9x^2+4x+7=3x^2-8 How do I solve this?
9x^2+4x+7=3x^2-8 How do I solve this?
9x^2+4x+7=3x^2-8
subtract 3x² from each side:
6x²+4x+7=-8
add 8 to each side:
6x²+4x+15=0
It turns out that the equation does not have real roots, just complex roots:
x=(2±sqrt(86))/6
It turns out that the equation does not have real roots, just complex roots:
x=(-2±sqrt(86))/6
To solve this equation, you need to gather all the terms involving x on one side of the equation and move all the constant terms to the other side. The first step is to subtract 3x^2 from both sides to eliminate the variable with the highest exponent:
9x^2 + 4x + 7 - (3x^2 - 8) = 0
Simplifying this equation, we get:
6x^2 + 4x + 15 = 0
Now, we have a quadratic equation in the form of ax^2 + bx + c = 0, where a = 6, b = 4, and c = 15.
To solve this quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we have:
x = ( -4 ± √(4^2 - 4*6*15) ) / (2*6)
Simplifying further:
x = ( -4 ± √(16 - 360) ) / 12
x = ( -4 ± √(-344) ) / 12
Here, we encounter a problem because the square root of a negative number is not a real number. This means that the given equation has no real solutions. Therefore, the equation is unsolvable in real numbers.