Two numbers whose sum is $6$ and the absolute value of whose difference is $8$ are roots of the equation:

x^{2} - 6 x + 7 = 0

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

x^{2} - 6 x - 7 = 0

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

x^{2} + 6 x - 8 = 0

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

x^{2} - 6 x + 8 = 0

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

x^{2} + 6 x - 7 = 0

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

Open in App

Solution

The correct option is B $x^{2} - 6 x - 7 = 0$
Let's assume that these two numbers are root of a equation $α, β$
so $α + β = 6$
$1 α - β 1 = 8$
If $α > β$ then $α - β = 8$
$α + β = 6$
$α - β = 8$
so $α = 7$ , $β = - 1$
if $α < β$ then $α - β = - 8$
$α + β = 6$
$α - β = - 8$
so $α = - 1, β = 7$
Now $α + β = 6$
$α β = - 7$
so equation $= x^{2} - (α + β) x + α β = 0$
$= x^{2} - 6 x - 7 = 0$

Suggest Corrections

Similar questions

x^{2} - 3 x + 2 = 0.

(3 + 2 \sqrt{3})

x^{2} + 6 x - 3 = 0

x^{2} - 6 x - 3 = 0

x^{2} + 6 x + 3 = 0

x^{2} - 6 x + 3 = 0

x^{2} - 6 x + 6 = 0

x^{2} + 6 x - 6 = 0

x^{2} - 6 x - 6 = 0

x^{2} + 6 x + 6 = 0

x^{2} + 6 x - 7 = 0

Join BYJU'S Learning Program