Question

Solve the following
kx²+(2k+4)x+9

Answer 1

Answer:

Here is ur answer↓

k=4and k=1

Step-by-step explanation:

Your Question- kx²+(2k+4)x+9 has equal roots find value of k?

Solution:-

Using Discriminant Formula↓

a=k b=(2k+4) c=9

d=b²-4ac

=>(2k+4)²-4×k×9=0

[°•° Roots Are Real d=0]

4k²+16k+16-36k=0

4k²-20k=-16

4k²-20k+16=0---÷by4

=>k²-5k+4=0

Sum=-5

Product=1×4=4

-4×-1=4 Also,-4-1=-5

k²-k-4k+4=0

k(k-1)-4(k-1)=0

k-1=0

k=1

k-4=0

k=4

Hope you Understood™✓

Answer 2

$\boxed{\blue{ \boxed{ \huge{ \red{Question}}}}}$

Find value of K for which the roots of the equation are equal

$\boxed{\blue{ \boxed{ \red{\huge{Answer}}}}}$

Given equation

$\green{kx ^{2} + (2k + 4)x + 9} = 0$

Here a = k ,b = 2k+4 , c = 9

So to find the value of K we can apply the descriminent formula that is

$\bold{ \boxed{ \blue{b ^{2} - 4ac}} }$

So the

Now putting the value of a.b and c in the discriminent formula

$= > (2k + 4) ^{2} - \: 4 \times 9 \times k = 0 \\ \\ = > 4k ^{2} \: + 16 + 16k - 36k = 0 \\ \\ = > 4k ^{2} - 20k + 16 = 0 \\ \\ = >k ^{2} - 5k + 4 = 0$

Now let us apply Sridhar acchariyas formula to find the possible value of k

$= > k = \frac{ - b \:plusminus\: \sqrt{b ^{2} - 4ac } }{2a}$

$= > k = \frac{5 \: plusminus\sqrt{25 - 16} }{2} \\ = > \boxed{ \red{ k = \frac{ 5 \: pluaminus\: 3}{2} }}$

Are the roots