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2 years ago

The difference between the roots of the quadratic equation x2 -14x +q =0 is 6. Find q

Mathematics

1 answer:

Irina-Kira [14]2 years ago

3 0

9514 1404 393

Answer:

q = 40

Step-by-step explanation:

When the quadratic has roots p and r, it can be factored as ...

(x -p)(x -r) = x² -(p+r)x +pr

So, the sum of the roots is 14, and their difference is 6. This lets us find the roots from ...

p + r = 14

p - r = 6

2p = 20 . . . add the two equations

p = 10

r = 14 -p = 4

The value of interest is then ...

q = pr = (10)(4)

q = 40

The graph shows the roots to be 4 and 10, as we found.

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Find the distance between the point (-2, -3) and the line with the equation y=4-2/3x

otez555 [7]

Answer:

$D = \frac{25\sqrt{13}}{13}$

Step-by-step explanation:

Given

$y = 4 - \frac{2}{3}x$

$(x_1,y_1) = (-2,-3)$

Required

Determine the distance

$y = 4 - \frac{2}{3}x$

Write the above equation in standard form:

$Ax + By + C = 0$

So, we have:

$\frac{2}{3}x+y - 4 = 0$

By comparison:

$A = \frac{2}{3}$ $B = 1$ and $C = -4$

The distance is calculated using:

$D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

Where:

$(x_1,y_1) = (-2,-3)$

$A = \frac{2}{3}$ $B = 1$ and $C = -4$

This gives:

$D = \frac{|\frac{2}{3} * (-2) + 1 * (-3) - 4|}{\sqrt{(\frac{2}{3})^2 + 1^2}}$

$D = \frac{|-\frac{4}{3} -3 - 4|}{\sqrt{\frac{4}{9} + 1}}$

Take LCM

$D = \frac{|\frac{-4-9-12}{3}|}{\sqrt{\frac{4+9}{9}}}$

$D = \frac{|\frac{-25}{3}|}{\sqrt{\frac{13}{9}}}$

$D = |\frac{-25}{3}|/\sqrt{\frac{13}{9}}$

$D = \frac{25}{3}/\sqrt{\frac{13}{9}}$

Split the square root

$D = \frac{25}{3}/\frac{\sqrt{13}}{\sqrt{9}}$

Change / to *

$D = \frac{25}{3}*\frac{\sqrt{9}}{\sqrt{13}}$

$D = \frac{25}{3}*\frac{3}{\sqrt{13}}$

$D = \frac{25}{\sqrt{13}}$

Rationalize

$D = \frac{25}{\sqrt{13}} * \frac{\sqrt{13}}{\sqrt{13}}$

$D = \frac{25\sqrt{13}}{13}$

Hence, the distance is:

$D = \frac{25\sqrt{13}}{13}$

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