Question

Sum and product of the Zeroes of a quadratic polynomial.
p(x) =ax²+bx-4 are 1/4 and _1 respectively .then find the values of a and b.​

Answer 1

p(x) = ax² + bx - 4

$let \: \alpha \: and \: \beta \: be \: the \: roots \: of \: the \: \\</p><p>given \: equation$

According to given condition

$\alpha = \frac{1}{4} \: and \: \beta = - 1 \\ \\ \alpha \beta = \frac{c}{a} \\ (\frac{1}{4} )( - 1) = \frac{ - 4}{a} \\ \frac{1}{4} = \frac{4}{a} \\ a = 16 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ... \: (i)$

$\alpha + \beta = \frac{ - b}{a} \\ \frac{1}{4} - 1 = \frac{ - b}{16} \\ \frac{ - 3}{4} = \frac{ - b}{16} \\ \frac{3}{4} = \frac{b}{16} \\ 3 = \frac{b}{4} \\ b = 12$

Hence, values of a and b are 16 and 12 respectively.

Answer 2

Step-by-step explanation:

Given: p(x)= ax^2 + bx - 4..........(p)

$\alpha + \beta = \frac{1}{4}$

$\alpha \times \beta = 1$

To Find : values of a and b

solutions :-

formula of quadratic equation:-

p(x) = a[x^2 - (sum of zeros)x + product of zeros]

Now:-

p(x) = a[x^2 - (1/4)x + 1]

p(x) = a[x^2 - 1/4x + 1]..........(q)

Now:- by comparing (p) and (q)

a = 1

b = -1/4