Question

Assuming that √3 is an irrational number, prove that 5 √3 – 7 is an irrational number.

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Answer 1

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.

Answer 2

This contradicts our assumption that p and q have no common factors other than 1. Therefore, our initial assumption that 5√3 - 7 is a rational number must be false. Therefore, 5√3 - 7 is an irrational number.

To prove that 5√3 - 7 is irrational, we will use the proof by contradiction. We will assume that 5√3 - 7 is a rational number and then show that it leads to a contradiction.

So, let's assume that 5√3 - 7 is a rational number. This means that we can write 5√3 - 7 as a ratio of two integers p and q, where q is not equal to 0 and p and q have no common factors other than 1. We can express this as:

5√3 - 7 = p/q where p and q are integers with no common factors other than 1.

Now, let's isolate √3 in this equation and square both sides of the equation:

5√3 = p/q + 7 (adding 7 to both sides)

25 x 3 = (p/q)^2 + 2(7)(p/q) + 49 (squaring both sides)

75 = p^2/q^2 + 14p/q + 49

Subtracting 49 from both sides, we get:

26 = p^2/q^2 + 14p/q

Now, let's consider the first term p^2/q^2. Since p and q have no common factors other than 1, both p and q must be relatively prime to 3 (otherwise, 3 would be a common factor). Therefore, p^2 and q^2 are also relatively prime to 3. This means that p^2/q^2 cannot be equal to 3 or any multiple of 3.

However, the left-hand side of the equation is equal to 26, which is not divisible by 3. Therefore, the right-hand side must also not be divisible by 3. But the only way that p^2/q^2 + 14p/q could be not divisible by 3 is if p and q are both not divisible by 3.

So, we have shown that p and q have no common factors other than 1 and are both not divisible by 3. But now we can use this information to reach a contradiction.

Recall that 5√3 - 7 = p/q. We can multiply both sides of this equation by q and get:

5√3q - 7q = p

Since q is not divisible by 3, 7q is also not divisible by 3. But 5√3q is clearly divisible by 3, since √3 is irrational and cannot cancel out the factor of 3. Therefore, p is the sum of a number that is divisible by 3 (5√3q) and a number that is not divisible by 3 (7q), which means that p must be divisible by 3.

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