Question

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

Answer 1

Answer:

$k=-5$

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$6x^{2} +7x+k=0$ 

so

$a=6\\b=7\\c=k$

substitute in the formula

$x=\frac{-7\pm\sqrt{7^{2}-4(6)(k)}} {2(6)}\\\\x=\frac{-7\pm\sqrt{49-24k}} {12}$

so

$x_1=\frac{-7+\sqrt{49-24k}} {12}\\\\x_2=\frac{-7-\sqrt{49-24k}} {12}$

Remember that

$2x_1+3x_2=-4$

substitute

$2(\frac{-7+\sqrt{49-24k}} {12})+3(\frac{-7-\sqrt{49-24k}} {12})=-4\\\\(\frac{-14+2\sqrt{49-24k}} {12})+(\frac{-21-3\sqrt{49-24k}} {12})=-4$

Multiply by 12 both sides

$(-14+2\sqrt{49-24k})+(-21-3\sqrt{49-24k})=-48\\\\-35-\sqrt{49-24k}=-48\\\\\sqrt{49-24k}=48-35\\\\\sqrt{49-24k}=13$

squared both sides

$49-24k=169\\24k=49-169\\24k=-120\\k=-5$

therefore

The equation is

$6x^{2} +7x-5=0$

The roots are

$x=\frac{-7\pm\sqrt{49-24(-5)}} {12}\\\\x=\frac{-7\pm\sqrt{169}} {12}\\\\x=\frac{-7\pm13} {12}\\\\x_1=\frac{-7+13} {12}=\frac{1} {2}\\\\x_2=\frac{-7-13} {12}=-\frac{5} {3}$