If roots of equation of x2- x- 1-0 are a - b and roots of x2-px- q-0 are ab- ba then value of p-q is

Consider the first equation, we can write, a+b= 1, ab=1 Now, p = ab + ba=a^2+b^2ab ⇒ nbsp;p= (a+b)^2 2ab1 ⇒ nbsp;p = ( 1)^2 2· 11 ⇒ nbsp;p = 1 q= ab· ...

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Byju's Answer

Standard XI

Mathematics

Relations Between Roots and Coefficients

If roots of e...

Question

If roots of equation of $x^{2} + x + 1 = 0$ are $a, b$ and roots of $x^{2} + p x + q = 0$ are $\frac{a}{b}, \frac{b}{a}$ then value of $p + q$ is

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\frac{\sqrt{2} + 1}{2}

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Solution

The correct option is C $2$
Consider the first equation, we can write,
$a + b = - 1, a b = 1$
Now,
$- p = \frac{a}{b} + \frac{b}{a} = \frac{a^{2} + b^{2}}{a b} \Rightarrow p = - \frac{(a + b)^{2} - 2 a b}{1} \Rightarrow p = - \frac{(- 1)^{2} - 2 \cdot 1}{1} \Rightarrow p = 1$
$q = \frac{a}{b} \cdot \frac{b}{a} = 1$

Hence $p + q = 2$

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Similar questions

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If roots of equation of x2+x+1=0 are a, b and roots of x2+px+q=0 are ab,ba then value of p+q is

The correct option is C 2Consider the first equation, we can write, a+b=−1, ab=1 Now, −p=ab+ba=a2+b2ab⇒p=−(a+b)2−2ab1⇒p=−(−1)2−2⋅11⇒p=1 q=ab⋅ba=1 Hence p+q=2

If roots of equation of $x^{2} + x + 1 = 0$ are $a, b$ and roots of $x^{2} + p x + q = 0$ are $\frac{a}{b}, \frac{b}{a}$ then value of $p + q$ is