Math, asked by aathithnaturelover, 5 hours ago

if 2x^2-3x + k =0, x^2-9x + 20 =0 have a common root then k

Answers

Answered by negivardhan993
3

Explanation:

x^2 - 9x + 20= 0

==> x^2 - 4x - 5x + 20=0

==> x(x-4)-5(x-4)=0

==> (x-4)(x-5)=0

Hence, the roots for this equation are x = 4 or x = 5.

It is given that equations 2x^2 - 3x + k = 0 and x^2 - 9x + 20 =0 share a common root.

2x^2 - 3x + k=0

==> 2(x^2 - \frac{3}{2}x+\frac{k}{2})=0

==> x^2 - \frac{3}{2}x + \frac{k}{2}=0

Let the one root be 4 and the other root be r.

We know that in a quadratic polynomial of form ax^2 + bx + c with roots \alpha and \beta,

\alpha +\beta =\frac{-b}{a}, meaning

4+r=\frac{-(-3)}{2}=\frac{3}{2}

r = \frac{3 - 8}{2} = \frac{-5}{2}

We also know that

\alpha \beta =\frac{c}{a}, meaning

4\times \frac{-5}{2}=\frac{k}{2}

==> \frac{k}{2}=-2 \times 5=-10

k = -10\times2=-20

Verification:

2x^2 - 3x + k =0

==> 2(4)^2-3(4)-20=0

==> 32 - 12 - 20 = 0

==> 0 = 0

∴ LHS = RHS

Answer: k = -20

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