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3 years ago

(3x+4)^2=14 what are the solutions?

1 answer:

7 0

$\bf (3x+4)^2=14\implies \sqrt{(3x+4)^2}=\pm\sqrt{14}\implies 3x+4=\pm\sqrt{14}
\\\\\\
3x=\pm \sqrt{14}-4\implies x=\cfrac{\pm \sqrt{14}-4}{3}\implies x=
\begin{cases}
\cfrac{\sqrt{14}-4}{3}\\\\
\cfrac{- \sqrt{14}-4}{3}
\end{cases}$

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3 years ago

Let R1 and R2 are the remainders when the polynomials x^3 + 2x^2 − 5ax − 7 and x^3 + ax^2 − 12x + 6 are divided by x + 1 and x −

Andru [333]

Answer:

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Step-by-step explanation:

Given: x³ + 2x² - 5ax - 7 and x³ + ax² - 12x + 6

Also, R1 + R2 = 6

in order to find the value of a:

Let p(x) = x³ + 2x² - 5ax - 7 and q(x) = x³ + ax² - 12x + 6

Using remainder theorem i.e if a polynomial p(x) is divisible by polynomial of form x - a then remainder is given by p(a).

Then,

R1 = p( -1 ) = (-1)³ + 2(-1)² - 5a(-1) - 7 = -1 + 2 + 5a - 7 = 5a - 6

R2 = q( 2 ) = 2³ + a(2)² - 12(2) + 6 = 8 + 4a - 24 + 6 = 4a - 10

Now,

R1 + R2 = 6

5a - 6 + 4a - 10 = 6

9a = 22

a=2.44

Therefore, Value of a is 2.44

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