Question

If one zero of the polynomial x^-3kx+4k is twice the other, find the value of k.

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Answer 1

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

If one zero of the polynomial x² - 3kx + 4k is twice the other.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The value of k.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

Let the one zero be r

Let the other zero be 2r

As the given quadratic polynomial as we compared with ax² + bx + c

• a = 1
• b = -3k
• c = 4k

So;

$\dag\underline{\underline{\bf{Sum\:of\:zeroes\::}}}}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x}{Coefficient\:of\:(x)^{2} } }\\\\\\\mapsto\sf{2r+r=\dfrac{-(-3k)}{1} }\\\\\\\mapsto\sf{\cancel{3}r=\cancel{3}k}\\\\\\\mapsto\sf{r=k.........................(1)}$

$\dag\underline{\underline{\bf{Product\:of\:zeroes\::}}}}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term}{Coefficient\:of\:(x)^{2} } }\\\\\\\mapsto\sf{2r\times r=\dfrac{4k}{1} }\\\\\\\mapsto\sf{\cancel{2}r^{2} =\cancel{4}k}\\\\\\\mapsto\sf{r^{2} =2k}\\\\\\\mapsto\sf{(k)^{2} =2k\:\:\:\:[from(1)]}\\\\\\\mapsto\sf{k(k-2)=0}\\\\\\\mapsto\sf{k=0\:\:Or\:\:k-2=0}\\\\\\\mapsto\sf{\pink{k=0\:\:\:Or\:\:\:k=2}}$

Thus;

The value of k is 0 and 2 .

Answer 2

Solution:

◕ f(x) = x² - 3kx +4k

let first zero of polynomial is q and ans second is 2q

✧compare it with ax²+bx+c

→a= 1

→b= -3k

→c= 4k

☞Sum of zeros (α+β)=-coefficient ofx/coefficient ofx²

→ α+β =- (-3k/1)

→q+2q = -(-3k)

→3q = -(-3k)

→q= k.....(1)

☞Product of its zeros = constant term/coefficient ofx²

→αβ = 4k/1

→(q)(2q) = 4k

→2q² = 4k

→2(k)²=4k

→2k²-4k=0

→2k(k-2)=0

→k(k-2)=0

→k = 0. , k = 2

Hence the value of k=0,2.

___________________________

✿If α and β are the zeros of the quadratic polynomial ax²+bx+c then,

◕α+β = -b/a

◕αβ = c/a