Question

find the value of k such that sum of zeroes is equal to the product of zeroes of the following quadratic polynomial (k+1)x²+(2k+1)x-9​

Answer 1

$\huge\sf\pink{Answer}$

☞ Value of k = 4

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$\huge\sf\blue{Given}$

✭ Sum of zeroes is equal to the product of the zeroes of the quadratic polynomial (k+1) x² + (2k+1)x - 9.

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$\huge\sf\gray{To \:Find}$

☆ Value of k?

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$\huge\sf\purple{Steps}$

$\leadsto { \sf{(k + 1) {x}^{2} + (2k + 1)x - 9 = 0 }}$

$\sf\underline{We \ know \ that}$

$\dashrightarrow { \sf{sum \: \: of \: \: roots = - \dfrac{coffieciant \: \: of \: \: x}{coffieciant \: \: of \: \: {x}^{2} } }}$

$\dashrightarrow{ \sf{sum \: \: of \: \: roots = - \dfrac{(2k + 1)}{(k + 1)} }}$

$\dashrightarrow{ \sf{sum \: \: of \: \: roots = - \bigg\lgroup \: \dfrac{2k + 1}{k + 1}\bigg\rgroup }}$

$\dashrightarrow { \sf{ \: Product \: \: of \: \: roots = \dfrac{constant \: \: term}{coffieciant \: \: of \: \: {x}^{2} } }}$

$\dashrightarrow{ \sf{Product \: \: of \: \: roots = \dfrac{( - 9)}{(k + 1)} }}$

$\dashrightarrow{ \sf{Product \: \: of \: \: roots = - \bigg\lgroup \dfrac{9}{k + 1} }}\bigg\rgroup$

$\sf\underline{As \ per \ the \ Question}$

$\twoheadrightarrow{ \sf{Sum \: of \: Roots = Product \: of \: Roots}}$

$\twoheadrightarrow{ \sf{- \bigg\lgroup \: \dfrac{2k + 1}{k + 1}\bigg\rgroup = - \bigg\lgroup\dfrac{9}{k + 1}\bigg\rgroup}}$

$\twoheadrightarrow{ \sf{(2k + 1)(k + 1) = (k + 1)9}}$

$\twoheadrightarrow{ \sf{(2k + 1)(k + 1) - (k + 1)9 = 0}}$

$\twoheadrightarrow{ \sf{(2k + 1 - 9)(k + 1) = 0}}$

$\twoheadrightarrow{ \sf{(2k - 8)(k + 1) = 0}}$

$\twoheadrightarrow{ \sf{(k - 4)(k + 1) = 0}}$

$\twoheadrightarrow \color{lime}{ \sf{k = 4 \: , \: - 1}}$

But whena k = -1 leading coffieciant of quadratic equation will be zero , but we know that coffieciant of x² can't be zero.

$\sf\color{aqua}{\therefore K = 4}$

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Answer 2

Step-by-step explanation:

ANSWER:-

We know that:-

$\alpha + \beta = \frac{ - b}{a}$

and

$\alpha \beta = \frac{c}{a}$

So it is given that sum of zeroes is equal to product of the same

ie

$\alpha + \beta = \alpha \beta$

ie

$\frac{ - b}{a} = \frac{c}{a}$

$- b = c$

We know that :-

$b = 2k + 1$

$c = - 9$

Placing the following

$- (2k + 1) = - 9$

$2k + 1 = 9$

$2k = 8$

$k = 4$

k = 4 is the answer....

THINGS TO NOTE :-

$\alpha + \beta = \frac{ - b}{a}$

$\alpha \beta = \frac{c}{a}$

$\alpha \beta \gamma = \frac{-d}{a}$

$\alpha + \beta + \gamma = \frac{ - b}{a}$

$\alpha \beta + \beta \gamma + \gamma \alpha = \frac{ c}{a}$