Question

Find the sum and the product of the zeros of the quadratic polynomial \[5{x^2} - 8x + 3\]

Answer 1

Hint: To solve this type of problems, first calculate the sum of the roots by finding the roots directly or by using coefficient relation \[\alpha + \beta = - \dfrac{b}{a}\]. Secondly calculate the product of the roots by multiplying the roots of the equation or by using \[\alpha \beta = \dfrac{c}{a}\] the relation between the product of the roots with the coefficient of the equation.

Complete Step-by-step solution
Let the roots or zeros of the quadratic equation is \[\alpha \] and \[\beta \]
On comparing \[5{x^2} - 8x + 3 = 0\] with the standard form of quadratic equation \[a{x^2} + bx + c = 0\]
We get
\[a = 5\]
\[b = - 8\]
\[c = 3\]
Now by applying the sum of the zeros relation with the ratios of the coefficient of the quadratic equation
\[\alpha + \beta = - \dfrac{b}{a}\]
On putting the value of a and b in the above equation
We get
Sum of the roots \[ = - \left( {\dfrac{{ - 8}}{5}} \right)\]
Or,
Sum of the roots =\[\dfrac{8}{5}\]
Similarly we know that the product of the roots of the quadratic equation\[a{x^2} + bx + c = 0\] is
\[\alpha \beta = \dfrac{c}{a}\]
On putting the value of a and c on the above equation
We get,
Product of the roots of the equation \[ = \dfrac{3}{5}\]
Hence the sum and the product of the zeros of the given quadratic equation is \[\dfrac{8}{5}\] and \[\dfrac{3}{5}\] respectively.

Note:The subtraction of zeros of the quadratic equation is \[\dfrac{{\sqrt {{b^2} - 4ac} }}{a}\]
And the division of the zeros of the quadratic equation is \[\dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{ - b - \sqrt {{b^2} - 4ac} }}\]
We can also calculate the sum and product of the zeros of the quadratic equation \[a{x^2} + bx + c = 0\]. By directly calculating the zeros by using the formula \[\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] and then finding their sum and products.
>The highest power of the variable in any polynomial is the degree of the equation and the number of roots of an equation is exactly the same as the degree of the equation.