Answer:
k=-5
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]6x^{2} +7x+k=0[/tex]
so
[tex]a=6\\b=7\\c=k[/tex]
substitute in the formula
[tex]x=\frac{-7\pm\sqrt{7^{2}-4(6)(k)}} {2(6)}[/tex]
[tex]x=\frac{-7\pm\sqrt{49-24k}} {12}[/tex]
so
[tex]x_1=\frac{-7+\sqrt{49-24k}} {12}[/tex]
[tex]x_2=\frac{-7-\sqrt{49-24k}} {12}[/tex]
Remember that
[tex]2x_1+3x_2=-4[/tex]
substitute
[tex]2(\frac{-7+\sqrt{49-24k}} {12})+3(\frac{-7-\sqrt{49-24k}} {12})=-4[/tex]
[tex](\frac{-14+2\sqrt{49-24k}} {12})+(\frac{-21-3\sqrt{49-24k}} {12})=-4[/tex]
Multiply by 12 both sides
[tex](-14+2\sqrt{49-24k})+(-21-3\sqrt{49-24k})=-48[/tex]
[tex]-35-\sqrt{49-24k}=-48[/tex]
[tex]\sqrt{49-24k}=48-35[/tex]
[tex]\sqrt{49-24k}=13[/tex]
squared both sides
[tex]49-24k=169\\24k=49-169\\24k=-120\\k=-5[/tex]
therefore
The equation is
[tex]6x^{2} +7x-5=0[/tex]
The roots are
[tex]x=\frac{-7\pm\sqrt{49-24(-5)}} {12}[/tex]
[tex]x=\frac{-7\pm\sqrt{169}} {12}[/tex]
[tex]x=\frac{-7\pm13} {12}[/tex]
[tex]x_1=\frac{-7+13} {12}=\frac{1} {2}[/tex]
[tex]x_2=\frac{-7-13} {12}=-\frac{5} {3}[/tex]
