The roots of a quadratic equation a x +b x +c =0 are (2+i √2)/3 and (2−i √2)/3 . Find the values of b and c if a = −1.

Answers

Answer 1

[tex]\begin{cases} x=\frac{2+i\sqrt{2}}{3}\implies 3x=2+i\sqrt{2}\implies 3x-2-i\sqrt{2}=0\\\\ x=\frac{2-i\sqrt{2}}{3}\implies 3x=2-i\sqrt{2}\implies 3x-2+i\sqrt{2}=0 \end{cases} \\\\\\ \stackrel{ \textit{original polynomial} }{a(3x-2-i\sqrt{2})(3x-2+i\sqrt{2})=\stackrel{ 0 }{y}} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{ \textit{difference of squares} }{[(3x-2)-(i\sqrt{2})][(3x-2)+(i\sqrt{2})]}\implies (3x-2)^2-(i\sqrt{2})^2 \\\\\\ (9x^2-12x+4)-(2i^2)\implies 9x^2-12x+4-(2(-1)) \\\\\\ 9x^2-12x+4+2\implies 9x^2-12x+6 \\\\[-0.35em] ~\dotfill\\\\ a(9x^2-12x+6)=y\hspace{5em}\stackrel{\textit{now let's make}}{a=-\frac{1}{9}} \\\\\\ -\cfrac{1}{9}(9x^2-12x+6)=y\implies \boxed{-x^2+\cfrac{4}{3}x-\cfrac{2}{3}=y}[/tex]


Related Questions

Which expressions are equivalent to 8(3/4y -2)+6(-1/2+4)+1

Answers

Answer: 6y + 6

Step-by-step explanation:

To simplify the expression 8(3/4y -2) + 6(-1/2+4) + 1, we can follow the order of operations (PEMDAS):

First, we simplify the expression within parentheses, working from the inside out:

6(-1/2+4) = 6(7/2) = 21

Next, we distribute the coefficient of 8 to the terms within the first set of parentheses:

8(3/4y -2) = 6y - 16

Finally, we combine the simplified terms:

8(3/4y -2) + 6(-1/2+4) + 1 = 6y - 16 + 21 + 1 = 6y + 6

Therefore, the expression 8(3/4y -2) + 6(-1/2+4) + 1 is equivalent to 6y + 6.

P, Q, R, S, T and U are different digits.
PQR + STU = 407

Answers

Step-by-step explanation:

There are many possible solutions to this problem, but one possible set of values for P, Q, R, S, T, and U is:

P = 2

Q = 5

R = 1

S = 8

T = 9

U = 9

With these values, we have:

PQR = 251

STU = 156

And the sum of PQR and STU is indeed 407.

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