Solve a quadratic equation by factoring j2–30j=0
1 answer:
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Answer:
Our system of equations is:
- y+2x+1=0
- 4y-4x²-12x = -7
We are looking for x
Let's express y using x
- y+2x+1=0
- y= -2x-1
Replace x in the second equation with the result
- 4y-4x²-12x = -7
- 4(-2x-1)-4x²-12x = -7
- -8x-4-4x²-12x = -7
- -8x-4x²-12x = -7+4
- -4x²-20x = -3
- -4x²-20x+3 = 0 multiply by -1 to get rid of the - signs with x
- 4x²+20x-3=0
4x²+20x+3=0 is a quadratic equation
Let Δ be our discriminant
- a= 4
- b= 20
- c= -3
Δ= 20²-4*4*(-3)
Δ=448 > 0 so we have two solutions for x
let x and x' be the solutions
- x =
= -5.145 ≈ -5.15
- x'=
= 0.145≈ 0.15
so the solutions are:
-5.15 and 0.15
Step-by-step explanation:
I hope this helps you answer your question and sorry if its wrong.
