1. ## solve for k

i don't know how to solve this, any help is appreciated.

1. find the values of k such that $\displaystyle 3x^2+kx+11=0$

2. find the value of so that the equation $\displaystyle 6x^2-(k+4)x-9=8x$ has roots that are numerically equal but opposite in sign.

thank you so much.

2. Originally Posted by princess_21
i don't know how to solve this, any help is appreciated.

1. find the values of k such that $\displaystyle 3x^2+kx+11=0$

2. find the value of so that the equation $\displaystyle 6x^2-(k+4)x-9=8x$ has roots that are numerically equal but opposite in sign.

thank you so much.
ax^2 + bx +c =0

1. Is it the exact question, anything about x ?? is it variable?

For having real roots what's the condition??
....
b^2 >4ac

2. sum of roots of a quadratic is -b/a
In our case -b/a= 0 ,

so -k-4 -8= 0 ,solve!!

3. Hello, princess_21!

Another approach to #2 . . .

2. Find the value of $\displaystyle k$ so that the equation $\displaystyle 6x^2-(k+4)x-9\:=\:8x$
has roots that are numerically equal but opposite in sign.
The equation simplifies to: .$\displaystyle 6x^2 - (k+12)x - 9 \:=\:0$

Consider the Quadratic Formula: .$\displaystyle x \;=\;\frac{-b \pm\sqrt{b^2-4ac}}{2a}$

. . The roots will be equal and opposite in sign when $\displaystyle b = 0.$

Hence: .$\displaystyle -(k+12) \:=\:0 \quad\Rightarrow\quad\boxed{ k \:=\:\text{-}\,12}$

ax^2 + bx +c =0

1. Is it the exact question, anything about x ?? is it variable?

For having real roots what's the condition??

....
b^2 >4ac

2. sum of roots of a quadratic is -b/a
In our case -b/a= 0 ,

so -k-4 -8= 0 ,solve!!
this is the exact question..
1. the values of k such that 3x^2+kx+11=0 has a unique real solution are _____ and _______.

5. Originally Posted by princess_21
this is the exact question..
1. the values of k such that 3x^2+kx+11=0 has a unique real solution are _____ and _______.

For having a single(real) root , we must have both the roots equal

So as the given roots by Soroban

Both the roots will be equal if

b^2 = 4ac

Put that thing in your equation

So
k^2 = 4*3 * 11

$\displaystyle k = \pm\sqrt{4*3*11}$