3x^2- 27 … you divide the whole thing by 3 to get x^2- 9 and that’s the furthest you can simplify
In a quadratic equation
q(x) = ax^2 + bx + c
The discriminant is = b^2 - 4ac
We have that discriminant = 3
If
b^2 - 4ac > 0, then the roots are real.
If
b^2 - 4ac < 0 then the roots are imaginary
<span>In
this problem b^2 - 4ac > 0 3 > 0 </span>
then
the two roots must be real
Answer:
see explanation
Step-by-step explanation:
Given
4
- 5a² + 1 = 0
Use the substitution u = a², then equation is
4u² - 5u + 1 = 0
Consider the product of the coefficient of the u² term and the constant term
product = 4 × 1 = 4 and sum = - 5
The factors are - 4 and - 1
Use these factors to split the u- term
4u² - 4u - u + 1 = 0 ( factor the first/second and third/fourth terms )
4u(u - 1) - 1(u - 1) = 0 ← factor out (u - 1) from each term
(u - 1)(4u - 1) = 0
Equate each factor to zero and solve for u
u - 1 = 0 ⇒ u = 1
4u - 1 = 0 ⇒ 4u = 1 ⇒ u = 
Convert u back into terms of a, that is
a² = 1 ⇒ a = ± 1
a² =
⇒ a = ± 
Solutions are a = ± 1 , a = ± 
Answer:
243 or 3^5
Step-by-step explanation:
You can do the product by multiplying starting from the left.
3 * 3 * 3 * 3 * 3 =
= 9 * 3 * 3 * 3
= 27 * 3 * 3
= 81 * 3
= 243
You can also change it to an exponential form.
3 * 3 * 3 * 3 * 3 = 3^5