If the discriminant of equation $6 x^{2} + b x + 2 = 0$ is $1$ then find the value of $^{'} b^{'}$ .

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Solution

Discriminant $Δ = b^{2} - 4 a c$
Given:Discriminant $= 1$
$\Rightarrow b^{2} - 4 a c = 1$ where $a = 6, b = b, c = 2$
$\Rightarrow b^{2} - 4 \times 6 \times 2 = 1$
$\Rightarrow b^{2} - 48 = 1$
$\Rightarrow b^{2} = 1 + 48 = 49$
$∴ b = \pm 7$

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