Question

heya!!!❤️❤️
pls solve this question ✌️✌️​

Answer 1

We have been given that the zeros of the equation $x^{2} - kx + 6$ are in the ratio 2:3.

We are required to find the value of k.

Let the zeroes be 2y and 3y.

Sum of roots of a quadratic equation is  $=\dfrac{-b}{a}$

Here,

$2y + 3y = \dfrac{-(-k)}{1} \\\\5y = k$

And the product of the roots is $=\dfrac{c}{a}$

Here,

$(2y)(3y)=\dfrac{6}{1} \\\\6y^{2}= 6\\\\y^{2} = 1\\\\y=\pm 1$

Since, both 2y and 3y are roots of the equation. Both of them should give us zero when substituted into the equation.

$(3y)^2 - k(3y) + 6 = 0 \ \ ...(1)\\\\(2y)^2 - k(2y) + 6 = 0 \ \ ...(2)\\\\From \ (1) \ and \ (2) \\\\(3y)^2 - k(3y) + 6 = (2y)^2 - k(2y) + 6 = 0\\\\9y^{2} - 3ky = 4y^{2} - 2ky\\\\5y^{2}=ky\\\\5 = ky$

We have $y=\pm1$

So, $k = 5(\pm1)$

Therefore, the value of k would be $=\pm5$ !

Answer 2

AnswEr :

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

If the zeroes of x² - kx + 6 are in the ratio 3:2.

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

The value of k

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

We have, x² - kx + 6

Let the α & β be the zeroes of given equation.

$\leadsto\sf{\alpha \:and\: \beta \:are\:in\:the\:ratio\:=3:2}$

$\bf{\red{\underline{\underline{\tt{A.T.Q\::}}}}}}$

$\mapsto\sf{\alpha :\beta =3:2}\\\\\\\mapsto\sf{\dfrac{\alpha }{\beta } =\dfrac{3}{2} }\\\\\\\mapsto\sf{2\alpha =3\beta }\\\\\\\mapsto\sf{\orange{\alpha =\dfrac{3\beta }{2} ............(1)}}$

$\bf{\purple{\underline{\underline{\bf{\blacksquare{Sum\:of\:the\:zeroes\::}}}}}}$

$\leadsto\tt{\alpha +\beta =\dfrac{-b}{a} }\\\\\\\leadsto\tt{\alpha +\beta =\dfrac{-k}{1} }\\\\\\\leadsto\tt{\alpha +\beta =-k}\\\\\\\leadsto\tt{\dfrac{3\beta }{2} +\beta =-k\:\:\:\:\: \big[from(1)\big]}}\\\\\\\leadsto\tt{\dfrac{3\beta +2\beta }{2} =-k}\\\\\\\leadsto\tt{\dfrac{5\beta }{2} =-k}\\\\\\\leadsto\tt{\red{5\beta =-2k................(2)}}$

$\bf{\purple{\underline{\underline{\bf{\blacksquare{Product\:of\:zeroes\::}}}}}}$

$\leadsto\tt{\alpha \times \beta =\dfrac{c}{a} }\\\\\\\leadsto\tt{\alpha \times \beta =\dfrac{6}{1} }\\\\\\\leadsto\ttt{\frac{3\beta }{2} \times \beta =6\:\:\:\:\:\:\:\bg[from(1)\big]}\\\\\\\leadsto\tt{\dfrac{3\beta^{2} }{2} =6}\\\\\\\leadsto\tt{3\beta ^{2} =12}\\\\\\\leadsto\tt{\beta ^{2} =\cancel{\dfrac{12}{3} }}\\\\\\\leadsto\tt{\beta ^{2} =4}\\\\\\\leadsto\tt{\beta =\sqrt{4} }\\\\\\\leadsto\tt{\orange{\beta =\pm2}}$

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

$\longmapsto\tt{5\beta =-2k}\\\\\\\longmapsto\tt{5(\pm2)=-2k}\\\\\\\longmapsto\tt{\pm10=-2k}\\\\\\\longmapsto\tt{k=\mp\cancel{\dfrac{10}{2} }}\\\\\\\longmapsto\tt{\orange{k=\mp5}}$

∴ The value of k is $\mp$5.