Find the roots of the following quadratic equ | vedclass

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

Question

(A) Comparing the given equation $x^{2}-5x-1=0$ with the standard form $ax^{2}+bx+c=0$,we get $a=1, b=-5, c=-1$.First,calculate the discriminant $D =

Vedclass · Accepted Answer

$\frac{5+\sqrt{29}}{2}$ and $\frac{5-\sqrt{29}}{2}$

Vedclass · Answer

(A) Comparing the given equation $x^{2}-5x-1=0$ with the standard form $ax^{2}+bx+c=0$,we get $a=1, b=-5, c=-1$.First,calculate the discriminant $D = b^{2}-4ac$:$D = (-5)^{2} - 4(1)(-1) = 25 + 4 = 29$.Since $D > 0$,the equation has two distinct real roots.The quadratic formula is given by $x = \frac{-b \pm \sqrt{D}}{2a}$.Substituting the values:$x = \frac{-(-5) \pm \sqrt{29}}{2(1)} = \frac{5 \pm \sqrt{29}}{2}$.Thus,the roots are $\frac{5+\sqrt{29}}{2}$ and $\frac{5-\sqrt{29}}{2}$.

Find the roots of the following quadratic equation by using the quadratic formula,if they exist: $x^{2}-5x-1=0$

Similar Questions

Find the roots of the quadratic equation by applying the quadratic formula:
$2 x^{2}-7 x+3=0$

Find the roots of the following quadratic equation by the method of completing the square: $x^{2}-8x+12=0$

Verify whether the given value of $x$ is a solution of the quadratic equation or not: $x + \frac{1}{x} = 3 \frac{1}{3}$; $x = \frac{1}{3}$.

The quadratic equation having roots $-1$ and $2$ is .... .

Solve the following quadratic equation using the quadratic formula,if the equation has a solution in $R$: $5x^{2} + 8x + 3 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module