Question

NEED HELP PLEASE ANSWER ASAP! (41)

Answer 1

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 sum to 3 and multiply to -4, 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:  $x^2-3x-4=0$

Proof

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.

Answer 2

Answer:

   41. (2) x² - 3x - 4 =0

Step-by-step explanation:

Question 41

Standard form of polynomial

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