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kap26 [50]
1 year ago
12

Write the equation in slope-intercept form for the line that passes through the given pointand is perpendicular to the given equ

ation.x + 3y = 6and passes through (-1, -6)
Mathematics
1 answer:
Bas_tet [7]1 year ago
5 0

When 2 lines are perpendicular, the product of the slopes is equal to -1

Given x + 3y = 6

3y = -x + 6

y = -x/3 + 2

The slope is -1/3

Hence the slope of the line perpendicular to this

= -1/-1/3

= 3

Hence the equation of the line

y - -6 = 3(x - -1)

y + 6 = 3(x + 1)

y + 6 = 3x + 3

y = 3x + 3 - 6

y = 3x -3

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Find the value(s) of k such that 4x² + kx + 4 = 0 has 1 rational solution
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Answer:

If you mean only one rational solution, the answer is

k_1 = 8, k_2 = -8

If you mean at least 1 rational solution, the answer is

k\in (-\infty, -8]\cup[8, \infty)

Step-by-step explanation:

4x^2 + kx + 4 = 0

Let's calculate the discriminant.

\Delta = b^2 - 4ac

\Delta = k^2 -4 \cdot 4 \cdot 4

\Delta = k^2 -64

Now, remember that:

\text{If } \Delta > 0 : \text{2 Real solutions}

\text{If } \Delta = 0 : \text{1 Real solution}

\text{If } \Delta < 0 : \text{No Real solution}

Therefore, I will just consider the first two cases.

k^2 - 64 > 0

and

k^2 -64 = 0

\boxed{\text{For } k^2 - 64 > 0}

k^2 > 64 \Longleftrightarrow k>\pm\sqrt{64}   \Longleftrightarrow k\in (-\infty, -8)\cup(8, \infty)

\boxed{\text{For } k^2 - 64 = 0}

k^2 = 64 \Longleftrightarrow k=\pm\sqrt{64} \implies k_1 = 8, k_2 = -8

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