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kumpel [21]
2 years ago
7

NEED HELP PLEASE ANSWER ASAP! (41)

Mathematics
2 answers:
Cerrena [4.2K]2 years ago
7 0

Answer:

(2)  x^2-3x-4=0

Step-by-step explanation:

Standard form of a quadratic equation:  ax^2+bx+c=0

When factoring a quadratic (finding the roots) we find two numbers that multiply to ac and sum to b, then rewrite b as the sum of these two numbers.

So if the roots <u>sum to 3</u> and <u>multiply to -4</u>, then the two numbers would be 4 and -1.

\implies b=1+-4=-3

\implies ac=1 \cdot -4

As there the leading coefficient is 1, c=-4.  

Therefore, the equation would be:  x^2-3x-4=0

<u>Proof</u>

Factor  x^2-3x-4=0

Find two numbers that multiply to ac and sum to b.

The two numbers that multiply to -4 and sum to -3 are:  -4 and 1.

Rewrite b as the sum of these two numbers:

\implies x^2-4x+x-4=0

Factorize the first two terms and the last two terms separately:

\implies x(x-4)+1(x-4)=0

Factor out the common term  (x-4):

\implies (x+1)(x-4)=0

Therefore, the roots are:  

(x+1)=0 \implies x=-1

(x-4)=0 \implies x=4

So the sum of the roots is:  -1 + 4 = 3

And the product of the roots is:  -1 × 4 = -4

Thereby proving that  x^2-3x-4=0  has roots whose sum is 3 and whose product is -4.

ExtremeBDS [4]2 years ago
3 0

Answer:

   41. <u>(2) x² - 3x - 4 =0</u>

Step-by-step explanation:

<u>Question 41</u>

<u>Standard form of polynomial</u>

  • x² - (sum of roots) + product of roots = 0
  • <u>(2) x² - 3x - 4 =0</u>
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