Question

give an example of quadratic equation which has no real roots give an example of quadratic equation which have not real root ​

Answer 1

f(x) =

f(x) ={x}^{2}x

f(x) ={x}^{2}x 2

f(x) ={x}^{2}x 2

f(x) ={x}^{2}x 2 _3x+4 it has not real roots because discriminat is Negative

Answer 2

The quadratic equation x² + 2x + 2 = 0 has no real roots.

Explanation:

→ For a quadratic equation ax² + bx + c = 0, the roots are given as:

                                     $x=\frac{(-b\pm\sqrt{b^{2} -4ac}) }{2a}$

→ Here (b² - 4ac) is known as Discriminant (D). Depending upon the value of 'D', the roots can be real and distinct, real and equal or imaginary:

• If 'D' is greater than zero, then the roots are real and distinct.
• If 'D' is equal to zero, then the roots are real and equal.
• If 'D' is less than zero, then the roots are imaginary (non-real).

→ For the Quadratic equation: x² + 2x + 2 = 0:

• The values of coefficients are: a = 1, b = 2 and c = 2.

       $Discriminant(D)=b^{2} -4ac\\\\Discriminant (D)=(2)^{2}-4\times(1)\times(2) }\\\\Discriminant (D)=-4$

→ The discriminant (D) comes out to be less than zero. Hence the quadratic equation x² + 2x + 2 = 0 has no real roots.

Therefore the Quadratic equation x² + 2x + 2 = 0 has no real roots it only has imaginary roots.

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