Which two values of x are roots of the polynomial below 4x^2-6x+1

Mathematics

1 answer:

goblinko [34]2 years ago

7 0

Step-by-step explanation:

Just apply the quadratic formula $x=\frac{-b \left \ {{+} \atop {-}} \right. \sqrt{b^{2}-4ac} }{2a}$

Your equation is $4x^2-6x+1$
Every quadratic equation follows the general form $ax^2+bx+c$

In this case $a=4,b=-6,c=1$
Substitute into the quadratic formula:
$x=\frac{-(-6) \left \ {{+} \atop {-}} \right. \sqrt{(-6)^{2}-4(4)(1)} }{2(4)}\\x=\frac{6 \left \ {{+} \atop {-}} \right. \sqrt{36-16} }{8}\\x=\frac{6 \left \ {{+} \atop {-}} \right. \sqrt{20} }{8}$ Now take note that the square root of any number gives a (+) and (-) result, so two different answers
$x=\frac{6 + \sqrt{20} }{8} ,and\\x=\frac{6 - \sqrt{20} }{8}$ are the answers

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How do you solve this problem?

enot [183]

Easy, it says "for ANY positive integer" so just test any positive integer

remember that n! means times all integers from 1 to that number n

lets try 1

(1+1)!/(1!)-1=
(2!)/(1!)-1=
2/1-1=
2-1=
1

if you don't believe me, try 2
(2+1)!/(2!)-2=
(3!)/(2!)-2=
(6)/(2)-2=
3-2=
1

te answer is 1, B

and number 9
easy, remember the exponential law
(x^m)(x^n)=x^(m+n)
jsut add the exponents
first gropu like bases
(r^2r^2/3)(t^1/2t^-3/2)
add bases
(r^2 and 2/3)(t^-1)=
(r^2∛(r^2))(1/t)=
$\frac{r^{2} \sqrt[3]{r^{2}} }{t}$