Question

√3x²+10x-8√3=0 solve the x

Answer 1

here you go ☆☆

let's first verify if it has roots,

▪${b}^{2} - 4ac$

▪${10}^{2} - 4 \times \sqrt{3} \times ( - 8 \sqrt{3} )$

▪$100 + 32 \times 3$

▪$100 + 96$

▪196>0

so roots are real

•as we know,

▪$x = \frac{ - b ( + - )\sqrt{ {b}^{2} - 4 ac} }{2a}$

▪$x = \frac{ - 10( + - ) \sqrt{196} }{2 \times \sqrt{3} }$

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

▪$x = \frac{4}{2 \sqrt{3} } .. \frac{ - 24}{2 \sqrt{3} }$

▪$x = \frac{2}{ \sqrt{3} } ... \frac{ - 12}{ \sqrt{3} }$


hope it helps you..........✌✌

Answer 2

√3x²+10x-8√3=0

√3x²+12x-2x-8√3=0

√3x(x+4√3)-√2(x+4√3)=0

(x+4√3)(√3x-√2)=0

x+4√3=0 ... x= -4√3

√3x-√2=0 ... x=√2/√3