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

Let $b$ be a non-zero real number. Suppose the quadratic equation $2x^2 + bx + \frac{1}{b} = 0$ has two distinct real roots. Then:

The roots of $4x^2 + 6px + 1 = 0$ are equal,then the value of $p$ is

The sum of all the real roots of the equation $(e^{2x} - 4)(6e^{2x} - 5e^x + 1) = 0$ is

If $\alpha$ and $\beta$ are the roots of ${x^2} + px + q = 0$ and $\alpha + h$ and $\beta + h$ are the roots of ${x^2} + rx + s = 0$,then

Let $p, q \in \{1, 2, 3, 4\}$. The number of equations of the form $px^2 + qx + 1 = 0$ having real roots is

Vedclass Test Series

Exam Paper Generator

Online Exam Module