Quadratic Equations

Quadratic Equation Practice Problem 1

The roots of the quadratic equation ( a + b -2c )x² – ( 2a – b – c)x +( a – 2b + c) = 0 are-

A. (a + b + c ) and (a – b- c)

B.  \( \frac{1}{2} \) and a – 2b + c

C. a – 2b + c and  \( \frac{1}{(a + b – 2c)} \)

D. None of the above.

Answer

Clearly we see that x=1 satisfies the equation

⇒x=1 is a root of the that equation

Let α be other root

Product of roots =1 × α =  \( \frac{a – 2b + c}{a + b – 2c} \)

other root =  α = \( \frac{a – 2b + c}{a + b – 2c} \)

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vee kwari

how do we state the nature of quadratics

Karthik
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Karthik

A polynomial can have real numbers as zero(ie rational and irrational)to decide it’s nature we can use the relationship between the coefficients and zero (by comparing sum and product of zeroes)

Vikas
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Vikas

HI

HELLOOO
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HELLOOO

Hi

Bhargav
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Bhargav

Hi

Hjgv
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If d 0 – real & distinct roots
d =0 – real & equal roots ….

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