It is known that x1 and x2 are roots of the equation [tex]6x^{2} +7x+k=0[/tex]
, where [tex]2x_{1} +3x_{2} =-4[/tex]. Find k.

Respuesta :

Answer:

[tex]k=-5[/tex]

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)}\\\\x=\frac{-7\pm\sqrt{49-24k}} {12}[/tex]

so

[tex]x_1=\frac{-7+\sqrt{49-24k}} {12}\\\\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\\\\(\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\\\\-35-\sqrt{49-24k}=-48\\\\\sqrt{49-24k}=48-35\\\\\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}\\\\x=\frac{-7\pm\sqrt{169}} {12}\\\\x=\frac{-7\pm13} {12}\\\\x_1=\frac{-7+13} {12}=\frac{1} {2}\\\\x_2=\frac{-7-13} {12}=-\frac{5} {3}[/tex]

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