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miv72 [106K]
2 years ago
11

Solve the radical equation the square root of 8x+9=x+2, how would you start it?

Mathematics
1 answer:
kobusy [5.1K]2 years ago
7 0

Answer:  I will start it by squaring both sides of the equation.  

The solution is  x = 5 or x=-1

Step-by-step explanation:

\sqrt{8x+9} =x+2     Square the left and right sides. The square root of 8x+9 squared is 8x+9  and the x+2 squared is x^2 + 4x +4. We will now have the new equation,

8x + 9 = x^{2} + 4x + 4   now set them equal zero first sutract x^2 from both sides

-x^2       -x^2

-x^2 + 8x + 9 = 4x +4      subtract 4x from both sides      

          -4x            -4x

-x^2 +4x + 9 = 4        Finally subtract 4 from both sides

               -4    -4

-x^2 + 4x +5 = 0    Remember the largest degree has a negative coefficient so divide them all by -1.

x^2 - 4x -5 = 0        Now find two numbers that their product is -5 and their sum is -4.     the numbers  1 and -5 works out.

Now rewrite the whole equation as,

x^2+ 1x -5x -5=0      Factor by grouping on the left side

x(x+1) -5(x+1)= 0      Factor out x+1

(x+1)(x-5) = 0         Now apply the zero product

x+1 = 0     or x-5 = 0

   -1    -1          +5   +5

x = -1   or  x = 5

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The quadratic trinomials in x and y on the left side of the equations are perfect squares.

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3 years ago
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3 years ago
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Given

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5 0
2 years ago
The equation $a^7xy-a^6y-a^5x=a^4(b^4-1)$ is equivalent to the equation $(a^mx-a^n)(a^py-a^2)=a^4b^4$ for some integers $m$, $n$
Ksivusya [100]
Let <span>simplify the equations a^7xy-a^6y-a^5x=a^4(b^4-1) and (a^mx-a^n)(a^py-a^2)=a^4b^4:
</span>
<span>1) a^7xy-a^6y-a^5x+a^4=a^4b^4
</span>
<span>and
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<span>2) a^{m+p}xy-a^{n+p}y-a^{m+2}x+a^{n+2}=a^4b^4.
</span>
<span>Equate the coefficients:
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</span><span>xy: m+p=7  \\ y: n+p=6 \\ x:  m+2=5 \\ 1: n+2=4.</span>
Then n=2 \\ p=4 \\ m=3 and mnp=24.
<span />
3 0
3 years ago
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