Question

If the zeros of x
2 + ax + b are two consecutive integers, then prove that a
2 = 1 + 4b.

Answer 1

$\large\underline{\sf{Given- }}$

The zeroes of the polynomial x² + ax + b are two consecutive integers.

$\large\underline{\sf{To\:prove - }}$

$\rm :\longmapsto\: {a}^{2} = 1 + 4b$

$\large\underline{\sf{Solution-}}$

Given that

The zeroes of the polynomial x² + ax + b are two consecutive integers.

Let assume that zeroes of the polynomial be y and y + 1.

We know that,

In a quadratic polynomial,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

So,

Using this result we have

$\rm :\longmapsto\:y + y + 1 = - \: \dfrac{a}{1}$

$\rm :\longmapsto\:2y + 1 = a$

$\rm :\longmapsto\:y = \dfrac{a - 1}{2} - - - (1)$

Also, we know that,

In quadratic polynomial,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\rm :\longmapsto\:y( y + 1) = \: \dfrac{b}{1}$

On substituting the value of y from equation (1), we get

$\rm :\longmapsto\:\bigg(\dfrac{a - 1}{2} \bigg)\bigg(\dfrac{a - 1}{2} + 1 \bigg) = b$

$\rm :\longmapsto\:\bigg(\dfrac{a - 1}{2} \bigg)\bigg(\dfrac{a - 1 + 2}{2} \bigg) = b$

$\rm :\longmapsto\:\bigg(\dfrac{a - 1}{2} \bigg)\bigg(\dfrac{a + 1}{2} \bigg) = b$

$\rm :\longmapsto\: {a}^{2} - 1 = 4b$

$\rm :\longmapsto\: {a}^{2} = 4b + 1$

Hence, Proved.

Additional Information :-

$\rm :\longmapsto\: \alpha , \beta , \gamma \: are \: zeroes \: of \: a {x}^{3} + b {x}^{2} + cx + d, \: then$

$\red{\boxed{ \bf{ \: \alpha + \beta + \gamma = - \: \dfrac{b}{a} }}}$

$\red{\boxed{ \bf{ \: \alpha \beta + \beta \gamma + \gamma \alpha = \: \dfrac{c}{a} }}}$

$\red{\boxed{ \bf{ \: \alpha \beta \gamma = - \: \dfrac{d}{a} }}}$