Question

Find the quadratic equation whose sum and product of zeroes are $\dfrac{21}{8}$ and $\dfrac{5}{16}$ respectively.

Answer 1

Hint: Every quadratic equation is of the form ${{x}^{2}}-Sx+P=0$(where S is the sum of the zeroes of the quadratic equation and P is the product of the quadratic equation). If $\alpha $ and $\beta $are the zeroes of a quadratic equation, then the equation can be represented as ${{x}^{2}}-(\alpha +\beta )+\alpha \beta $.

Complete step by step answer:
We are given the sum of the zeroes is equal to $\dfrac{21}{8}$and the product of the zeroes is equal to $\dfrac{5}{16}$.
We are asked to find this particular quadratic equation.
We know that every quadratic equation is of the form,
 ${{x}^{2}}-Sx+P=0.................(1)$
(Where S is the sum of the zeroes of the quadratic equation and P is the product of the quadratic equation)
The sum of the zeroes, S = $\dfrac{21}{8}$
The product of the zeroes, P = $\dfrac{5}{16}$
Hence, equation (1) becomes,
${{x}^{2}}-\left( \dfrac{21}{8} \right)x+\dfrac{5}{16}=0$
Taking LCM,
$16{{x}^{2}}-(2\times 21)x+5=0$
$\therefore 16{{x}^{2}}-42x+5=0$
$\therefore $ The required quadratic equation is$16{{x}^{2}}-42x+5=0$.
We can find the zeroes of this quadratic equation using three different methods namely factorizing or splitting the middle term, completing the square method and the quadratic formula (Using discriminant). Here we are solving the quadratic equation using factorizing method.
Using splitting the middle term method, as the name suggests we will split the middle term into two, whose sum is equal to the middle term and their product is equal to the product of the first and third term.
In the quadratic equation $16{{x}^{2}}-42x+5=0$, the product is equal to $80{{x}^{2}}$ and the sum is equal to $-42x$.
Hence, we split the equation as,
$16{{x}^{2}}-40x-2x+5=0$
$8x(2x-5)-1(2x-5)=0$
$(8x-1)(2x-5)=0$
Hence, zeroes of x are $\dfrac{1}{8}$ and $\dfrac{5}{2}$.

Note:
We can also find the quadratic equation using a different method
Sum of the zeroes $=\dfrac{21}{8}$
Product of the zeroes $=\dfrac{5}{16}$
Let $\alpha $ and $\beta $are the zeros of the quadratic equation. Then,
$\alpha +\beta =\dfrac{21}{8}$
$\alpha \beta =\dfrac{5}{16}$
$\alpha =\dfrac{5}{16\beta }$
$\therefore \dfrac{5}{16\beta }+\beta =\dfrac{21}{8}$
Taking LCM,
$\dfrac{5+16{{\beta }^{2}}}{16\beta }=\dfrac{21}{8}$
$5+16{{\beta }^{2}}=\dfrac{21\times 16\beta }{8}$
$16{{\beta }^{2}}+5=42\beta $
Substituting $\beta $ with $x$, we get,
 $16{{x}^{2}}-42x+5=0$, which is indeed the required quadratic equation.