Question

Determine whether the given quadratic equations have real roots and if so, find the roots.
√3x^2 + 10x - 8√3 = 0​

Answer 1

EXPLANATION.

• GIVEN

√3x^2 + 10x - 8√3 quadratic equation have

real roots , then find Root's

solutions

√3x^2 + 10x - 8√3 = 0

D = 0

b^2 - 4ac = 0

(10)^2 - 4 ( √3)(-8√3) = 0

100 + 96 = 0

196

Therefore,

equation has real roots

$\bigstar\boxed{\large{\bold{x = \frac{ - b \: \pm \: \sqrt{d} }{2a}}}}$

case = 1

$x \: = \frac{ - b + \sqrt{d} }{2a}$

$x \: = \frac{ - 10 + \sqrt{196} }{2 \sqrt{3} }$

$x = \frac{ - 10 + 14}{2 \sqrt{3} }$

$x = \frac{4}{2 \sqrt{3} }$

$x = \frac{2}{ \sqrt{3} }$

case = 2

$x = \frac{ - b - \sqrt{d} }{2a}$

$x = \frac{ - 10 - 14}{2 \sqrt{3} }$

$x = \frac{ - 24}{2 \sqrt{3} }$

$x = \frac{ - 12}{ \sqrt{3} }$

$x = \frac{2 \sqrt{3} }{ \sqrt{3} } = 2$

Answer 2

Determine whether the given quadratic equations have real roots and if so, find the roots.

√3x^2 + 10x - 8√3 = 0