Question

If sum of zeroes of the quadratic polynomial kx2 – 2x - 3k is equal to their product, find value of k.
​

Answer 1

Given:

Sum of zeroes of the quadratic polynomial

kx² - 2x - 3k is equal to their product.

To Find:

Value of K

Solution:

General quadratic equation is given by

ax² + bx + c = 0

Comparing it to above equation

a = k

b = -2

c = -3k

As we know

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

Product of zeros = c/a = -3k/k = -3

ATQ

Sum of zeros = Product of zeros

2/k = -3

k = -2/3

Hence, value of k is -2/3.

Answer 2

$\large{\boxed{\boxed{\sf Let's \: Understand \: Question \: F1^{st}}}}$

Here, we have given a quadratic polynomial and said that it's sum of zeroes and product of zeroes is equal and have to find the value of k.

$\large{\boxed{\boxed{\sf How \: To \: Do \: It?}}}$

Here, f1st we simply find the sum of zeroes and product of zeroes then going according to question we simply place them equal to one another and by solving we will get our required answer.

Let's Do It ⚡

$\huge{\underline{\boxed{\sf AnSwer}}}$

______________________________

Given:-

• Quadratic eq. = kx² - 2x -3k = 0
• Sum of zeroes = Product of zeroes

Find:-

• Value of k

Solution:-

F1st compare the given eq. with ax² + bx + c = 0

We, get

a = k

b = -2

c = -3k

Now, using

$\huge{\underline{\boxed{\sf Sum \: of \: zeroes = \dfrac{ - b}{a} = \dfrac{- (coefficient \: of \: x)}{coefficient \: of \: {x}^{2} }}}}$

$\sf where \small{\begin{cases} \sf b = - 2 \\ \sf a = k\end{cases}}$

$\red\bigstar$ Substituting these values:

$\sf :\implies Sum \: of \: zeroes = \dfrac{ - b}{a}$

$\sf :\implies Sum \: of \: zeroes = \dfrac{ - ( - 2)}{k}$

$\sf :\implies Sum \: of \: zeroes = \dfrac{ 2}{k} \quad \big\lgroup \because (- )(-) = + \big\rgroup$

$\sf :\implies Sum \: of \: zeroes = \dfrac{ 2}{k}$

Now, Using

$\huge{\underline{\boxed{\sf Product \: of \: zeroes = \dfrac{c}{a} = \dfrac{constant \: term}{coefficient \: of \: {x}^{2} }}}}$

$\sf where \small{\begin{cases} \sf c = - 3k \\ \sf a = k\end{cases}}$

$\pink\bigstar$ Substituting these values:

$\implies\sf Product \: of \: zeroes = \dfrac{c}{a}$

$\implies\sf Product \: of \: zeroes = \dfrac{ - 3k}{k}$

$\implies\sf Product \: of \: zeroes = - 3$

$\blue\bigstar\pink{\textsf{According To Question}}$

$\sf\dashrightarrow Sum \: of \: zeroes = Product \: of \: zeroes$

$\sf where \small{\begin{cases} \sf \sf\dashrightarrow Sum \: of \: zeroes = \dfrac{2}{k} \\ \sf Product \: of \: zeroes = - 3\end{cases}}$

$\green\bigstar$ Substituting these values:

$\sf\dashrightarrow Sum \: of \: zeroes = Product \: of \: zeroes$

$\sf\dashrightarrow \dfrac{2}{k} = - 3$

$\bigstar$ Cross-multiplication

$\sf\dashrightarrow - 3k = 2$

$\sf\dashrightarrow k = \dfrac{2}{ - 3}$

$\sf\dashrightarrow k = - \dfrac{2}{3}$

$\underline{ \boxed{\sf \therefore Value\:of\:k\:is\: - \dfrac{2}{3}}}$