Question

help please anybody @ganeshie8 please solve it

Answer 1

\[-b \pm \sqrt{b^2 - 4ac}/2a\]
the 2a is under the whole equation not just the last part

Answer 2

hurry up please

Answer 3

no this eq can't help me ,, this question is way more than that ,, quadratic equation cant help

Answer 4

the resulting equations you get will be
\[\alpha = some-function-of-a \]
and
\[\beta = some-other-function-of-a\]
then you can equate the two equations to find a

Answer 5

well the two equations and the \[2*\alpha-\beta=7\] that they give you. Combine them to find 'a'

Answer 6

Just don't mix up the a in the quadratic equation with the 'a' that you are trying to find. Wasn't great of them to use the letter 'a' for the unknown

Answer 7

@aa123 Do not post your question in somebody elses.

Answer 8

@Dom89 u don't even know what this Question is about,, seriously it can't b solved by quadretic equation

Answer 9

yeah @YanaSidlinskiy thats right

Answer 10

If the roots of the equation are \(\alpha\) and \(\beta\), then the equation is:

\((x - \alpha)(x - \beta) = 0 \)

\(x^2 - \beta x - \alpha x + \alpha \beta = 0 \)

\(x^2 - (\alpha + \beta)x + \alpha \beta = 0\)

You are given

\(x^2 - 3x + a = 0\)

That means

\(\alpha + \beta = -3 \)

You are also given \(2\alpha - \beta = 0\)

You have two equations and two variables.
Solve the system of equations for \(\alpha\) and \(\beta\).
Once you know those values, substitute them in
\((x - \alpha)(x - \beta) = 0 \).
Then multiply it out to find a.

Answer 11

thanks ,, alpha +beta is not -3 it's 3 but so much thanks it gave me right way to solve ,, thanks

Answer 12

Good point. Thanks for pointing that out.