants in those two formulas mentioned above, where the coefficients are defined as a constant by which an algebraic term is multiplied.

To get the required answer do remember those things. This will also help us to verify the relationship.

Last updated date: 28th May 2023

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