If Equations x 2 + a x + b = 0 & x 2 + c x + d = 0 have a common root and first equation have equal roots,solve Then prove that 2(b+d)=ac How to prove?solve

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Answer 1 · 2018-06-28T14:10:24

Please refer to a Proof in the Explanation.

Since, the eqn. $x^2+ax+b=0$ has equal roots, its discriminant

must be zero.

$:. a^2-4b=0, or, b=a^2/4............(star)$ .

Hence, the eqn. becomes, $x^2+ax+a^2/4=0$ .

$:. (x+a/2)^2=0 , or, x=-a/2$ .

This has to be a root of the second eqn. : $x^2+cx+d=0$ .

Subst.ing $x=-a/2$ in the second eqn., we have,

$a^2/4+c(-a/2)+d=0," or, by (star), "b-(ac)/2+d=0$ .

This gives, $2(b+d)=ac$ , as desired!

Answer 2 · 2018-06-28T14:13:38

Please see the explanation below

The roots of the first equation $x^2+ax+b=0$ are

$alpha$ and $alpha$

Therefore,

$2alpha=-a$ ...................... $(1)$

and

$alpha^2=b$ ............................ $(2)$

The roots of the second equation $x^2+cx+d=0$ are

$alpha$ and $beta$

Therefore,

$alpha+beta=-c$ ............ $(3)$

$alphabeta=d$ ............... $(4)$

From the last equation,

$=>$ , $beta=d/alpha$

Substituting in equation $(3)$

$alpha+d/alpha=-c$

$=>$ , $(alpha^2+d)/(alpha)=-c$

Plugging in the values from equations $(1)$ and $(2)$

$(b+d)/(-a/2)=-c$

$=>$ , $2(b+d)=ac$