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03 Feb 2023, 02:30
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B
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Difficulty:
75%
(hard)
Question Stats:
60% (02:17) correct
40%
(02:32)
wrong
based on 139
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What is the equation where roots are \((α^2 + 1)\) and \((β^2 + 1)\), if α and β are roots of \(ax^2 - bx + a = 0\)?
A. \(a^2x^2 - b^2x - b^2 = 0\)
B. \(a^2x^2 - b^2x + b^2 = 0\)
C. \(a^2x^2 + b^2x - b^2 = 0\)
D. \(a^2x^2 + b^2x + b^2 = 0\)
E. \(a^2x^2 + b^2x + 2b^2 = 0\)
A. \(a^2x^2 - b^2x - b^2 = 0\)
B. \(a^2x^2 - b^2x + b^2 = 0\)
C. \(a^2x^2 + b^2x - b^2 = 0\)
D. \(a^2x^2 + b^2x + b^2 = 0\)
E. \(a^2x^2 + b^2x + 2b^2 = 0\)
07 Feb 2023, 04:26
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What is the equation where roots are \((α^2+1)\) and \((β^2+1)\), if α and β are roots of \(ax^2−bx + a = 0\) ?
We know that if roots of the equation is m, n then the equation can be rewritten as: (x - m) * (x - n) = 0
= \(x^2 - (m+n)x + m*n = 0\)
= \(x^2 \) - (Sum of roots) x + Product of roots = 0
Now compare with above given equation \(ax^2−bx + a = 0\)
Since α and β are roots of above equation => Sum of roots, \(α + β = -(-\frac{b}{a}) = \frac{b}{a }\) and product of roots is \(α*β = \frac{a}{a}\) = 1.
Squaring both sides of α + β = (b/a)
\(α^2 + β^2 +2α*β = (\frac{b}{a})^2\)
or \(α^2 + β^2 +2 = (\frac{b}{a})^2\) .............eqn 1
Now, If roots are \(α^2 + 1\) and \(β^2 + 1 \)
then sum of roots = \(α^2 +1 + β^2 +1\)
= \(α^2 + β^2 + 2 = (\frac{b}{a})^2\) {from eqn 1}
and product of roots = \((α^2 +1)*(β^2 +1) = α^2*β^2 + α^2 + β^2 + 1 \\
= 1 + α^2 + β^2 + 1\\
= α^2 + β^2 + 2 = (\frac{b}{a})^2\) {from eqn 1}
therefore, the required equation is \(x^2 \) - (Sum of roots) x + Product of roots = 0
= \(a^2x^2 - b^2y^2 + b^2\) = 0
Answer B
We know that if roots of the equation is m, n then the equation can be rewritten as: (x - m) * (x - n) = 0
= \(x^2 - (m+n)x + m*n = 0\)
= \(x^2 \) - (Sum of roots) x + Product of roots = 0
Now compare with above given equation \(ax^2−bx + a = 0\)
Since α and β are roots of above equation => Sum of roots, \(α + β = -(-\frac{b}{a}) = \frac{b}{a }\) and product of roots is \(α*β = \frac{a}{a}\) = 1.
Squaring both sides of α + β = (b/a)
\(α^2 + β^2 +2α*β = (\frac{b}{a})^2\)
or \(α^2 + β^2 +2 = (\frac{b}{a})^2\) .............eqn 1
Now, If roots are \(α^2 + 1\) and \(β^2 + 1 \)
then sum of roots = \(α^2 +1 + β^2 +1\)
= \(α^2 + β^2 + 2 = (\frac{b}{a})^2\) {from eqn 1}
and product of roots = \((α^2 +1)*(β^2 +1) = α^2*β^2 + α^2 + β^2 + 1 \\
= 1 + α^2 + β^2 + 1\\
= α^2 + β^2 + 2 = (\frac{b}{a})^2\) {from eqn 1}
therefore, the required equation is \(x^2 \) - (Sum of roots) x + Product of roots = 0
= \(a^2x^2 - b^2y^2 + b^2\) = 0
Answer B
15 Feb 2023, 06:55
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Bunuel
for \((α^2 + 1) and (β^2 + 1)\), product of the root is:
\((α^2 + 1)*(β^2 + 1)=a^2b^2+a^2+b^2+1\)
The sum of the roots is \(a^2+1+b^2+1=a^2+b^2+2\)
then we have a and b as roots of the equation in the question, so product of the root =\( a*b=\frac{a}{a}=1\)
The sum of the roots is \(a+b=\frac{b}{a}\)
now we know a*b=1 we can replace the value in the product of the roots of the first equation
\((α^2 + 1)*(β^2 + 1)=a^2b^2+a^2+b^2+1\)
\(=1+a^2+b^2+1\)
\(=a^2+b^2+2\)
with that we can see that the sum and product of the root are equal, so that we need to find an equation from the answer choices that match that condition.
B is the only one that fulfill that requirement (try all options using sum=c/a and product=-b/a)
IMO B
