Let b be a non-zero real number- Suppose the quadratic equation 2 x 2- bx -1- b -0 has two distinct real roots- Then b-1-b-5-2B- b2-3 b-2C- b 2-1- b 2-4b-1-b-5-2

The correct option is B b^2 3b gt; 2If the equation 2x^2+bx+1b=0 has two distinct real roots, then Δ gt;0 ⇒ b^2 8b gt;0 ⇒b^3 8b gt;0 ⇒ b ∈ ( ∞,0) ∪ (2 ...

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Standard XII

Mathematics

Relation between Roots and Coefficients for Quadratic

Let b be a no...

Question

Let $b$ be a non-zero real number. Suppose the quadratic equation $2 x^{2} + b x + \frac{1}{b} = 0$ has two distinct real roots. Then

b + \frac{1}{b} > \frac{5}{2}

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b + \frac{1}{b} < \frac{5}{2}

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b^{2} - 3 b > - 2

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b^{2} + \frac{1}{b^{2}} < 4

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Solution

The correct option is C $b^{2} - 3 b > - 2$
If the equation $2 x^{2} + b x + \frac{1}{b} = 0$ has two distinct real roots, then $Δ > 0$

$\Rightarrow b^{2} - \frac{8}{b} > 0 \Rightarrow \frac{b^{3} - 8}{b} > 0 \Rightarrow b \in (- \infty, 0) \cup (2, \infty)$

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Let b be a non-zero real number. Suppose the quadratic equation 2x2+bx+1b=0 has two distinct real roots. Then

The correct option is C b2−3b>−2If the equation 2x2+bx+1b=0 has two distinct real roots, then Δ>0 ⇒b2−8b>0⇒b3−8b>0⇒b∈(−∞,0)∪(2,∞)

Let $b$ be a non-zero real number. Suppose the quadratic equation $2 x^{2} + b x + \frac{1}{b} = 0$ has two distinct real roots. Then

The correct option is C $b^{2} - 3 b > - 2$
If the equation $2 x^{2} + b x + \frac{1}{b} = 0$ has two distinct real roots, then $Δ > 0$

$\Rightarrow b^{2} - \frac{8}{b} > 0 \Rightarrow \frac{b^{3} - 8}{b} > 0 \Rightarrow b \in (- \infty, 0) \cup (2, \infty)$