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Brilliant_brown [7]
3 years ago
15

Without solving determine the number of real solutions for each quadratic equation

Mathematics
1 answer:
snow_lady [41]3 years ago
8 0

Step-by-step explanation:

b^2-4b+3=0

b²-3x-b+3=0

b(b-3)-1(b-3)=0

(b-3)(b-1)=0

either

b=3 or b=1

.

2n^2 + 7 = -4n + 5

2n²+4n+7-5=0

2n²+4n+2=0

2(n²+2n+1)=0

(n+1)²=0/2

:.n=-1

.

x - 3x^2 = 5+ 2x - x^2

0=5+ 2x - x^2-x +3x^2

0=5+x+2x²

2x²+x+5=0

comparing above equation with ax²+bx +c we get

a=2

b=1

c=5

x={-b±√(b²-4ac)}/2a ={-1±√(1²-4×2×5)}/2×1

={-1±√-39}/2

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Answer:

we conclude that the only option (a) is true.

Step-by-step explanation:

As we know that the multiples of 5 are the numbers which we get when we multiply by 5.

i.e.

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5×2=10

Here, 5 and 10 are multiples of 5.

Let p and q are integers that are multiples of 5.

Let us consider

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A number is divisible by 5 if it ends in 5 or 0.

i.e. 15/5 = 3

so p+q is divisible by 3 as there is no remainder left.

Therefore, option (a) is true.

Checking the other options:

(b) P –q is divisible by 10

As

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p-q=5-10

     = -5

Numbers that are divisible by 10 need to be even or divisible by 2 and divisible by 5.

As -5 is not divisible by 2.

So, option b is NOT true.

(c) P +q is divisible by 20

As

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q=10

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p-q=5+10

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Divisibility rule of  20  implies that the last two digits of the number are either  00  or divisible by  20 .

Therefore, P + q= 20 is not divisible by 20 as we don't get the whole number.

(d) P + q is divisible by 25​

As

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q=10

so

p-q=5+10

     = 15

p+q=15 is not divisible by 25 as it does not end with 00, 25, 50, or 75.

so, option d is NOT correct.

Therefore, we conclude that the only option (a) is true.

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