The roots of the equation x^2 - 2sqrt{2}x + 1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For the quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.Here,$a = 1$,$b = -2\sqrt{2}$,and $c = 1$.$D = (-2

Vedclass · Accepted Answer

Real and distinct

Vedclass · Answer

(B) For the quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.Here,$a = 1$,$b = -2\sqrt{2}$,and $c = 1$.$D = (-2\sqrt{2})^2 - 4(1)(1) = 8 - 4 = 4$.Since $D > 0$,the roots are real and distinct.Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:$x = \frac{2\sqrt{2} \pm \sqrt{4}}{2} = \frac{2\sqrt{2} \pm 2}{2} = \sqrt{2} \pm 1$.Since $\sqrt{2} \pm 1$ are irrational,the roots are real and distinct.

The roots of the equation $x^2 - 2\sqrt{2}x + 1 = 0$ are:

Similar Questions

The product of all the real roots of $x^2-8x+9-\frac{8}{x}+\frac{1}{x^2}=0$ is

If $2 + i\sqrt{3}$ is a root of the equation $x^2 + px + q = 0$,where $p$ and $q$ are real,then $(p, q) = $

$4+\frac{1}{4+\frac{1}{4+\frac{1}{4+\ldots \infty}}} = $

If $x$ is real,then the minimum value of ${x^2} - 8x + 17$ is

Solve the equation $x^{2}+3x+9=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module