If the roots of x2-a x-b-0 are real and differ by a quantity which is less than cc-0 then b -A- a2-c2-4- a2-4B- -a2-c2-4- a2-4C- a2-4- a2-c2-40D- -a2-4- a2-c2-4

The correct option is A ( a^2 c^24, a^24 )Let the roots be α, β and α gt;β ⇒α + β = a, αβ = b ⇒ (α β)^2=a^2 4b ∵α β lt;c ⇒√(a^2 4b) lt;c ⇒ c^2 gt; a ...

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Byju's Answer

Standard XII

Mathematics

Cube Root of a Complex Number

If the roots ...

Question

If the roots of $x^{2} - a x + b = 0$ are real and differ by a quantity which is less than $c (c > 0)$ then $b \in$

(\frac{a^{2} - c^{2}}{4}, \frac{a^{2}}{4})

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(\frac{- a^{2} - c^{2}}{4}, \frac{a^{2}}{4})

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(\frac{a^{2}}{4}, \frac{a^{2} + c^{2}}{4})

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(- \frac{a^{2}}{4}, \frac{a^{2} + c^{2}}{4})

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Solution

The correct option is A $(\frac{a^{2} - c^{2}}{4}, \frac{a^{2}}{4})$
Let the roots be $α, β$ and $α > β$
$\Rightarrow α + β = a, α β = b$
$\Rightarrow (α - β)^{2} = a^{2} - 4 b$
$∵ α - β < c$
$\Rightarrow \sqrt{a^{2} - 4 b} < c$
$\Rightarrow c^{2} > a^{2} - 4 b$
$\Rightarrow b > \frac{a^{2} - c^{2}}{4} \dots (1)$
Now roots are real and distinct
$a^{2} - 4 b \geq 0$
$\Rightarrow b < \frac{a^{2}}{4} \dots (2)$
From equation $(1)$ and $(2)$
$b \in (\frac{a^{2} - c^{2}}{4}, \frac{a^{2}}{4})$

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Similar questions

Q. If the roots of the equation

x^{2} - a x + b = 0

are real and differ by a quantity which is less than c(c > 0), then b lies between

\frac{a^{2} - c^{2}}{4}

and

\frac{a^{2}}{4} .

Q. If each pair of equations

a_{1} x^{2} + b_{1} x + c_{1} = 0, a_{2} x^{2} + b_{2} x + c_{2} = 0

and

a_{3} x^{2} + b_{3} x + c_{3} = 0

has a common root, then show that
(i)

\frac{c_{1} a_{2} + c_{2} a_{1}}{c_{1} a_{2} - c_{2} a_{1}} + \frac{a_{1} a_{2} b_{3}}{a_{3} (a_{1} b_{2} - a_{2} b_{1})} = 0

(ii)

{(\frac{a_{1} b_{2} - a_{2} b_{1}}{a_{1} c_{2} - a_{2} c_{1}})}^{2} = \frac{a_{1} a_{2} c_{3}}{c_{1} c_{2} a_{3}}

Q. If

\frac{a}{b} = \frac{4}{5}

and

\frac{b}{c} = \frac{15}{16}

, then

\frac{c^{2} - a^{2}}{c^{2} + a^{2}}

is equal to __________.

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

Q. If

a + b + c = 0

then

\frac{1}{b^{2} + c^{2} - a^{2}} + \frac{1}{c^{2} + a^{2} - b^{2}} + \frac{1}{a^{2} + b^{2} - c^{2}}

is equal to

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If the roots of x2−ax+b=0 are real and differ by a quantity which is less than c(c>0) then b∈

The correct option is A (a2−c24, a24)Let the roots be α,β and α>β ⇒α+β=a, αβ=b ⇒(α−β)2=a2−4b ∵α−β<c ⇒√a2−4b<c ⇒c2>a2−4b ⇒b>a2−c24⋯(1) Now roots are real and distinct a2−4b≥0 ⇒b<a24⋯(2) From equation (1) and (2) b∈(a2−c24, a24)

If the roots of $x^{2} - a x + b = 0$ are real and differ by a quantity which is less than $c (c > 0)$ then $b \in$