SOLUTION: One root of {{{ x^2-3x+k-2=0 }}} is -4. Find the value of k.
Here is what I tried at the start:
{{{ -4=-3+- sqrt(9-4(k-2))/2 }}}
But I am stuck from there on. Thank you for
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Polynomials-and-rational-expressions
-> SOLUTION: One root of {{{ x^2-3x+k-2=0 }}} is -4. Find the value of k.
Here is what I tried at the start:
{{{ -4=-3+- sqrt(9-4(k-2))/2 }}}
But I am stuck from there on. Thank you for
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Question 1143372: One root of is -4. Find the value of k.
Here is what I tried at the start:
But I am stuck from there on. Thank you for your help! Found 4 solutions by ankor@dixie-net.com, rothauserc, MathTherapy, ikleyn:Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! One root of is -4. Find the value of k.
:
continuing what you have
b becomes +3
multiply both sides by 2
subtract 3 from both sides
square both sides
9 - 4(k-2) = 121
-4k + 8 = 121 - 9
-4k = 112 - 8
-4k = 104
k = 104/-4
k = -26
:
:
check: the equation then is: y = x^- 3x - 28
factors to (x+4)(x-7)
You can put this solution on YOUR website! Since one root is -4, then x+4 must divide the polynomial
:
(x^2 -3x +k -2)/(x+4) = x-7 with remainder k +26
:
we know the remainder must be 0, then
:
k +26 = 0
:
k = -26
:
You can put this solution on YOUR website! One root of is -4. Find the value of k.
Here is what I tried at the start:
But I am stuck from there on. Thank you for your help!
You could use the quadratic equation formula, but I prefer the method below:
You may remember that:
Let's call the other root: R2. We can see from the quadratic equation that: a = 1; b = - 3, and c = k - 2
We then get:
We also get:
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AND an even easier method is the REMAINDER THEOREM method used by TUTOR @Rothauserc, which should actually be:
The EASIEST way to solve it is to substitute x= -4 into the equation and use the fact that "-4" is the root:
4^2 - 3*(-4) + k - 2 = 0
16 + 12 + k - 2 = 0
k + 26 = 0
k = -26. ANSWER