Question

if alpha and beta are the zeros of the quadratic polynomial X square + 5 x + b and alpha minus beta equals to 1 then the value of B is​

Answer 1

Solution :

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

If α and β are the zeroes of the quadratic polynomial x² + 5x + b and α - β = 1.

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

The value of b.

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

We compared given quadratic equation with ax² + bx + c = 0

• a = 1
• b = 5
• c = b

So;

$\blacksquare\bf{\large{\green{\underline{\underline{\tt{Sum\:of\:the\:zeroes\::}}}}}}$

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} }\\\\\mapsto\sf{\alpha +\beta=\dfrac{-5}{1} }\\\\\mapsto\sf{\red{\alpha +\beta =-5}}$

$\blacksquare\bf{\large{\green{\underline{\underline{\tt{Product\:of\:the\:zeroes\::}}}}}}$

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} }\\\\\mapsto\sf{\alpha \times \beta=\dfrac{b}{1} }\\\\\mapsto\sf{\red{\alpha \times \beta =b}}$

We have already given;

$\mapsto\sf{\alpha -\beta =1}\\\\\mapsto\bf{\alpha =1+\beta .....................(1)}$

Putting the value of α in sum of the zeroes,we get;

$\mapsto\sf{1+\beta +\beta =-5}\\\\\mapsto\sf{1+2\beta =-5}\\\\\mapsto\sf{2\beta =-5-1}\\\\\mapsto\sf{2\beta =-6}\\\\\mapsto\sf{\beta =\cancel{\dfrac{-6}{2} }}\\\\\mapsto\sf{\red{\beta =-3}}$

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

$\mapsto\sf{\alpha =1+(-3)}\\\\\mapsto\sf{\alpha =1-3}\\\\\mapsto\sf{\red{\alpha =-2}}$

Putting the value of α & β in product of the zeroes,we get;

$\mapsto\sf{b=(-2)\times (-3)}\\\\\mapsto\sf{\red{b=6}}$

Thus;

The value of b is 6 .

Answer 2

$\tt{\huge{\orange {----------}}}M. V$

QUESTION :

if alpha and beta are the zeros of the quadratic polynomial X square + 5 x + b and alpha minus beta equals to 1 then the value of B is......

SOLUTION :

Given Polynomial :

X^2 + 5X + b

We know that sum of Zeroes is equal to { - b / a }

=> Alpha + Beta = - 5...... [ 1 ]

We know that the product of Zeroes is equal to { c / a }

=> Alpha × Beta = b....... [ 2 ]

Squareing,

( Alpha + Beta ) ^ 2 = 25

=> ( Alpha - Beta ) ^ 2 + 4 Alpha × Beta = 25

=> Substituting The required Values,

=> ( 1 ) ^ 2 + 4 b = 25

=> 4 b = 24

=> b = 4

Hence, the value of B is 4.

The required Quadratic Equation thus becomes :

X^2 + 5 X + 4 = 0

ADDITIONAL INFORMATION :

Factorising the resulting quadratic equation,

X ^ 2 + 5 X + 4

=> X ^ 2 + 4 X + X + 4

=> X ( X + 4 ) + ( X + 4 ) = 0

=> ( X + 1 ) ( X + 4 ) = 0

Roots of the equation are :

X = -1 and -4.

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