Find the quadratic equation in $x$ , whose roots are $(m + n)$ and $(m - n)$

x^{2} + 2 m x + (m^{2} + n^{2}) = 0

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

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

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

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

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

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

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

Open in App

Solution

The correct option is D $x^{2} - 2 m x + (m^{2} - n^{2}) = 0$
Roots are (m+n) & (m-n)
sum of roots = m+n+m-n=2m
product of roots = (m+n)(m-n)
= $m^{2} - n^{2}$
equation
$x^{2}$ -(sum of roots)x+product of roots = 0
$x^{2} - 2 m x + m^{2} - n^{2} = 0$

Suggest Corrections

Similar questions

^{2}

^{2}

^{2}

m, n

x^{2} - 3 x + 5 = 0,

(m^{2} - 3 m + 7) & (n^{2} - 3 n + 7) =

x^{2} - 2 m x + m^{2} + 1 = 0

m

x^{2} - 2 m x + m^{2} - 1 = 0

(- 2, 4)

Values of 'm' for which both the roots of equation $x^{2}$ - 2mx + $m^{2}$ - 1=0 are less than 4 are

Join BYJU'S Learning Program