Let - 1- -2 and -1- -2 be the roots of ax2 - bx - c - 0 and px2 - qx - r - 0 respectively- If the system of equations -1 y - -2 z - 0 and -1 y - -2 z - 0 has non-trivial solution- then prove that b2q2 - acpr-

(a) We have the following relations α_1+α_2 = ba, α_1, α_2 = ca #160; #160; #160;........(1) β_1 + β_2 = qp, β_1, β_2 = rp #160; #160; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Consistency of Linear System of Equations

Let α 1, α2...

Question

Let $α_{1}, α_{2}$ and $β_{1}, β_{2}$ be the roots of $a x^{2} + b x + c = 0$ and $p x^{2} + q x + r = 0$ respectively. If the system of equations $α_{1} y + α_{2} z = 0$ and $β_{1} y + β_{2} z = 0$ has non-trivial solution, then prove that $\frac{b^{2}}{q^{2}} = \frac{a c}{p r}$ .

Open in App

Solution

(a) We have the following relations
$α_{1} + α_{2} = \frac{- b}{a}, α_{1}, α_{2} = \frac{c}{a}$ ........(1)

$β_{1} + β_{2} = - \frac{q}{p}, β_{1}, β_{2} = \frac{r}{p}$ ........(2)

Since the given system has non-trivial solution

$△ = ∣ ∣ ∣ \begin{matrix} α_{1} & α_{2} β_{1} & β_{2} \end{matrix} ∣ ∣ ∣ = 0$

or $\frac{b^{2}}{q^{2}} = \frac{a c}{p r}$

Suggest Corrections

Similar questions

Let $α_{1}, α_{2}, β_{1}, β_{2}$ be the roots of $a x^{2} + b x + c = 0 a n d p x^{2} + q x + r = 0$ , respectively. If the system of equations $α_{1} y + α_{2} z = 0 a n d β_{1} y + β_{2} z = 0$ has a non - trivial solution, then

Q. Let

α_{1}, α_{2}

and

β_{1}, β_{2}

be the roots of

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

and

p x^{2} + q x + r = 0

respectively. If the system of equations

α_{1} y + α_{2} z = 0

and

β_{1} y + β_{2} z = 0

has a non-trivial solution, then which of the following options is CORRECT ?

Q. if

α_{1}, α_{2}

are the roots of the equation

x^{2} + p x + q = 0

and

β_{1}, β_{2}

are those of the equation

x^{2} + r x + s = 0

and the system of the equation

α_{1} y + α_{2} z = 0

and

β_{1} y + β_{2} z = 0

has non trivial solution, then

Q. Let

- \frac{π}{6} < θ < - \frac{π}{12}

. Suppose

α_{1}

and

β_{1}

are the roots of the equation

x^{2} - 2 x s e c θ + 1 = 0

α_{2}

and

β_{2}

are the roots of the equation

x^{2} + 2 x t a n θ - 1 = 0

. If

α_{1} > β_{1}

and

α_{2} > β_{2}

, then

α_{1} + β_{2}

equals

Join BYJU'S Learning Program

Let α1,α2 and β1,β2 be the roots of ax2+bx+c=0 and px2+qx+r=0 respectively. If the system of equations α1y+α2z=0 and β1y+β2z=0 has non-trivial solution, then prove that b2q2=acpr.

(a) We have the following relationsα1+α2=−ba,α1,α2=ca ........(1)β1+β2=−qp,β1,β2=rp ........(2)Since the given system has non-trivial solution△=∣∣∣α1α2β1β2∣∣∣=0or b2q2=acpr

Let $α_{1}, α_{2}$ and $β_{1}, β_{2}$ be the roots of $a x^{2} + b x + c = 0$ and $p x^{2} + q x + r = 0$ respectively. If the system of equations $α_{1} y + α_{2} z = 0$ and $β_{1} y + β_{2} z = 0$ has non-trivial solution, then prove that $\frac{b^{2}}{q^{2}} = \frac{a c}{p r}$ .