State whether x = − k/2 is the root of the quadratic equation 2x² + (k-6)x - 3k = 0
Answers
Explanation:
Now putting both values of k in the equation if both values satisfy the equation means by putting these values if the equation equals to 0 then we can say that x is the root of the given equation.
At k =0
At k=2
Here ,both values of k satisfy the equation.
Hence ,x=-k/2 is the roots of the given equation
Answer:
:
Explanation:
= > 2 {x}^{2} + (k - 6)x - 3k = 0=>2x
2
+(k−6)x−3k=0
= > 2 {( - \frac{k}{2} )}^{2} + (k - 6) \times - \frac{k}{2} - 3k = 0=>2(−
2
k
)
2
+(k−6)×−
2
k
−3k=0
= > \frac{2k}{2} \frac{ - {k}^{2} + 6k}{2} - 3k = 0=>
2
2k
2
−k
2
+6k
−3k=0
= > k - \frac{ {k}^{2} + 6k }{2} - 3k=>k−
2
k
2
+6k
−3k
= > \frac{2k - {k}^{2} + 6k - 6k }{2} = 0=>
2
2k−k
2
+6k−6k
=0
= > 2k - {k}^{2} = 0=>2k−k
2
=0
= > k(2 - k) = 0=>k(2−k)=0
\bold{k = 0}k=0
\bold{k = 2}k=2
Now putting both values of k in the equation if both values satisfy the equation means by putting these values if the equation equals to 0 then we can say that x is the root of the given equation.
At k =0
2 {( \frac{0}{2} )}^{2} + (0 - 6) \times - \frac{k}{2} - 3k2(
2
0
)
2
+(0−6)×−
2
k
−3k
= > 0 + \frac{6k}{2} - 3k = 3k - 3k = 0=>0+
2
6k
−3k=3k−3k=0
At k=2
2 { (\frac{2}{2} )}^{2} + (2 - 6) \times - \frac{k}{2} - 3(2)2(
2
2
)
2
+(2−6)×−
2
k
−3(2)
= > 2 + ( - 4) \times - \frac{2}{2} - 3(2)=>2+(−4)×−
2
2
−3(2)
= > 2 + 4 - 6 = 6 - 6 = 0=>2+4−6=6−6=0
Here ,both values of k satisfy the equation.
Hence ,x=-k/2 is the roots of the given equation