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2 years ago

What is the solution for 5x^2-x+6=0

Mathematics

1 answer:

kati45 [8]2 years ago

7 0

The easiest way to solve this problem is by using the Pythagorean theorem :D

Pythagorean theorem: [-b +- sqrt(b^2 - 4ac)]/2a

5 = a

-1 = b

6 = c

therefore, by plugging these values in!

[-(-1) +- sqrt((-1)^2 - 4(5)(6))]/2(5)

[1 +- sqrt(1 - 120)]/10

Oh so we're using imaginary numbers, so in this case you'll need to know that i = sqrt(-1), so keep this in mind (I can see why this was so difficult)

[1 +- sqrt(119)*i]10

answers:

( 1 + i*sqrt(119) ) / 10

( 1 - i*sqrt(119) ) / 10

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Given that P = (-7, 16) and Q = (-8, 7), find the component form and magnitude of vector QP .

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a)The component form is

$\vec {QP}=\binom{ - 1}{ - 8}$

b)The magnitude is √65

c) <2,14>

Step-by-step explanation:

Recall that:

$\vec {QP}=\vec {OP}-\vec{OQ}$

We substitute the position vectors to get:

$\vec {QP}=\binom{ - 8}{7} - \binom { - 7}{15}$

We subtract the corresponding components to obtain:

$\vec {QP}=\binom{ - 8 - - 7}{7 - 15}$

This gives:

$\vec {QP}=\binom{ - 8 + 7}{7 - 15}$

This simplifies to:

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$\binom{x}{y}$

$\sqrt{ {x}^{2} + {y}^{2} }$

This means:

$|\vec {QP}|= \sqrt{ {( - 1)}^{2} + {( - 8)}^{2} }$

This simplifies to:

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c) We have the vectors u = <4, 8>, v = <-2, 6>.

We want to find:

u+v

This implies that:

u+v=<4,8>+<-2,6>

We add the corresponding components to get;

u+v=<4+-2,8+6>

This simplifies to:

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