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

If $P(x) = ax^2 + bx + c$ and $Q(x) = -ax^2 + dx + c$ where $ac \neq 0$, then $P(x) \cdot Q(x) = 0$ has $(a, b, c, d \in \mathbb{R})$:

If $(x^2+5x+5)^{x+5}=1$,then the number of integers satisfying this equation is

How many real roots does the equation $e^{\sin x} - e^{-\sin x} - 4 = 0$ have?

If the roots of the quadratic equation $ax^2 - bx - c = 0$ are shifted by a constant value,which of the following expressions involving $a, b, c$ remains unchanged?

The number of distinct real solutions of the equation $x|x+4|+3|x+2|+10=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module