Question

if one of zeroes of the polynomial 2x2-5x-(2k+1) is twice the other, find the zeroes of the polynomial and the value ' k'​

Answer 1

Solution :

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

If one of zeroes of the polynomial 2x² - 5x - (2k+1) is twice the other.

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

The zeroes of the polynomial and the value of k.

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

Let the one zeroes be $\sf{\alpha }$

Let the other zeroes be $\sf{2\alpha }$

As the given quadratic polynomial as we compared with ax²+bx+c=0

• a = 2
• b = -5
• c = -(2k+1)

We know that, sum of the zeroes :

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} }\\\\\\\mapsto\sf{\alpha +2\alpha =\dfrac{-(-5)}{2} }\\\\\\\mapsto\sf{3\alpha =\dfrac{5}{2} }\\\\\\\mapsto\sf{6\alpha =5}\\\\\\\mapsto\sf{\pink{\alpha =\dfrac{5}{6} }}$

Thus;

$\sf{The\:one\:zeroes\:=\alpha =\pink{\dfrac{5}{6}}}}\\\\\ \sf{The\:other\:zeroes\:=2\alpha =\frac{5*2}{6}=\cancel{\dfrac{10}{6}} =\pink{\dfrac{5}{3}}}}$

We know that, product of the zeroes :

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} }\\\\\\\mapsto\sf{\alpha \times 2\alpha =\dfrac{-(2k+1)}{2} }\\\\\\\mapsto\sf{2\alpha ^{2} =\dfrac{(-2k+1)}{2} }\\\\\\\mapsto\sf{4\alpha ^{2} =(-2k+1)}\\\\\\\mapsto\sf{4*\bigg(\dfrac{5}{6} \bigg)^{2} =(-2k+1)\:\:\:\:\:[\alpha =5/6]}\\\\\\\mapsto\sf{\cancel{4}*\dfrac{25}{\cancel{36}} =(-2k+1)}\\\\\\\mapsto\sf{\dfrac{25}{9} =-2k-1}\\\\\\\mapsto\sf{-2k=1+\dfrac{25}{9} }\\\\\\\mapsto\sf{-2k=\dfrac{9+25}{9} }$

$\mapsto\sf{-2k=\dfrac{34}{9} }\\\\\\\mapsto\sf{-18k=34}\\\\\\\mapsto\sf{k=-\cancel{\dfrac{34}{18} }}\\\\\\\mapsto\sf{\red{k=-\dfrac{17}{9} }}$

Thus;

The value of k is -17/9.

Answer 2

Answer:

Step-by-step explanation:

Let, p(x)=2x²-5x-(2k+1)--------(1)

Also let, α be one root of (1).

Then the other root will be 2α.

α+2α=-(-5)/2

or, 3α=5/2

or, α=5/6

Again, α×2α=-(2k+1)/2

or, 2α²=-(2k+1)/2

or, -(2k+1)/2=2×(5/6)²

or, -(2k+1)=4×25/36

or, -2k-1=25/9

or, -2k=25/9+1

or, -2k=34/9

or,  k=-34/9×1/2

or, k=-17/9 Ans.

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