The number of values k for which $[x^{2} - (k - 2) x + k^{2}$ ][ $x^{2} + k x + (2 k - 1)]$ is a perfect square is

2

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

1

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

0

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

None of these

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

Open in App

Solution

The correct option is B $1$
$[x^{2} - (k - 2) x + k^{2}] [x^{2} + k x + (2 k - 1)]$

For the expression to be a perfect square ,ther are two possible way
$(i)$ When both the quadratic expression are perfect square for a particular value.For this to happen,
$k - 2 = 2 k \to k = - 2$
Now,for $k = - 2$ in the second expression ,we get,

$x^{2} - 2 x - 5$ ,which is not a perfect square.

$(i i)$ The other way is when both the quadratic equations are same.
$- (k - 2) = k \to k = 1 + k^{2} = 2 k - 1 \to k = 1$
At $k = 1$ ,the expression is a perfect square.

Thus, minimum values of $k i s$ 1

Suggest Corrections

Similar questions

x^{2} + x + 1 + 2 k (x^{2} - x - 1) = 0

k

k, (4 - k) x^{2} + (2 k + 4) x + (8 k + 1) = 0

x^{2} - 2 (k + 1) x + k^{2} = 0

k^{2} x^{2} - 2 (2 k - 1) x + 4 = 0

(k + 1) x^{2} - 2 (k - 1) x + 1 = 0

x^{2} + k (2 x + k - 1) + 2 = 0

k

x^{2} + k x + (k + 2) = 0 and x^{2} + (1 - k) x + 3 - k = 0

k

∣ ∣ ∣ \frac{x^{2} + k x + 1}{x^{2} + x + 1} ∣ ∣ ∣ < 2

x,

Join BYJU'S Learning Program