Question

if the sum of zeores of 3x²-k+6 is 3 find the k​

Answer 1

Answer:

Step-by-step explanation:

Hey mate here is your answer:

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→Compare 3x² - k x + 6

→with ax² + b x + c,

→a = 3 ,

→b = - k ,

→c = 6 ,

According to the problem given ,

→Sum of the zeroes = 3 ---( 1 ) ,

→- b / a = 3 ,

→- ( - k ) / 3 = 3 ,

→k / 3 = 3 ,

→k = 9 .

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Hope it helps you.

Answer 2

S O L U T I O N :

We have quadratic polynomial p(x) = 3x² - kx + 6 is 3 & zero of the polynomial p(x) = 0.

Let the zeroes be α & β.

As we know that given quadratic equation compared with ax² + bx + c;

• a = 3
• b = -k
• c = 6

A/q

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha + \beta = \dfrac{-b}{a} =\bigg\lgroup \dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}} \bigg\rgroup}$

$\mapsto\tt{\alpha + \beta =3}$

$\mapsto\tt{\alpha = 3-\beta ..............(1)}$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha \times \beta = \dfrac{c}{a} =\bigg\lgroup \dfrac{Constant\:term}{Coefficient\:of\:x^{2}} \bigg\rgroup}$

$\mapsto\tt{(3-\beta )\times \beta = \dfrac{6}{3}\:\:\:[from(1)]}$

$\mapsto\tt{3\beta -\beta^{2} = \cancel{\dfrac{6}{3}}}$

$\mapsto\tt{3\beta- \beta^{2} = 2}$

$\mapsto\tt{ \beta^{2} - 3\beta +2 = 0 }$

$\mapsto\tt{ \beta^{2} - 2\beta-\beta +2 = 0 }$

$\mapsto\tt{\beta(\beta -2) -1(\beta -2) = 0}$

$\mapsto\tt{(\beta -2)(\beta -1) = 0}$

$\mapsto\tt{\beta-2 = 0\:\:\:Or\:\:\:\beta - 1 = 0}$

$\mapsto\tt{\beta=2\:\:\:Or\:\:\:\beta =1}$

∴ Putting the value of β (2) in equation (1),we get;

$\mapsto\tt{\alpha = 3 - 2}$

$\mapsto\bf{\alpha = 1}$

Now,

$\longrightarrow\sf{\alpha +\beta =\dfrac{-b}{a} }$

$\longrightarrow\sf{1+ 2= \dfrac{k}{3} }$

$\longrightarrow\sf{3=\dfrac{k}{3} }$

$\longrightarrow\sf{k=3\times 3}$

$\longrightarrow\bf{k=9}$

Thus,

The value of k will be 9 .