In the quadratic equation x^2-2kx+3k+4=0, k is a real number.a)If the

1.	In the quadratic equation x^2-2kx+3k+4=0, k is a real number.a)If the equation has equal roots, find the values of k and solve the equationfor each value of kb)if the equation has no real roots, find the range of values of k.
Answer» Answer: hey MATE here is your answer PLS mark as brilliant Step-by-step EXPLANATION: ANSWER x2+kx+k=0 has only one SOLUTION. ∴ The discriminant b2−4ac=0 ∴k2−4k=0 ⇒k(k−4)=0 ∴k=0 or k=4 So for k=0 or k=4 the equation x2+4x+4=0 has only one solution.

In the quadratic equation x^2-2kx+3k+4=0, k is a real number.a)If the equation has equal roots, find the values of k and solve the equationfor each value of kb)if the equation has no real roots, find the range of values of k.

Answer»

Answer:

hey MATE here is your answer PLS mark as brilliant

Step-by-step EXPLANATION:

ANSWER

x2+kx+k=0 has only one SOLUTION.

∴ The discriminant b2−4ac=0

∴k2−4k=0

⇒k(k−4)=0

∴k=0 or k=4

So for k=0 or k=4 the equation x2+4x+4=0 has only one solution.

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In the quadratic equation x^2-2kx+3k+4=0, k is a real number.a)If the equation has equal roots, find the values of k and solve the equationfor each value of kb)if the equation has no real roots, find the range of values of k.

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