Which quadratic equation is equivalent to (x+2)2+5(x+2)-6=0

Mathematics

1 answer:

professor190 [17]3 years ago

6 0

$(x+2)^2+5(x+2)-6=0\\\\x^2+2\cdot x\cdot2+2^2+5\cdot x+5\cdot2-6=0\\\\x^2+4x+4+5x+4-6=0\\\\x^2+9x+2=0$

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2^5×8^4/16=2^5×(2^a)4/2^4=2^5×2^b/2^4=2^c A= B= C= Please I'm gonna fail math

aleksley [76]

9514 1404 393

Answer:

a = 3, b = 12, c = 13

Step-by-step explanation:

The applicable rules of exponents are ...

(a^b)(a^c) = a^(b+c)

(a^b)/(a^c) = a^(b-c)

(a^b)^c = a^(bc)

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You seem to have ...

$\dfrac{2^5\times8^4}{16}=\dfrac{2^5\times(2^3)^4}{2^4}\qquad (a=3)\\\\=\dfrac{2^5\times2^{3\cdot4}}{2^4}=\dfrac{2^5\times2^{12}}{2^4}\qquad (b=12)\\\\=2^{5+12-4}=2^{13}\qquad(c=13)$

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Additional comment

I find it easy to remember the rules of exponents by remembering that an exponent signifies repeated multiplication. It tells you how many times the base is a factor in the product.

$2\cdot2\cdot2 = 2^3\qquad\text{2 is a factor 3 times}$