Math

Question Find the value of kk that makes x2kx+121=0x^2 - kx + 121 = 0 a perfect square trinomial.

Studdy Solution

STEP 1

Assumptions
1. We are given a quadratic equation in the form x2kx+121=0x^{2} - kx + 121 = 0.
2. We are looking for a positive value of kk that makes the left side a perfect square trinomial.
3. A perfect square trinomial is of the form (ax+b)2=a2x2+2abx+b2(ax + b)^{2} = a^{2}x^{2} + 2abx + b^{2}.

STEP 2

Identify the structure of a perfect square trinomial.
A perfect square trinomial can be written as (xp)2(x - p)^{2}, where pp is a constant. Expanding this, we get:
(xp)2=x22px+p2(x - p)^{2} = x^{2} - 2px + p^{2}

STEP 3

Compare the given quadratic equation with the perfect square trinomial structure.
We have x2kx+121x^{2} - kx + 121 and we want it to match the form x22px+p2x^{2} - 2px + p^{2}.

STEP 4

From the comparison in STEP_3, we see that p2=121p^{2} = 121. We need to find the value of pp.

STEP 5

Take the square root of 121 to find pp.
p=121p = \sqrt{121}

STEP 6

Calculate the value of pp.
p=121=11p = \sqrt{121} = 11

STEP 7

Now, we need to find the value of kk that makes the middle term of the perfect square trinomial equal to kx-kx.

STEP 8

From the structure of the perfect square trinomial, we know that the middle term is 2px2px. Therefore, we have 2p=k2p = k.

STEP 9

Substitute the value of pp into the equation 2p=k2p = k.
k=211k = 2 \cdot 11

STEP 10

Calculate the value of kk.
k=211=22k = 2 \cdot 11 = 22
The positive value of kk that makes the left side of the equation a perfect square trinomial is k=22k = 22.

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