Question

Find the value of a for which the quadratic equation (a+3) x2

– 4 (a+3)x +16 = 0

has two equal roots. ​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

a=1

Step-by-step explanation:

for real and equal roots b²-4ac=0

now substitute the values a, b and c

(-4(a+3))²-4(a+3)(16)=0

16(a+3)²-64(a+3)=0

let us say that a+3 be some y

16y²-64y=0

16y²=64y

y²=4y

y²/y=4

∴y=4

∵y=a+3=4

a=4-3

∴∴a=1

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