Question

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

Answer 1

Step-by-step explanation:

We are going to express

y in terms of x.

=> 2x + 3y = -4, y = (-4 - 2x)/3.

By Vieta's Formula,

We have SOR = -b/a and POR = c/a.

=> x + y = -7/6, xy = k/6.

When x + y = -7/6,

x + (-4 - 2x)/3 = -7/6.

=> (x - 4)/3 = -7/6

=> (x - 4) = -7/2

=> x = 4 - 7/2

=> x = 0.5.

When xy = k/6,

x(-4 - 2x)/3 = k/6.

Hence k = 6[x(-4 - 2x)/3]

= 2x(-4 - 2x) = 2(0.5)(-4 - 2(0.5)) = -5.

The value of k is -5.

Answer 2

Answer:

$k = -5$

Step-by-step explanation:

We have the quadratic equation:

$6x^2+7x+k=0$

Where x and y are the roots of the equation and:

$2x+3y=-4$

First, using the quadratic formula with a = 6, b = 7, and c = k, we can find the roots to be:

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

Simplify:

$\displaystyle x=\frac{-7\pm\sqrt{49-24k}}{12}$

So, our two roots are:

$\displaystyle x=\frac{-7+\sqrt{49-24k}}{12}\text{ and } y=\frac{-7-\sqrt{49-24k}}{12}$

For our first root, we can multiply both sides by 2.

And for our second root, we can multiply both sides by 3.

So, this produces:

$\displaystyle 2x=\frac{-7+\sqrt{49-24k}}{6}\text{ and } 3y=\frac{-7-\sqrt{49-24k}}{4}$

Since we are given that:

$2x+3y=-4$

By substitution:

$\displaystyle \Big(\frac{-7+\sqrt{49-24k}}{6}\Big)+\Big(\frac{-7-\sqrt{49-24k}}{4}\Big)=-4$

Remove the fractions by multiplying both sides by 12:

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

Combine like terms:

$-35-\sqrt{49-24k}=-48$

Adding 35 to both sides produces:

$-\sqrt{49-24k}=-13$

So:

$\sqrt{49-24k}=13$

Squaring produces:

$49-24k=169$

Therefore:

$-24k=120\Rightarrow k=-5$