Question

.Find the value of k for which the quadratic equation has equal roots 3x^2-5x+k=0

Answer 1

Answer:

$\large{k = \frac{25}{12} }$

Step-by-step explanation:

$\text{for \: }a {x}^{2} + bx + c = 0 \text{\: to \: have \: equal \: roots} \\ \\ {b}^{2} = 4ac \\ \\ \text{given} \quad \: 3 {x}^{2} - 5x + k = 0 \\ \\ = > a = 3 \quad \: b = - 5 \quad \: c = k \\ \\ = > {( - 5)}^{2} = 4(3)(k) \\ \\ = > 12k = 25 \\ \\ = > k = \frac{25}{12}$

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Answer 2

Answer:

\large{k = \frac{25}{12} }k=

12

25

Step-by-step explanation:

\begin{gathered}\text{for \: }a {x}^{2} + bx + c = 0 \text{\: to \: have \: equal \: roots} \\ \\ {b}^{2} = 4ac \\ \\ \text{given} \quad \: 3 {x}^{2} - 5x + k = 0 \\ \\ = > a = 3 \quad \: b = - 5 \quad \: c = k \\ \\ = > {( - 5)}^{2} = 4(3)(k) \\ \\ = > 12k = 25 \\ \\ = > k = \frac{25}{12}\end{gathered}

for ax

2

+bx+c=0to have equal roots

b

2

=4ac

given3x

2

−5x+k=0

=>a=3b=−5c=k

=>(−5)

2

=4(3)(k)

=>12k=25

=>k=

12

25

hope it helps you mark brainliest

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