Question

solve by using quadratic formula

a) z² - 32z - 105 = 0
b) 40 + 3x - x² = 0
c) 6 - x - x² = 0
d) 7x² + 49x + 84 = 0
e) m² + 17mn -84n² = 0​

Answer 1

therefore this is ur answer

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Answer 2

Answer:

Solution :

We have ,

\sf\:f(x)={x}^{2}-p(x+1)-cf(x)=x

2

−p(x+1)−c

\sf \implies \: f(x) = x{}^{2} - px - p- c⟹f(x)=x

2

−px−p−c

\sf \implies \: f(x) = {x}^{2} - px - (p + c)⟹f(x)=x

2

−px−(p+c)

We know that ,

\sf \alpha + \beta = \dfrac{ - cofficient \: of \: x}{cofficient \: of \: x {}^{2} }α+β=

cofficientofx

2

−cofficientofx

\sf \implies \alpha + \beta = \dfrac{ - ( - p)}{ 1} = p...(1)⟹α+β=

1

−(−p)

=p...(1)

\sf \: and \: \alpha \beta = \dfrac{constant}{cofficient \: of \: x {}^{2} }andαβ=

cofficientofx

2

constant

\sf \implies \alpha \beta = \dfrac{ - (p + c)}{1} = - (p + c)..(2)⟹αβ=

1

−(p+c)

=−(p+c)..(2)

We have to find the value of

\sf(\alpha+1)(\beta+1)(α+1)(β+1)

\sf = \alpha \beta + \alpha + \beta + 1=αβ+α+β+1

\sf = \alpha \beta + ( \alpha + \beta ) + 1=αβ+(α+β)+1

Now put the values of equation (1) and (2)

\sf = - (p + c) + p + 1=−(p+c)+p+1

\sf = - p - c + p + 1=−p−c+p+1

\sf = 1 - c=1−c