CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Quantitative Aptitude

Equations

The polynomia...

Question

The polynomial $({a x}^{2} + b x + c) ({- a x}^{2} + d x + c) = 0, a c \neq 0$ has how many roots?

Open in App

Solution

$p (x) = a x^{2} + b x + c$

$q (x) = - a x^{2} + d x + c$

$p (x) * q (x) = 0$

$(a x^{2} + b x + c) * (- a x^{2} + d x + c) = 0$

Considering ${A x}^{2} + B x + C = 0$

Discriminating ${= B}^{2} - 4 A C$

In above equation $A$ and $C$ are negative.

$B^{2} - 4 A C$ will be positive and have real roots.

$p (x) = a x^{2} + b x + c$

Product of co-efficient of $x^{2}$ and constant term is $a c$

$q (x) = - a x^{2} + d x + c$

Product of co-efficient of $x^{2}$ and constant term is $- a c$

If $a c$ is positive then $- a c$ will be negative and vice-versa

$(a x^{2} + b x + c) * (- a x^{2} + d x + c) = 0$ will have atleast two real roots.

Suggest Corrections

thumbs-up

1

Similar questions

Q. The polynomial

(a x^{2} + b x + c) (a x^{2} - d x - c), a c \neq 0,

P (x) = a x^{2} + a x^{2} + b x + c

Q (x) = - a x^{2} + d x + c

a c \neq 0

then P(x) Q(x)=0 has at least two real roots.

P (x) = a x^{2} + b x + c

Q (x) = - a x^{2} + d x + c

a c \neq 0

, then the equation

P (x) . Q (x) = 0

P (x) = a x^{2} + b x + c

Q (x) = - a x^{2} + d x + c

\neq 0

p (x) Q (x) = 0

Zeroes of the quadratic polynomial ax² + bx + c and roots of the quadratic equation ax² + bx + c = 0 are__________

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Equations

QUANTITATIVE APTITUDE

Watch in App

explore_more_icon

Explore more

Other Quantitative Aptitude

Join BYJU'S Learning Program

CrossIcon