Problem 1 :
If the equations x^{2} - ax + b = 0 and x^{2} - ex + f = 0 have one root in common and if the second equation has equal roots, then prove that ae = 2(b + f).
Solution :
Let α be the common root for both quadratic equations
Let β be the other root of the quadratic equations
Since the roots of the second equation will be same, α and α are the roots of the second equation.
x^{2} - ax + b = 0 Sum of roots = a α + β = a ----(1) Product of roots = b αβ = b β = b/α ----(2) |
x^{2} - ex + f = 0 Sum of roots = e α + α = e 2α = e α = e/2 ----(3) Product of roots = f α(α) = f α^{2} = f ----(4) |
Problem 2 :
Discuss the nature of roots of
−x^{2} + 3x + 1 = 0
Solution :
To find the nature of roots, we have to use the formula for discriminant.
Discriminant = b^{2} - 4ac
a = -1, b = -3 and c = 1
= (-3)^{2} - 4(-1) (1)
= 9 + 4
= 13 > 0
So, the roots are real and distinct.
Problem 3 :
Discuss the nature of roots of
4x^{2} − x − 2 = 0
Solution :
To find the nature of roots, we have to use the formula for discriminant.
Discriminant = b^{2} - 4ac
a = 4, b = -1 and c = -2
= (-1)^{2} - 4(4) (-2)
= 1 + 64
= 65 > 0
So, the roots are real and distinct.
Problem 4 :
Discuss the nature of roots of
9x^{2} + 5x = 0.
Solution :
To find the nature of roots, we have to use the formula for discriminant.
Discriminant = b^{2} - 4ac
a = 9, b = 5 and c = 0
= (5)^{2} - 4(9) (0)
= 25 > 0
So, the roots are real and distinct.
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