Question

If alfa,bita are zeroes of the polynomial p(x)= 3x^2-12x+k such that alfa-bita=2 then find the value of k.​

Answer 1

Answer:

✩ Given Polynomial : 3x² – 12x + k

where, a = 3 ,⠀ b = – 12 ,⠀ c = k

⠀

Here, $\alpha$ and $\beta$ are zeroes of p(x).

$\rule{100}{0.8}$

☢ $\underline{\textsf{Let's Consider a Formula :}}$

$:\implies\sf \bigg(\alpha-\beta\bigg)^2= \bigg(\alpha + \beta \bigg)^2 - 4\alpha\beta\\\\\\:\implies\sf(2)^2 =(Sum\:of\:Zeroes)^2-4Product\:of\: Zeroes\\\\\\:\implies\sf 4 =\bigg( \dfrac{-\:b}{a} \bigg)^2 - \bigg(4 \times \dfrac{c}{a} \bigg)\\\\\\:\implies\sf 4 =\bigg( \dfrac{-\:( - 12)}{3} \bigg)^2 - \bigg(4 \times \dfrac{k}{3}\bigg)\\\\\\:\implies\sf 4 =(4)^2 -\dfrac{4k}{3}\\\\\\:\implies\sf 4 =16 - \dfrac{4k}{3}\\\\\\:\implies\sf 4 \times 3 = (16 \times 3) - \bigg(\dfrac{4k}{3} \times 3\bigg)\\\\\\:\implies\sf 12 = 48 - 4k\\\\\\:\implies\sf 4k = 48 - 12\\\\\\:\implies\sf 4k = 36\\\\\\:\implies\sf k = \dfrac{36}{4}\\\\\\:\implies\underline{\boxed{\sf k = 9}}$

⠀

$\therefore\:\underline{\textsf{Hence, required value of k is \textbf{9}}}.$

Answer 2

Answer:

k = 9

Step-by-step explanation:

There is an Easier way to solve this problem...

we know that,

α + β = -b/a

α + β = - ( -12 ) / 3

α + β = 4 ---- eq 1

α - β = 2 ---- eq 2 (given)

add eq 1 and eq 2

α + β = 4

α - β = 2

----------------

2 α = 6

α = 3

using eq 1 and eq 2 you can find the value of β

and you will get

β = 1

now

αβ = c/a

(3)(1) = k / 3

3 = k / 3

k = (3)(3)

k = 9

we DID IT...