If the equations $a x^{2} + 2 b x + c = 0, d x^{2} + 2 b^{'} x + c^{'} = 0$ have a common root, prove that the equation $(b^{2} - a c) x^{2} + (2 b b^{'} - a c^{'} - d c) x + ({b^{'}}^{2} - d c^{'}) = 0$ has equal roots.

Open in App

Solution

As in Q. 1, the condition for common roots is
${(c d - c^{'} a)}^{2} = 4 (b c^{'} - b^{'} c) (a b^{'} - d b)$ .........(1)
For equality of roots of 2nd equation its discriminant
$B^{2} - 4 A C = 0$
${(2 b b^{'} - a c^{'} - d c)}^{2} - 4 (b^{2} - a c) ({b^{'}}^{2} - d c^{'}) = 0$
or $4 b^{2} {b^{'}}^{2} + {(a c^{'} + d c)}^{2} - 4 b b^{'} (a c^{'} + d c)$
$= 4 b^{2} {b^{'}}^{2} - 4 b^{2} d c^{'} - 4 {b^{'}}^{2} a c + 4 a a^{'} c c^{'}$
or ${(a c^{'} + d c)}^{2} - 4 a a^{'} c c^{'} = 4 b b^{'} (a c + d c) - 4 b^{2} d c^{'} - 4 {b^{'}}^{2} a c$
or ${(c d - c^{'} a)}^{2} = 4 [b c^{'} (a b^{'} - d b) - b^{'} c (a b^{'} - d b)] = 4 [(a b^{'} - d b) (b c^{'} - b^{'} c)]$
Which is true by (1).

Suggest Corrections

Similar questions

If the roots of the equations $a x^{2} + 2 b x + c = 0$ and $b x^{2} - 2 \sqrt{a c} x + b = 0$ are simultaneously real, then prove that $b^{2} = a c$ .

Q. If the quadratic equations

a x^{2} + 2 c x + b = 0

and

a x^{2} + 2 b x + c = 0, (b \neq c)

have a common root, then

a + 4 b + 4 c

equals to

Q. Assertion :If

a > 0

and

b^{2} - a c + c < 0

,then domain of the function

f (x) = \sqrt{a x^{2} + 2 b x + c} is R

Reason: If

b^{2} - a c < 0

,then

a x^{2} + 2 b x + c = 0

has imaginary roots.

If $α, β$ are the roots of the equation a $x^{2}$ +2bx +c = 0 and $α + δ$ , $β + δ$ are the roots of A $x^{2}$ + 2Bx + C= 0 , then $\frac{b^{2} - a c}{B^{2} - A C}$ =

If a,b,c are in G.P., the equations $a x^{2} + 2 b x + c = 0$ and $d x^{2} + 2 e x + f = 0$ have a common root if $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in :

Join BYJU'S Learning Program

If the equations ax2+2bx+c=0,dx2+2b′x+c′=0 have a common root, prove that the equation (b2−ac)x2+(2bb′−ac′−dc)x+(b′2−dc′)=0 has equal roots.

If the equations $a x^{2} + 2 b x + c = 0, d x^{2} + 2 b^{'} x + c^{'} = 0$ have a common root, prove that the equation $(b^{2} - a c) x^{2} + (2 b b^{'} - a c^{'} - d c) x + ({b^{'}}^{2} - d c^{'}) = 0$ has equal roots.