If the roots of equation $x^{2} + p x + q = 0$ differ by 1, then

p^{2} = 4 q

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p^{2} = 4 q + 1

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p^{2} = 4 q - 1

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None of these

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Solution

The correct option is B $p^{2} = 4 q + 1$
Given equation $x^{2} + p x + q = 0$
Let the roots be $α, β$
$α + β = - p$ and $α β = q$
Now, $(α - β)^{2} = (α - β)^{2} - 4 α β$
$\Rightarrow (α - β)^{2} = p^{2} - 4 q$
Given $| α - β$ | $= 1$
$(α - β)^{2} = 1$
$p^{2} - 4 q = 1$

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State the following statement is True or False

If the roots of the equation $x^{2} + p x + q = 0$ differ by $1$ , then $p^{2} = 1 + 4 q$

p, q \in R

2 - \sqrt{3}

x^{2} + p x + q = 0

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