basic algebra

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vaibhav101: Master | Next Rank: 500 Posts; Posts: 138; Joined: Mon May 01, 2017 11:56 pm; Thanked: 4 times

by vaibhav101 » Wed May 16, 2018 7:53 am

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if the sum of the roots of the quadratic equation, $$ax^2+bx+c=0$$ , is equal to the sum of the squares of their reciprocals , then $$\left(\frac{b^2}{ac}+\frac{bc}{a^2}\right)$$ is equal to
A 2
B -2
C 1
D-1
E 0

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Source: Beat The GMAT — Problem Solving | Wed May 16, 2018 7:53 am

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

by deloitte247 » Sat May 19, 2018 2:57 pm

Let the roots be p and q
p + q = $$\frac{b}{a}$$
pq = $$\frac{c}{a}$$

but given = p +q = $$\frac{1}{p^2}+\frac{1}{q2}
$$ $$\frac{-b}{a}\ =\ \frac{p^{2\ +q^2}}{p2q^2}$$
$$\frac{-b}{a}\ =\ \left(p+q\right)^{2\ }-\frac{2pq}{p^2q2}
frac{-b}{a}\ =left(\frac{-b}{a}\right)^2\ -\ 2\left(\frac{c}{a}\right)2 $$
$$\frac{-b}{a}\ =\frac{b^2}{a^2}-\ 2\left(\frac{\frac{c}{a}}{c^2}\right):a^2\ \ \ $$
$$\frac{-b}{a}\left(\frac{c^2}{a^2}\ \ \right)=\ \frac{b^2}{a^2}-\ 2\ \left(\frac{c}{a}\right)$$
$$\frac{bc^2}{a^3}\ =\ \frac{b^2}{a^2}-2\left(\frac{2}{a}\right)$$
$$2\left(\frac{c}{a}\right)=\ \frac{b^2}{a^2}+\ \frac{bc^2}{a3}$$
$$2\left(\frac{c}{a}\right)=\ \frac{ab^{2\ +\ bc^2}}{a^3}$$
$$2\left(\frac{c}{a}\right)=\ b\left(\frac{ab+c^2}{a^3}\right)$$
$$2\left(\frac{c}{a}\right)\left(a^3\right)=\ ab^{2\ +\ bc^2}$$
$$2ca^2=ab^{2\ +\ }bc^2$$
$$2=\frac{ab^{2\ +\ }bc^2}{ca^2}$$
$$2=\frac{ab^2}{ca^2}+\frac{bc^2}{ca^2}$$
$$2=\frac{b^2}{ac}+\frac{bc^{ }}{a^2}$$
the answer is A =2

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