Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

If the coefficient of ${{x}^{2}}$ and the constant term of a quadratic equation have opposite signs, the quadratic equation has ________ roots.
( a ) Real and Distinct roots
( b ) real and equal roots
( c ) Imaginary roots
( d ) None of these

Answer
VerifiedVerified
579.3k+ views
Hint: We know that the quadratic equation $a{{x}^{2}}+bx+c=0$ can have real distinct roots, real repeated roots and imaginary roots. So, according to signs of a and c we will find out which condition we get as an outcome which will tell us about the nature of roots.

Complete step-by-step answer:
Let, quadratic equation be of form $a{{x}^{2}}+bx+c=0$ then, nature of roots are as follows
If \[{{b}^{2}}-4ac > 0\] then we have real and distinct roots.
If \[{{b}^{2}}-4ac=0\] then we have two same roots.
If \[{{b}^{2}}-4ac < 0\]then we have imaginary roots.
Now, in question it is given that coefficient of ${{x}^{2}}$and constant term of quadratic equation have opposite signs, then if coefficient of ${{x}^{2}}$is negative then constant term will be positive and if coefficient of ${{x}^{2}}$is positive then constant term will be negative that is if a > 0 then c < 0 and if a < 0 then c > 0.
So, quadratic equation will be of form either $-a{{x}^{2}}+bx+c=0$or $a{{x}^{2}}+bx-c=0$
Now, the product of coefficient of ${{x}^{2}}$ and constant will be equal to negative integers in case of both quadratic equations.
So, ac < 0
Multiplying both sides by 4, we get
\[4ac < 0\]
Multiplying both sides by -1, we get
\[-4ac > 0\] , as multiplying with negative terms will change the sign of terms on both sides as well as inequality.
Now, adding \[{{b}^{2}}\] on both sides we get
\[{{b}^{2}}-4ac > {{b}^{2}}\]
As, \[{{b}^{2}} > 0\]
So, \[{{b}^{2}}-4ac > 0\] …… ( i )
We above discussed that if coefficients of quadratic equation $a{{x}^{2}}+bx+c=0$ satisfies the condition \[{{b}^{2}}-4ac > 0\] then we have real and distinct roots.
And, for quadratic equations $-a{{x}^{2}}+bx+c=0$or $a{{x}^{2}}+bx-c=0$ we get condition …… ( i )
Hence, If the coefficient of ${{x}^{2}}$ and the constant term of a quadratic equation have opposite signs, the quadratic equation has real and distinct roots.

Note: For quadratic equation, cases to check nature of roots must be kept in mind while solving it gives an idea of relationship between coefficients of quadratic equation which helps in finding the nature roots . Do remember whenever you multiply an inequality by any negative integer, say –k always change signs on both sides and reverse the inequality also else the answer will get changed.