Question

please solve for me do correct please its class 9th

The roots of ax²+bx+c=0 are differ by unity,then b²-a²=​

fast please


now

Answer 1

Answer:

b2-a2 = 1 + 4b

Step-by-step explanation:

x2 - ax + b = 0 Let α,β be the roots of quadratic. So α + β = a and α,β  = b

Answer 2

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

Given that,

~ The roots of ax²+bx+c=0 are differ by unity

Let assume that,

$\rm :\longmapsto\: \alpha, \beta \: be \: the \: roots \: of \: {ax}^{2} + bx + c = 0$

So,

$\bf\implies \: \alpha - \beta = 1$

We know that,

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

$\bf\implies \: \alpha + \beta = \dfrac{-b}{a}$

and

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

$\bf\implies \: \alpha \beta = \dfrac{c}{a}$

Now, given that,

$\bf\implies \: \alpha - \beta = 1$

On squaring both sides, we get

$\bf\implies \: (\alpha - \beta)^{2} = 1$

can be rewritten as

$\bf\implies \: (\alpha + \beta)^{2} - 4 \alpha \beta = 1$

$\rm :\longmapsto\: {\bigg[\dfrac{ - b}{a} \bigg]}^{2} - 4 \times \bigg[\dfrac{c}{a} \bigg] = 1$

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

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

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

$\bf\implies \: {b}^{2} - {a}^{2} = 4ac$

Additional Information :-

For Cubic Polynomial

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

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

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

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