Question

if the sum of zeroes of the quadratic polynomial f(t)=kt square+2t+3k is equal to their product. find the value of k

Answer 1

Heya!

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f (t) = kt^2 + 2t + 3k [ Correct ques ]

Here,

a = k
b = 2
c = 3k

Let zeroes be @ (alpha) and ß (beta)

Sum of zeroes = -b/a = - 2 / k

Product of zeroes = c/a = 3k / k

Now,

Given,

@ + ß = @ × ß

-2 / k = 3k

k = -2/3


Value of k = -2/3

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Hope it helps...!!!

Answer 2

Heya !!!!

here is the answer,

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The given quadratic polynomial is kt^2+2t+3k.

Comparing the given quadratic expression with standard form of expression ax^2+bx+c, we have;

a = k, 
b = 2 and 
c = 3k

So, sum of zeroes of quadratic polynomial = 
$\frac{ - b}{a} = \frac{ - 2}{k}$


And product of zeroes of quadratic polynomial = 
$\frac{c}{a} = \frac{3k}{k} = 3$


And according to the question we have;

Sum of zeroes of quadratic polynomial = product of zeroes of quadratic polynomial
$= > \frac{ - 2}{k} = 3 \\ \\ = > k = \frac{ - 2}{3}$

Therefore the value of k is -2/3.

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Hope it helps u !!!!

Cheers :))

# Nikky