Question

The zeroes of the polynomial r(t) = -12t² + (k-3)t +48 are negative of each other. Then
k is

Answer 1

Answer:

Step-by-step explanation:

Let α and β be the zeroes of the polynomial.

Acc. To Qsn,

α = x, then  

β = -x

We Know That,

Sum of zeroes = $\frac{-b}{a}$

α+β= $\frac{k-3}{12}$=0

k-3 = 0

k=-3

There can be other zeroes top (may be) but u can find others by using other formula hope it helps.

Answer 2

SOLUTION

GIVEN

The zeroes of the polynomial

r(t) = - 12t² + ( k - 3 )t + 48

are negative of each other.

TO DETERMINE

The value of k

EVALUATION

Here the given Quadratic polynomial is

r(t) = - 12t² + ( k - 3 )t + 48

Comparing with the general quadratic polynomial

r(t) = at² + bt + c We get

a = - 12 , b = k - 3 , c = 48

Now it is given that the zeroes are negative of each other.

So sum of zeroes = 0

⇒ - b/a = 0

⇒ b = 0

⇒ k - 3 = 0

⇒ k = 3

FINAL ANSWER

Hence the required value of k = 3

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