Manipulation of equation into quadratic formcd1+d2-d1z1-δ(d2)P0P1d1+d2=-zβand&nbsp;(δ2p1p0)d22−d2[(c+zβ)2−2d1p1p0z1δ]+d1[z12−(c+zβ)2]=0

At first glance, it seems that the difficulty lies in the combination of terms that contain d2 and d1+d2 in the numerator, because if you simply square the expression, the two factors will mix, and you won't be able to get rid of the square root over d1+d2 in the mix term. However, note that the denominator itself is d1+d2, so you can split off the term cd1+d2 in the numerator, and then the left hand side becomes,cd1+d2−z1d1−d2δp1p0d1+d2=cd1+d2d1+d2−z1d1+d2δp1p0d1+d2 .This allows you to rearrange the equation intoz1d1+d2δp1p0d1+d2=c+zβand further intoz1d1+d2δp1p0=(c+zβ)d1+d2 .Now you can square both sides and rearrange it into a quadratic equation in d2.

Explained: "Manipulation of equation into quadratic formcd1+d2-d1z1-δ(d2)P0P1d1+d2=-zβand(δ2p1p0)d22−d2[(c+zβ)2−2d1p1p0z1δ]+d1[z12−(c+zβ)2]=0"

Achatesopo8

Answered question

2022-04-14

Manipulation of equation into quadratic form
$\frac{c \sqrt{d_{1} + d_{2}} - \sqrt{d_{1}} z_{1} - δ (d_{2}) \sqrt{P_{0} P_{1}}}{\sqrt{d_{1} + d_{2}}} = - z_{β}$
and $(δ^{2} p_{1} p_{0}) d_{2}^{2} - d_{2} [{(c + z_{β})}^{2} - 2 \sqrt{d_{1} p_{1} p_{0}} z_{1} δ] + d_{1} [z_{1}^{2} - {(c + z_{β})}^{2}] = 0$

Answer & Explanation

fallendreamsit2p

Beginner2022-04-15Added 14 answers

At first glance, it seems that the difficulty lies in the combination of terms that contain

d_{2} and \sqrt{d_{1} + d_{2}}

in the numerator, because if you simply square the expression, the two factors will mix, and you won't be able to get rid of the square root over

\sqrt{d_{1} + d_{2}}

in the mix term. However, note that the denominator itself is

\sqrt{d_{1} + d_{2}}

, so you can split off the term

c \sqrt{d_{1} + d_{2}}

in the numerator, and then the left hand side becomes,

\frac{c \sqrt{d_{1} + d_{2}} - z_{1} \sqrt{d_{1}} - d_{2} δ \sqrt{p_{1} p_{0}}}{\sqrt{d_{1} + d_{2}}} = \frac{c \sqrt{d_{1} + d_{2}}}{\sqrt{d_{1} + d_{2}}} - \frac{z_{1} \sqrt{d_{1}} + d_{2} δ \sqrt{p_{1} p_{0}}}{\sqrt{d_{1} + d_{2}}} .

This allows you to rearrange the equation into

\frac{z_{1} \sqrt{d_{1}} + d_{2} δ \sqrt{p_{1} p_{0}}}{\sqrt{d_{1} + d_{2}}} = c + z_{β}

and further into

z_{1} \sqrt{d_{1}} + d_{2} δ \sqrt{p_{1} p_{0}} = (c + z_{β}) \sqrt{d_{1} + d_{2}} .

Now you can square both sides and rearrange it into a quadratic equation in

d_{2}

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