# Quadratic equations again

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• Nov 11th 2007, 09:33 AM
Coach
Quadratic equations again
can comeone please tell me why does the solution of the following equation not use the standard quadratic formula?

the question is
Given that the equation $(4k+1)x^2-(k+10)x+2k=0$ has repeated root calculate the possible values of the constant k

I solved for the discriminant =0, but when i was applying the quadratic formula to the simplified equation of the discriminant, and then tried to use the quadratic formula to solve it, nothing sensible came out. I graphed the equation and found out, that it gives the correct roots only when divided by a and not by 2a as is in the quadratic formula?

Can someone please help me?

Thank you!
• Nov 11th 2007, 09:59 AM
Soroban
Hello, Coach!

Your game plan is excellent.
. . Must be your algebra . . .

Quote:

Given that the equation $(4k+1)x^2-(k+10)x+2k\:=\:0$ has a repeated root,
calculate the possible values of the constant $k$,

A quadratic has repeated roots if its discriminant is zero.
. . The discriminant is: . $D \:=\:b^2-4ac$

We have: . $a \:=\:4k+1,\;\;b \:=\:-(k+10),\;\;c \:=\:2k$

Hence: . $D \;=\;(k+10)^2 - 4(4k+1)(2k) \;=\;100 + 12k - 31k^2$

We have the quadratic:. . $31k^2 - 12k - 100\:=\:0$

. . which factors: . $(k - 2)(31k + 50) \:=\:0$

. . and has roots: . $k \;=\;2,\:-\frac{50}{31}$
• Nov 11th 2007, 10:18 AM
Ana
Hi
Hi

it is said in the problem that you need to find the value for of k such that the whole equation has repeated root or in other words only one solution.This situation is only when the value of discriminant is equal to zero.Once you have done that you will get another quadratic equation in terms of k which you need to solve.

Cheers
Ana