What values of a, b, and c would you use in the quadratic formula for the following equation? 6x^2+5x-4=0

Mathematics

1 answer:

Yanka [14]3 years ago

5 0

So remember that in a quadratic equation there is a type of quadratic equation of the form ax^2+bx+c=0 so to find what values of a, b, and c you need for the quadratic formula, you need to look at your equation and label a, b, and c. In this equation a=6, b=5 and c=-4. So for the quadratic formula you would use a=6, b=5 and C=-4. I hope this helps!!!!!

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NO LINKS!! Find an nth-degree polynomial function with real coefficient satisfying the given conditions.

Sati [7]

n=3; We need a third degree polynomials with the following given zero's: 2 and 5i are zeros; f(-1)=156.

Since these are solutions

x = 2 ; x = 5i. Since imaginaries travel in pairs, the other answer is x= -5i.

We have (x-2)(x-5i)(x+5i) = 0

Now,

f(-1) = (-1-2)(-1-5i)(-1+5i) = 156.

f(-1) = (-3)(26) = -78.

But -78 x -2 = 156, so our polynomial becomes

Y= -2x (x - 2 ) x (x to the power of 2 + 25) = 0

6 0

2 years ago

Read 2 more answers

Plz help me with this

Setler79 [48]

Answer: x = 8

(the correct answer was not provided as an option)

Step-by-step explanation:

$log_2(x-6)+log_2(x-4)=log_2(x)\\\\log_2[(x-6)(x-4)]=log_2(x)\\\\(x-6)(x-4)=x\\\\x^2-10x+24=x\\\\x^2-11x+24=0\\\\(x-3)(x-8)=0\\\\x=3\qquad x=8\\\\\\\text{The term after the log symbol (inside the parenthesis) must be greater than 0!}\\\\Check:\\3-6>0\ \text{FALSE --- so 3 is not a valid solution}\\\\8-6>0\ \text{TRUE}\\8-4>0\ \text{TRUE}\\8>0\ \text{TRUE --- 8 is a valid solution because it is true for all}$