Obtain the roots of the following quadratic e | vedclass

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

Question

(A) The given quadratic equation is $\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0$.Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = \

Vedclass · Accepted Answer

$-4 \sqrt{3}, \frac{2}{\sqrt{3}}$

Vedclass · Answer

(A) The given quadratic equation is $\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0$.Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = \sqrt{3}$,$b = 10$,and $c = -8 \sqrt{3}$.The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.First,calculate the discriminant $D = b^{2} - 4ac = (10)^{2} - 4(\sqrt{3})(-8 \sqrt{3}) = 100 + 32(3) = 100 + 96 = 196$.Now,substitute the values into the formula:$x = \frac{-10 \pm \sqrt{196}}{2 \sqrt{3}} = \frac{-10 \pm 14}{2 \sqrt{3}}$.Taking the positive sign: $x = \frac{-10 + 14}{2 \sqrt{3}} = \frac{4}{2 \sqrt{3}} = \frac{2}{\sqrt{3}}$.Taking the negative sign: $x = \frac{-10 - 14}{2 \sqrt{3}} = \frac{-24}{2 \sqrt{3}} = \frac{-12}{\sqrt{3}} = -4 \sqrt{3}$.Thus,the roots are $-4 \sqrt{3}$ and $\frac{2}{\sqrt{3}}$.

Obtain the roots of the following quadratic equation using the quadratic formula: $\sqrt{3} x^{2} + 10 x - 8 \sqrt{3} = 0$.

Similar Questions

If the roots of the quadratic equation $5x(2x - 3) - 4 = 0$ exist,find them using the method of completing the square.

At $t$ minutes past $2\, pm$, the time needed by the minute hand of a clock to show $3\, pm$ was found to be $3\, minutes$ less than $\frac{t^{2}}{4}$ $minutes$. Find $t$.

State whether the quadratic equation $x(1-x)-2=0$ has two distinct real roots. Justify your answer.

If the roots of the quadratic equation $3x^{2} + 2\sqrt{5}x - 5 = 0$ exist,find them using the method of completing the square.

The equation $9x^2 + 30x + 25 = 0$ has .............. in $R$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module