Question

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

Answer 1

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