For ax2-bx-c-0 - a0- b-0- roots are real if and only if

Given: nbsp;ax^2+bx+c=0 , a0, b=0 ⇒ ax^2+c=0 ⇒ x=±√( ca) Now, for real roots: ca gt;0 Which is possible if c & a have opposite signs. ⇒ c gt;0; a l ...

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Standard XII

Mathematics

Methods of Solving a Quadratic Equation

For ax2+bx+c=...

Question

For $a x^{2} + b x + c = 0, a \neq 0, b = 0$ , roots are real if and only if

$a < 0; c < 0$

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a > 0; c < 0

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a > 0; c > 0

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Solution

The correct option is B $a > 0; c < 0$
Given: $a x^{2} + b x + c = 0, a \neq 0, b = 0$
$\Rightarrow a x^{2} + c = 0 \Rightarrow x = \pm \sqrt{\frac{- c}{a}}$
Now, for real roots:
$\frac{- c}{a} > 0$
Which is possible if $c & a$ have opposite signs.
$\Rightarrow c > 0; a < 0 or c < 0; a > 0$

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For ax2+bx+c=0,a≠0,b=0, roots are real if and only if

The correct option is B a>0;c<0Given: ax2+bx+c=0,a≠0,b=0 ⇒ax2+c=0⇒x=±√−ca Now, for real roots: −ca>0 Which is possible if c & a have opposite signs. ⇒c>0;a<0 or c<0;a>0

For $a x^{2} + b x + c = 0, a \neq 0, b = 0$ , roots are real if and only if

The correct option is B $a > 0; c < 0$
Given: $a x^{2} + b x + c = 0, a \neq 0, b = 0$
$\Rightarrow a x^{2} + c = 0 \Rightarrow x = \pm \sqrt{\frac{- c}{a}}$
Now, for real roots:
$\frac{- c}{a} > 0$
Which is possible if $c & a$ have opposite signs.
$\Rightarrow c > 0; a < 0 or c < 0; a > 0$