The common root of two equations p2-p-q-r2 x2-p-q-r2-1 x-1-p2-0 and m2 x2-r2-m2 x-r2-0 isA- p2-r2-m2-p-qB- 0C- none of theseD- 1-p2-r2

The correct option is C none of these (p^2 √(p+q)+ r^2)x^2+(√(p+q) r^2 1)x+(1 p^2)=0 (p^2+r^2 √(p+q))x^2+( p^2 r^2+√(p+q)+p^2 1)x+(1 p^2)=0 (p^2+r^2 √(p+q))x[x ...

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

Byju's Answer

Standard XII

Mathematics

Fundamental Laws of Logarithms

The common ro...

Question

The common root of two equations $(p^{2} - \sqrt{p + q} + r^{2}) x^{2} + (\sqrt{p + q} - r^{2} - 1) x + (1 - p^{2}) = 0$ and $m^{2} x^{2} - (r^{2} + m^{2}) x + r^{2} = 0$ is

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is C none of these
$(p^{2} - \sqrt{p + q} + r^{2}) x^{2} + (\sqrt{p + q} - r^{2} - 1) x + (1 - p^{2}) = 0$
$(p^{2} + r^{2} - \sqrt{p + q}) x^{2} + (- p^{2} - r^{2} + \sqrt{p + q} + p^{2} - 1) x + (1 - p^{2}) = 0$
$(p^{2} + r^{2} - \sqrt{p + q}) x [x - 1] + (p^{2} - 1) (x - 1) = 0$
$(x - 1) [(p^{2} + r^{2} - \sqrt{p + q}) x + (p^{2} - 1)] = 0$
$m^{2} x^{2} - (r^{2} + m^{2}) x + r^{2} = 0$
$x m^{2} (x - 1) - r^{2} (x - 1) = 0$
$(m^{2} x - r^{2}) (x - 1) = 0$
Common root
x=1

Suggest Corrections

Similar questions

Q. The common root of two equations

(p^{2} - \sqrt{p + q} + r^{2}) x^{2} + (\sqrt{p + q} - r^{2} - 1) x + (1 - p^{2}) = 0

and

m^{2} x^{2} - (r^{2} + m^{2}) x + r^{2} = 0

Q. If

(p^{2} - q^{2}) x^{2} + (q^{2} - r^{2}) x + r^{2} - p^{2} = 0

and

(p^{2} - q^{2}) y^{2} + (r^{2} - p^{2}) y + q^{2} - r^{2} = 0

have a common root for all real values of p, q and r, then find the common root.

Q. If p, q, r are real, then roots of the equation

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

will always be?

Q. Let

p, q a n d r

be real numbers

(p \neq q, r \neq 0)

, such that the roots of the equation

\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}

are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :

Q. If

\frac{1}{q + r}

\frac{1}{r + p}

and

\frac{1}{p + q}

are in AP, then

p^{2}

q^{2}

and

r^{2}

are in

Join BYJU'S Learning Program

The common root of two equations (p2−√p+q+r2)x2+(√p+q−r2−1)x+(1−p2)=0 and m2x2−(r2+m2)x+r2=0 is

The common root of two equations $(p^{2} - \sqrt{p + q} + r^{2}) x^{2} + (\sqrt{p + q} - r^{2} - 1) x + (1 - p^{2}) = 0$ and $m^{2} x^{2} - (r^{2} + m^{2}) x + r^{2} = 0$ is