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Rzqust [24]
3 years ago
14

What values of b will cause 4x^2+bx+25=0 to have one real solution?

Mathematics
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

b=20 or b=-20

Step-by-step explanation:

Keep in mind the meanings of the values of the discriminant:

If b^2-4ac=0, then the quadratic will have only 1 solution (double root)

If b^2-4ac<0, then the quadratic will have no real solutions

If b^2-4ac>0, then the quadratic will have 2 unique solutions

In this case, to get one real solution, the discriminant must be set up as b^2-4ac=0. Setting up the equation we get:

b^2-4(4)(25)=0

b^2-400=0

b^2=400

b=20 or b=-20

So the values of b would have to be either 20 or -20

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