What are the solutions of the quadratic equation 4x^2-8x-12=0

Mathematics

1 answer:

Wewaii [24]3 years ago

6 0

Answer:

Step-by-step explanation:

$4x^2-8x-12=0\\{}\qquad\qquad^{\div4\qquad\div4}\\x^2-2x-3=0\\\\x^2+x-3x-3=0\\\\x(x+1)-3(x+1)=0\\\\(x+1)(x-3)=0\\\\x+1=0\quad\vee\quad x-3=0\\\\x=-1\quad\vee\quad x=3$

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How to solve equations with the substitution method

Paul [167]

Answer:

In a system, the substitution method is one of the 3 main ways to solve a system and can be very efficient at times.

<u>Skills needed: Systems, Algebra</u>

Step-by-step explanation:

1) Let's say we are given two equations below:

$y=4x \\ 4x-5y=-32$

We can use substitution here by substituting in for $y$ in the second equation. This means we put in $4x$ for $y$ in the 2nd equation so we only have $x$ variables in the equation, allowing us to solve for $x$ .

2) Solving it out:

$1) 4x-5(4x)=-32 \\ 2) 4x-20x=-32 \\3) -16x=-32 \\4)x=2\\5)y=4x = 4(2) = 8 \\ 6)y=8$

We essentially substitute in that value as seen in step 1. Steps 2 and 3 are just simplifying the left side and allowing for us to solve. Step 4 is where we divide by -16 on both sides to solve for x. Step 5 and 6 show us solving for y using the value for x. We get $x=2,y=8$