Question

find the value of k for which the quadratic equation (K + 1) x2 - 6 ( K + 1) X + 3 (K + 9 ) = 0,k=/ -1 has equal roots ​

Answer 1

Please ignore the messy work. Hope it helps you!

Answer 2

Discriminant of $ax^2+2b'x+c=0$

where $2b'$ is an even number of a linear term : $D/4=b'^2-ac$

Let discriminant $D/4$.

$D/4=3^2(k+1)^2-3(k+1)(k+9)$

$=9k^2+18k+9-3k^2-30k-27$

$=6k^2-12k-18$

As the quadratic equation has equal roots, $D/4=0$.

$6k^2-12k-18=0$

$k^2-2k-3=0$

∴ $k=3$ or $k=-1$

As $k\neq 1$, the value of $k$ is 3.

∴ $k=3$

(+ Did you know?)

------------------------------ But why is $D/4=b'^2-ac$? ------------------------------

Apply quadratic formula to equation $ax^2+2b'x+c=0$.

$x=\frac{-2b'\pm \sqrt{4b'^2-4ac} }{2a} =\frac{-b'\pm\sqrt{b'^2-ac} }{a}$

Two roots are $\frac{-b' \pm \sqrt{b'^2-ac} }{a}$.

We know that discriminant comes from the square root.

If the number '$b'^2-ac$' is 0, we will always have $-\frac{b'}{a}$ as double solution.

So in this situation, $b'^2-ac$ is the discriminant.

---------------------------------------------TIP---------------------------------------------

As mentioned above, if you want to get the double solution,

you don't have to solve the quadratic equation again.

Simply apply $-\frac{b'}{a}$ after finding $k$. This will save great amount of time.

In this question, $a=k+1$, $b'=-3(k+1)$ and $k+1\neq 0$.

Hence, the double solution is 3!!!

. A N S W E R S . W I T H . Q U A L I T Y .

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