Question

two zeros of a quadratic equation are 1/4 and -1 find the polynomial​

Answer 1

Answer:

= > 4 x² + 3 x - 1

Step-by-step explanation:

Given :

α = 1 / 4

b = - 1

Sum of zeroes = 1 / 4 + ( - 1 )

α + β = - 3 / 4

Product of zeroes = 1 / 4 × ( - 1 )

α β = - 1 / 4

Required polynomial = x² - ( α + β ) x + α β

  = x² - ( - 3 / 4 ) x + ( - 1 / 4 )

 = x² + 3 / 4 x - 1 / 4

Now dividing by 4 in whole polynomial

= > 4 x² + 3 x - 1

Answer 2

$\large{\underline{\underline{\mathfrak{\sf{\red{Answer-}}}}}}$

$\large{\underline{\boxed{\sf{\green{x^2+\dfrac{3}{4}x-\dfrac{1}{4}}}}}}$

$\large{\underline{\underline{\mathfrak{\sf{\red{Explanation-}}}}}}$

Given zeroes :

• $\sf{\dfrac{1}{4}}$
• -1

To find :

• Quadratic polynomial

Formula used :

• x² - Sx + p

$\begin{lgathered}\bold{Where} \begin{cases}\sf{\blue{S}\:refers\:to\:sum\:of\:zeroes} \\ \sf{\pink{P}\:refers\:to\:product\:of\:zeroes}\end{cases}\end{lgathered}$

Solution :

★ Sum of given zeroes = $\sf{\dfrac{1}{4}+(-1)}$

$\implies$ Sum of zeroes = $\sf{\dfrac{1}{4}-1}$

$\implies$ Sum of zeroes = $\sf{\dfrac{1-4}{4}}$

$\implies$ Sum of zeroes = $\sf{-\dfrac{3}{4}}$

★ Product of zeroes = $\sf{\dfrac{1}{4}×(-1)}$

$\implies$ Product of zeroes = $\sf{-\dfrac{1}{4}}$

$\rule{200}3$

Now, put the value of sum and product of zeroes in x² - Sx + P.

$\implies$ x² - $\sf{(-\dfrac{3}{4})x}$ + $\sf{(-\dfrac{1}{4})}$

We get,

$\large{\underline{\boxed{\sf{\green{x^2+\dfrac{3}{4}x-\dfrac{1}{4}}}}}}$