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Dafna1 [17]
3 years ago
13

If 2.5 is a root of the equation 2x^2+5x−q=0, what are the possible values of q?

Mathematics
2 answers:
bulgar [2K]3 years ago
4 0

Answer:

q = -2510 +5 = \pm \sqrt{5^2-4*2*q} \\ 225 = 25-8q \\\\8q = -200 \rightarrow q = -\frac {200}8 = -25

Step-by-step explanation:

Grab the generic solution of the equation. \frac 52 = \frac{-5 \pm \sqrt{5^2-4*2*q}}{2*2}. Let's isolate the radical, and square both sides.10 +5 = \pm \sqrt{5^2-4*2*q} \\ 225 = 25-8q \\\\8q = -200 \rightarrow q = -\frac {200}8 = -25

oksano4ka [1.4K]3 years ago
4 0

Answer:

since we know that x1=2.5

we can just just use vieta's theorem

x1+x2=-b/a

x1*x2=c/a

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Sergio [31]

Answer:

The values of p in the equation are 0 and 6

Step-by-step explanation:

First, you have to make the denominators the same. to do that, first factor 2p^2-7p-4 = \left(2p+1\right)\left(p-4\right)2p

2

−7p−4=(2p+1)(p−4)

So then the equation looks like:

\frac{p}{2p+1}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{5}{p-4}

2p+1

p

−

(2p+1)(p−4)

2p

2

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p−4

5

To make the denominators equal, multiply 2p+1 with p-4 and p-4 with 2p+1:

\frac{p^2-4p}{(2p+1)(p-4)}-\frac{2p^2+5}{(2p+1)(p-4)}=-\frac{10p+5}{(p-4)(2p+1)}

(2p+1)(p−4)

p

2

−4p

−

(2p+1)(p−4)

2p

2

+5

=−

(p−4)(2p+1)

10p+5

Since, this has an equal sign we 'get rid of' or 'forget' the denominator and only solve the numerator.

(p^2-4p)-(2p^2+5)=-(10p+5)(p

2

−4p)−(2p

2

+5)=−(10p+5)

Now, solve like a normal equation. Solve (p^2-4p)-(2p^2+5)(p

2

−4p)−(2p

2

+5) first:

(p^2-4p)-(2p^2+5)=-p^2-4p-5(p

2

−4p)−(2p

2

+5)=−p

2

−4p−5

-p^2-4p-5=-10p+5−p

2

−4p−5=−10p+5

Combine like terms:

-p^2-4p+0=-10p−p

2

−4p+0=−10p

-p^2+6p=0−p

2

+6p=0

Factor:

p=0, p=6p

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