Question

Find the quadratic polynomial whose product and sum of zeros are (-13)/5 and 3/5, respectively.
Here, 3/5 is POSITIVE.
Please could you write it fast? Please.

Answer 1

GiveN :

• Sum of zeros = 3/5
• Product of zeros = -13/5

To Find :

• Quadratic Equation

Solution :

let the zeroes be α and β ,

⇒ α + β = 3/5

⇒α × β = -13/5

We know that the general form of quadratic equation is :

➠ x² - (sum of zeros)x + Product

⇒x² - (α + β)x + α*β

⇒x² - (3/5)x + (-13/5)

⇒x² - 3/5x - 13/5

⇒(5x² - 5x - 13)/5

Now, we can remove 5 from denominator

$\therefore$ Equation is 5x² - 5x - 13

_______________________________

General Equation : ax² + bx + c

⇒ Sum of zeros = -b/a

⇒ Product of zeros = c/a

• Quadratic Formula :

x = -b± √(b² - 4ac) / 2a

Answer 2

$\blue{\underline{\underline{\tt{Given:}}}}$

The product of zeroes = $\red{\sf{\frac{-13}{5} }}$

The sum of the zeroes = $\red{\sf{\frac{3}{5} }}$

$\blue{\underline{\underline{\tt{To\:be\:found:}}}}$

The quadratic polynomial whose product of zeroes = $\red{\sf{\frac{-13}{5} }}$ and sum of the zeroes = $\red{\sf{\frac{3}{5} }}$.

$\yellow{\star} \pink{\underline{\sf{Here\:\: is\:\: the\:\: formula\:\: for\: \:a \:\:quadratic\:\: equation-}}}$

$\boxed{\bf x^{2}- (sum \:\:of \:\:zeroes)x+(product\:\:of\:\:zeroes)}$

$=x^{2} - \big( \frac{3}{5} \big) x+ \big( \frac{-13}{5} \big)$

$=x^{2} - \frac{3x}{5}+ \frac{-13}{5}$

$= \frac{5x^{2}-3x-13}{5}$

$= \bf 5x^{2}-3x-13$

we can consider $\red{\boxed{\bf{5x^{2}-3x-13} }}$ as required quadratic polynomial because it will also satisfy the given conditions.

$\violet{\underline{\sf{Check:}}}$

sum of zeroes =  $\bf \frac{- \: coefficient\:of\:x}{coefficient\:of\:x^{2}}=\frac{-b}{a} = \frac{-(-3)}{5} =\frac{3}{5}$

Product of zeroes = $\bf \frac{constant\:\:term}{coefficient\:of\:x^{2}}=\frac{c}{a} = \frac{(-13)}{5} =\frac{-13}{5}$