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lubasha [3.4K]
3 years ago
5

Use the discriminant to determine how many real number solutions exist for the quadratic equation –4j2 + 3j – 28 = 0. A. 3 B. 2

C. 0 D. 1
Mathematics
1 answer:
Zepler [3.9K]3 years ago
5 0

Answer:

C. 0

Step-by-step explanation:

–4j^2 + 3j – 28 = 0

The discriminant is b^2-4ac  if >0 we have 2 real solutions

                                                   =0  we have 1 real solutions

                                                   <0 we have 2 imaginary solutions

a = -4, b =3  c = -28

b^2 -4ac

(3)^2 - 4(-4)*(-28)

9 - 16(28)

9 -448

This will be negative so we have two imaginary solutions.

Therefore we have 0 real solutions

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Step-by-step explanation:

Fill in the number and evaluate:

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Factorise the given question: 3p^-7p-6​
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Answer:

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Step-by-step explanation:

STEP 1 :

Equation at the end of step 1

(3p2 -  7p) -  6

STEP 2 :

Trying to factor by splitting the middle term

2.1     Factoring  3p2-7p-6  

The first term is,  3p2  its coefficient is  3 .

The middle term is,  -7p  its coefficient is  -7 .

The last term, "the constant", is  -6  

Step-1 : Multiply the coefficient of the first term by the constant   3 • -6 = -18  

Step-2 : Find two factors of  -18  whose sum equals the coefficient of the middle term, which is   -7 .

-18 + 1  = -17  

-9 +2 = -7 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -9  and  2  

                    3p2 - 9p + 2p - 6

Step-4 : Add up the first 2 terms, pulling out like factors :

                   3p • (p-3)

             Add up the last 2 terms, pulling out common factors :

                   2 • (p-3)

Step-5 : Add up the four terms of step 4 :

                   (3p+2)  •  (p-3)

            Which is the desired factorization

Final result :

 (p - 3) • (3p + 2)

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