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How many real solutions does the equation \displaystyle -2x^2-6x+15=2x+5−2x 2 −6x+15=2x+5 have?

How many real solutions does the equation \displaystyle -2x^2-6x+15=2x+5−2x 2 −6x-example-1
User Jason Grife
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1 Answer

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13 votes

-2x² - 6x + 15 = 2x +5

Re-arrange the equation

-2x² - 6x -2x+ 15-5=0

-2x² -8x + 10 = 0

Multiply through by negative one

2x² + 8x - 10 =0

Now;

solve by factorization

Find two numbers such that its product give -20x² and its sum gives 8x and 8x by them

That is;

2x² + 10x - 2x - 10 = 0

2x(x+5) -2(x+5) = 0

(2x - 2) (x+5) = 0

Either 2x - 2 = 0

2x = 2

x= 1

Or

x+5 = 0

x=-5

Hence it has 2 real solutions

User Mike Lambert
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