Question

9. Find the values of k for which the quadratic equation (3k + 1) x2²+ 2 (k + 1) x + 1 = 0 hasequal roots. Also, find the roots.
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Answer 1

Answer:

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Answer 2

$\rm\underline\bold{Answer \purple{\huge{\checkmark}}}$

The given equation is (3k+1) x² + 2(k+1) x+1=0

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This is of the form ax²+bx+c=0,

where a=3k +1, b=2(k+1) and c=1

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$\sf \: D=b²-4ac$

$\sf \implies {[2(k+1)]}^{2}-4 \times (3k+1) \times 1$

$\sf \implies \: 4(k² +2k+1)− 4(3k+1)$

$\sf \implies \: 4k² +8k+4-12k-4$

$\sf \implies \: 4k²-4k$

The given equation will have real and equal roots if D = 0.

$\sf \therefore \: 4k²-4k=0$

$\sf \implies \: 4k (k-1)=0$

$\sf \implies \: k=0 \: or \: k-1=0$

$\sf \implies \: k=0 \: or \: k=1$

Hence, 0 and 1 are the required values of k.

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