Let \alphaandβ be the two distinct roots of the equation 2x2−6x+k=0,suchthat(\alpha + \beta)andαβ are the distinct roots of the equation x2+px+p=0. Then, the value of 8(k−p) is
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Official Correct Answer: 6. The question says: Let α and β be the two distinct roots of the equation 2x² - 6x + k = 0, such that (α + β) and αβ are the distinct roots of the equation x² + px + p = 0. Then, the value of 8(k - p) is... and the correct answer is 6.
Hmm, okay. So, We find the value of 8(k - p). Let's break this down.
First, for a quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a, and the product is c/a. That's from Vieta's formulas. So, maybe I can use that here.
We have the given equations:
1. 2x² - 6x + k = 0. The roots are α and β.
2. x² + px + p = 0. The roots are (α + β) and αβ.
So, for the first equation, the sum of the roots α + β = -(-6)/2 = 6/2 = 3. And the product αβ = k/2.
For the second equation, the sum of the roots is (α + β) + αβ = -p/1 = -p. And the product is (α + β)(αβ) = p/1 = p.
The second equation is x² + px + p = 0. So, the sum of the roots is -p/1 = -p, and the product is p/1 = p.
So, from the second equation, sum of roots = (α + β) + αβ = -p.
And product of roots = (α + β)(αβ) = p.
So, now I have two equations:
1. (α + β) + αβ = -p
2. (α + β)(αβ) = p
But from the first equation, we know α + β = 3 and αβ = k/2.
So, substituting these into the second equation:
Sum of roots for the second equation: 3 + (k/2) = -p.
Product of roots for the second equation: 3*(k/2) = p.
So, now I have:
Equation A: 3 + (k/2) = -p
Equation B: 3*(k/2) = p
Hmm, so I can substitute p from Equation B into Equation A.
From Equation B: p = (3k)/2
Substitute into Equation A:
3 + (k/2) = - (3k)/2
Let me solve for k.
So, 3 + (k/2) + (3k)/2 = 0
Combine like terms:
3 + (k/2 + 3k/2) = 0
Which is 3 + (4k/2) = 0
Simplify:
3 + 2k = 0
So, 2k = -3
Thus, k = -3/2
Wait, that seems a bit odd.
From Equation A: 3 + (k/2) = -p
From Equation B: 3*(k/2) = p
So, substituting p from Equation B into Equation A:
3 + (k/2) = - (3k/2)
Yes, that's correct.
So, 3 + (k/2) + (3k/2) = 0
Which is 3 + (4k/2) = 0
Simplify: 3 + 2k = 0 => 2k = -3 => k = -3/2
Okay, Now, let's find p.
From Equation B: p = 3*(k/2) = 3*(-3/2)/2 = 3*(-3/2)/2? Wait, no.
Wait, p = 3*(k/2). Since k = -3/2, then p = 3*(-3/2)/2? Wait, no, that's not right.
Wait, p = 3*(k/2). So, k is -3/2, so k/2 is (-3/2)/2 = -3/4.
So, p = 3*(-3/4) = -9/4.
Wait, that seems a bit messy, but let's go with it.
So, k = -3/2 and p = -9/4.
Now, the question asks for 8(k - p).
So, Computing k - p first.
k - p = (-3/2) - (-9/4) = (-3/2) + 9/4.
Convert to common denominator:
(-6/4) + 9/4 = (3/4)
So, 8*(3/4) = 6.
Ah, so 8(k - p) = 6. That's the answer.
First, from the first equation, sum of roots α + β = 3, product αβ = k/2.
Second equation: sum of roots is (α + β) + αβ = 3 + (k/2) = -p.
Product of roots is (α + β)(αβ) = 3*(k/2) = p.
So, 3 + (k/2) = -p and 3*(k/2) = p.
Substituting p from the second equation into the first:
3 + (k/2) = - (3k/2)
So, 3 = - (3k/2) - (k/2) = - (4k/2) = -2k
Thus, 3 = -2k => k = -3/2.
Then, p = 3*(k/2) = 3*(-3/2)/2? Wait, no, p = 3*(k/2) = 3*(-3/2)/2? Wait, no, that's not correct.
Wait, p = 3*(k/2). Since k = -3/2, then k/2 = (-3/2)/2 = -3/4.
So, p = 3*(-3/4) = -9/4.
So, k = -3/2, p = -9/4.
Then, k - p = (-3/2) - (-9/4) = (-3/2) + 9/4 = (-6/4 + 9/4) = 3/4.
Thus, 8*(3/4) = 6.
Yes, I think I did everything right. So, the