Question

x=-k/2 , 2x^2+(k-6)x-3k=0
with full explanation ​

Answer 1

Given: 2x2 + (k – 6)x – 3k = 0 and x =−k2 .

Given: 2x2 + (k – 6)x – 3k = 0 and x =−k2 .On substituting x =−k2 in L.H.S. of the given equation, we get:

Given: 2x2 + (k – 6)x – 3k = 0 and x =−k2 .On substituting x =−k2 in L.H.S. of the given equation, we get:

Given: 2x2 + (k – 6)x – 3k = 0 and x =−k2 .On substituting x =−k2 in L.H.S. of the given equation, we get: 2(−k2)2+(k−6)(−k2)−3k=2k24−k22+3k−3k=2k2−2k24=0

Given: 2x2 + (k – 6)x – 3k = 0 and x =−k2 .On substituting x =−k2 in L.H.S. of the given equation, we get: 2(−k2)2+(k−6)(−k2)−3k=2k24−k22+3k−3k=2k2−2k24=0

Given: 2x2 + (k – 6)x – 3k = 0 and x =−k2 .On substituting x =−k2 in L.H.S. of the given equation, we get: 2(−k2)2+(k−6)(−k2)−3k=2k24−k22+3k−3k=2k2−2k24=0 = >L.H.S. = R.H.S.

Given: 2x2 + (k – 6)x – 3k = 0 and x =−k2 .On substituting x =−k2 in L.H.S. of the given equation, we get: 2(−k2)2+(k−6)(−k2)−3k=2k24−k22+3k−3k=2k2−2k24=0 = >L.H.S. = R.H.S.Thus, x =−k2 satisfies the given equation.

Given: 2x2 + (k – 6)x – 3k = 0 and x =−k2 .On substituting x =−k2 in L.H.S. of the given equation, we get: 2(−k2)2+(k−6)(−k2)−3k=2k24−k22+3k−3k=2k2−2k24=0 = >L.H.S. = R.H.S.Thus, x =−k2 satisfies the given equation.Therefore, x =−k2 is a root of the given quadratic equation.

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Answer 2

Answer:

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Step-by-step explanation:

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