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# answer was not clearly Stated last Time!

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Let a be a real number for which there exists a unique value of b such that the quadratic equation x^2 + 2bx + (a-b) = 0 has one real solution. Find a.

Mar 25, 2020

#2
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I plan to use the quadratic formula on this equation:   x2 + 2bx + (a - b)  =  0

using     a = 1     b = 2b     c = a - b

For the equation to have only one real solution  b2 - 4ac  must be zero.

Replacing:     (2b)2 - 4(1)(a - b)  =  0

4b2 - 4a + 4b  =  0

-4a  =  -4b2 - 4b

Diviing by -4:                           a  =  b2 + b

I hope that the variables in the equation did not confuse you with the variables in the quadratic formula.

Mar 25, 2020

#1
+1

Hint:

One root means that the equation has to be in the form of (x+#)^2=0.

Hope this helped!

Mar 25, 2020
#2
+1

I plan to use the quadratic formula on this equation:   x2 + 2bx + (a - b)  =  0

using     a = 1     b = 2b     c = a - b

For the equation to have only one real solution  b2 - 4ac  must be zero.

Replacing:     (2b)2 - 4(1)(a - b)  =  0

4b2 - 4a + 4b  =  0

-4a  =  -4b2 - 4b

Diviing by -4:                           a  =  b2 + b

I hope that the variables in the equation did not confuse you with the variables in the quadratic formula.

geno3141 Mar 25, 2020