$I . x^{2} - 8 x + 15 = 0$
$I I . y^{2} - 3 y + 2 = 0$

x > y

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

\geq

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

x < y

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

\leq

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

x = y or the relationship cannot be established

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

Open in App

Solution

The correct option is A x > y
Equation I
$x^{2} - 8 x + 15 = 0$
$\Rightarrow x^{2} - 5 x - 3 x + 15 = 0$
$\Rightarrow x (x - 5) - 3 (x - 5) = 0$
$\Rightarrow (x - 3) (x - 5) = 0$
$∴ x = 5, 3$
Equation II
$y^{2} - 3 y + 2 = 0$
$\Rightarrow y^{2} - 2 y - y + 2 = 0$
$\Rightarrow y (y - 2) - 1 (y - 2) = 0$
$∴ y = 2, 1$
Hence, x > y

Suggest Corrections

Similar questions

I . x^{2} + 5 x + 6 = 0

I I . y^{2} + 3 y + 2 = 0

I . x^{2} + y^{2} = 1

a

) (0, 0)

x^{2} + y^{2} - 2 x = 1

x^{2} + y^{2} - 2 y = 1

I I . x^{2} + y^{2} - x + 3 y - 3 = 0, b) (2, 3)

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

x^{2} + y^{2} + 2 x + 3 y - 9 = 0

I I I . x^{2} + y^{2} - 8 x + 40 = 0, c) (8, - 15 / 2)

x^{2} + y^{2} - 5 x + 16 = 0

x^{2} + y^{2} - 8 x + 16 y + 160 = 0

I. $x^{2}$ + 9x + 14 = 0

II. $y^{2}$ + y - 2 = 0

I. $3 x^{2} - 15 x + 18 = 0$

II. $y^{2} - 3 y - 4 = 0$

I . x^{2} - x - 12 = 0

I I . y^{2} + 5 y + 6 = 0

Join BYJU'S Learning Program