Question

It is known that x1 and x2 are roots of the equation 6x^2+7x+k=0, where 2x1+3x2=−4.
Find k.

Answer 1

k=-5

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√(49-24k)} {12}`

so

`x_1=\frac{-7+√(49-24k)} {12}`

`x_2=\frac{-7-√(49-24k)} {12}`

Remember that

`2x_1+3x_2=-4`

substitute

`2(\frac{-7+√(49-24k)} {12})+3(\frac{-7-√(49-24k)} {12})=-4`

`(\frac{-14+2√(49-24k)} {12})+(\frac{-21-3√(49-24k)} {12})=-4`

Multiply by 12 both sides

`(-14+2√(49-24k))+(-21-3√(49-24k))=-48`

`-35-√(49-24k)=-48`

`√(49-24k)=48-35`

`√(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√(49-24(-5))} {12}`

`x=\frac{-7\pm√(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}`