Answer:
(2) [tex]x^2-3x-4=0[/tex]
Step-by-step explanation:
Standard form of a quadratic equation: [tex]ax^2+bx+c=0[/tex]
When factoring a quadratic (finding the roots) we find two numbers that multiply to [tex]ac[/tex] and sum to [tex]b[/tex], then rewrite [tex]b[/tex] 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.
[tex]\implies b=1+-4=-3[/tex]
[tex]\implies ac=1 \cdot -4[/tex]
As there the leading coefficient is 1, [tex]c=-4[/tex].
Therefore, the equation would be: [tex]x^2-3x-4=0[/tex]
Proof
Factor [tex]x^2-3x-4=0[/tex]
Find two numbers that multiply to [tex]ac[/tex] and sum to [tex]b[/tex].
The two numbers that multiply to -4 and sum to -3 are: -4 and 1.
Rewrite [tex]b[/tex] as the sum of these two numbers:
[tex]\implies x^2-4x+x-4=0[/tex]
Factorize the first two terms and the last two terms separately:
[tex]\implies x(x-4)+1(x-4)=0[/tex]
Factor out the common term [tex](x-4)[/tex]:
[tex]\implies (x+1)(x-4)=0[/tex]
Therefore, the roots are:
[tex](x+1)=0 \implies x=-1[/tex]
[tex](x-4)=0 \implies x=4[/tex]
So the sum of the roots is: -1 + 4 = 3
And the product of the roots is: -1 × 4 = -4
Thereby proving that [tex]x^2-3x-4=0[/tex] has roots whose sum is 3 and whose product is -4.
