Question Bank - Mathematics

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x2 - 6x + 9 = 0. The roots of this quadratic equation are

A.
Real and different
B.
Real and equal
C.
Imaginary
D.
Cannot be predicted

Solution:

Concept used: Consider a quadratic equation: \({\rm{a}}{{\rm{x}}^2} + {\rm{bx\;}} + {\rm{\;c}} = {\rm{\;}}0\)Discriminant = D = b2 “ 4ac1. ?D > 0 The roots are real and distinct(different)2. ?D = 0 The roots are real and equal? 3. D < 0The roots are Imaginary. Calculation:x2 - 6x + 9 = 0Comparing this with the standard form ax2 + bx + c = 0, we get a = 1, b = -6 and c = 9.D = b2 “ 4ac? D = (-6)2 - 4 × 1 × 9? D = 36 - 36 = 0D = 0? The roots of this quadratic equation are real and equal.Additional InformationConsider a quadratic equation: \({\rm{a}}{{\rm{x}}^2} + {\rm{bx\;}} + {\rm{\;c}} = {\rm{\;}}0\)Roots of quadratic equation are given by, ? \({\rm{x}} = \frac{{ - {\rm{b}} \pm \sqrt {{{\rm{b}}^2} - 4{\rm{ac}}} }}{{2{\rm{a}}}}\) The term \(({{\rm{b}}^2} - 4{\rm{ac}}\)) is called discriminant, it is important as it tells about the nature of roots.\({\rm{\;}}{{\rm{b}}^2} - 4{\rm{ac}} < 0\)There are no real roots\({\rm{\;}}{{\rm{b}}^2} - 4{\rm{ac}} = 0\)There is one real root\({\rm{\;}}{{\rm{b}}^2} - 4{\rm{ac}} > 0\)There are two real rootsA quadratic equation is graphically represented by a parabola with vertex located at the origin, below the x-axis, or above the x-axis. Therefore, a quadratic function may have one, two, or zero roots.If the quadratic function is always more than zero, that means, when graphed on coordinate system, the function never intersects the X-axis. That means that the quadratic equation has no real roots.

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