Answer:
Step-by-step explanation:
Factor out the GCF of each term. It looks like as far as the coefficients go, you can factor out a 4 from each, but nothing greater than a 4. As far as the x-terms go, you can factor out x-squared from each, but nothing greater than that. Factoring out 4 x-squared:
which is the last choice there. You cannot factor the quadratic over the real number solutions, because there are none. They are both imaginary. So you're done!
3x^2 + 9x + 6 = 0
3x^2 + 3x + 6x + 6 = 0
3x(x + 1) + 6(x + 1) = 0
(3x + 6)(x + 1) = 0
3x + 6 = 0 and x + 1 = 0
3x = -6 and x = -1
x = -2 and x = -1
<span>Can someone help me!!!
</span>
7.8
So, making x subject of the formula, x = [m - 2pt³ ±√(m² - 4pt²m)]/{2t⁵}
<h3>How to make x subject of the formula?</h3>
Since p = √(mx/t) - t²x
So, p + t²x = √(mx/t)
Squaring both sides, we have
(p + t²x)² = [√(mx/t)]²
p² + 2pt²x + t⁴x² = mx/t
Multiplying through by t,we have
(p² + 2pt²x + t⁴x²)t = mx/t × t
p²t + 2pt³x + t⁵x² = mx
p²t + 2pt³x + t⁵x² - mx = 0
t⁵x² + 2pt³x - mx + p²t = 0
t⁵x² + (2pt³ - m)x + p²t = 0
Using the quadratic formula, we find x.
where
- a = t⁵,
- b = (2pt³ - m) and
- c = p²t
Substituting the values of the variables into the equation, we have
So, making x subject of the formula,
Learn more about subject of formula here:
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The answer is:
a=5
b=9
c=0
Explanation:
1. You can use the quadratic formula to solve quadratic equations with the following standard form:
ax²+bx+c=0
2. The quadratic formula is:
x= (-b±√(b^2-4ac))/2a
3. You have the following quadratic equation:
<span>
5x</span>²<span>+9x=0
</span>
4. Then:
<span>
5x</span>²<span>+9x+0=0
a=5
b=9
c=0
5. When you substitute these values into the quadratic formula, you obtain:
x1=-9/5
x2=0</span>