Question

if a and a+1 are the roots of the equation X2 - 7x+a=0, then a equal​

Answer 1

Answer:

Question

if a and a+1 are the roots of the equation X2 - 7x+a=0, then a equal

To Find

value of a

solution

⊙ The general form of quadratic equation is

${\sf{\underline{ax²+by +c=0}}}$

╬ if alpha (α) & Beta (β) are the zeros of the above quadratic equation then

$\sf{ \to \color{green}\: Sum \: Of \: Zeros}$

$\sf{\to (β) +(α)= \frac{-b}{a}}$

$\sf{\to \color{magenta}{Product \: of \: Zeros}}$

$\sf{(β) ×(α) = \frac{c}{a}}$

Now,

Given quadratic equation is

$\sf{x²-7x+a= 0}$

Here Alpha & Beta are α & α+1.

$\sf{ \to \color{purple}\: sum \: of \: zeros}$

$\sf{α+α+1= \frac{-(-7)}{1}} \\ \\ \sf{2 α+ 1 = 7} \\ \\ \sf{α = 3}$

$\sf{ :\implies \: so \: the \: value \: of \: (α+1 ) \: is \: 4}$

$\sf{\to \color{orange}{Product \: of \: Zeros}}$

$\sf{:\implies α×(α+1)=\frac{a}{1}} \\ \\\sf{3×4=a} \\ \\ \sf{:\implies a= 12}$

Therefore the value of a is 12.

____________________________

Answer 2

$\sf{ \to \color{lime}\:Question }$

if a and a+1 are the roots of the equation X2 - 7x+a=0, then a equal

$\sf{ \to \color{lime}\:to \: find }$

value of a

$\sf{ \to \color{cyan}\:Solution }$

⊙ The general form of quadratic equation is

$</h2><h2>{\sf{\underline{ax²+by +c=0}}}$

ax²+by+c=0

╬ if alpha (α) & Beta (β) are the zeros of the above quadratic equation then

$\sf{ \to \color{green}\: Sum \: Of \: Zeros} \\ →SumOfZeros$

$\sf{\to (β) +(α)= \frac{-b}{a}}→(β)+(α)= </h2><h2>a</h2><h2>−b$

$\sf{\to \color{cyan}{Product \: of \: Zeros}} \\ →ProductofZeros$

$\sf{(β) ×(α) = \frac{c}{a}}(β)×(α)=$

a

c

Now,

Given quadratic equation is

\sf{x²-7x+a= 0}x²−7x+a=0

Here Alpha & Beta are α & α+1.

$\sf{ \to \color{lime}\: sum \: of \: zeros} \\ →sumofzeros$

$</h2><h2>\begin{gathered} \sf{α+α+1= \frac{-(-7)}{1}} \\ \\ \sf{2 α+ 1 = 7} \\ \\ \sf{α = 3}\end{gathered} </h2><h2>α+α+1= </h2><h2>1$

−(−7)

2α+1=7

α=3

$\sf{ :\implies \: so \: the \: value \: of \: (α+1 ) \: is \: 4}:⟹sothevalueof(α+1)is \\ \\ </h2><h2>\sf{\to \color{orange}{Product \: of \: Zeros}}→ProductofZeros$

$\begin{gathered}\sf{:\implies α×(α+1)=\frac{a}{1}} \\ \\\sf{3×4=a} \\ \\ \sf{:\implies a= 12} \end{gathered}$

:⟹α×(α+1)=

1

a

3×4=a

:⟹a=12

Therefore the value of a is 12.