Question

Show that p(x)=x²-2x+5 has no real roots.

Answer 1

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p(x)=x^{2}-2 x+5

For finding roots to the polynomial, we need to equate it to 0.

p(x) = 0

x^{2}-2x+5=0

For a quadratic polynomial a x^{2}+b x+c=0 to have roots, the discriminant needs to be examined.

i.e., If b^{2}-4 a c>0, the roots are real.

If b^{2}-4 a c, the roots are imaginary.

If b^{2}-4 a c=0, the roots are equal.

In the given problem, x^{2}-2x+5=0 in which a=1, b=-2, c=5

b^{2}-4a c=2^{2}-4(1)(5)=4-20=-16

The roots are imaginary

There are no real roots for the given quadratic equation.

Hence, proved.

Answer 2

x2 - 2x + 5, when compared with the standard form of the quadratic equation,

a = 1, b = - 2, c = 5

Value of the discriminant

= b2 - 4ac= (-2)2 - 4(1) (5) = 4 - 20 = -16 The value of the discriminant is negative i.e less than zero. Hence, the quadratic equation x2 - 2x + 5 has no real zeroes or roots.

(☆☆Remember that zeroes and roots means the same here. ☆☆

• In this type of question first u need to find a, b, c.
• Then find the value of discriminant.
• If the value is negative means leass then zero then it will have no real roots.
• If the value of discriminant is more than zero then it will have 2 real roots or distinct or different roots.
• If the value of discriminant is equal to zero then it will have 2 real identical zeroes.) for ur knowledge.