Question

heya!!!
Maths AryaBhatta...
Solve the question in the attachment ​

Answer 1

Correct Question :-- For what value of k , the roots of Equation x²-4x + k = 0 are real and Equal ?

Concept used :--

If A•x^2 + B•x + C = 0 ,is any quadratic equation,

then its discriminant is given by;

D = B^2 - 4•A•C

• If D = 0 , then the given quadratic equation has real and equal roots.

• If D > 0 , then the given quadratic equation has real and distinct roots.

• If D < 0 , then the given quadratic equation has unreal (imaginary) roots...

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Solution :--

In Given Equation, x² - 4x + k = 0, we have ,

→ A = 1

→ B = (-4)

→ C = k

Since, roots Are real and Equal, we can say that ,

→ D = B² - 4AC = 0

Putting values now , we get,

→ (-4)² - 4*1*k = 0

→ 16 = 4k

Dividing both sides by 4, we get,

→ k = 4 .

Hence, value of k is 4, than Roots of Equation are real and Equal.

Answer 2

Correct question: For what value of k, will the roots of the equation $\sf x^2\ -\ 4x\ +\ k\ =\ 0$, will the equation have real and equal roots.

Given: $\sf x^2\ -\ 4x\ +\ k\ =\ 0$

To find: The value of k.

Answer:

For finding the value of k, we can choose to solve it by equating it to its discriminant (D).

D =  $\sf b^2\ -\ 4ac$.

There are three conditions before doing so.

• If the equation has real and equal roots, D = 0.
• If the equation has real and distinct roots, D > 0.
• If the equation has imaginary roots, D < 0.

Back to the question, it says that the roots are real and equal. So the first condition (D = 0) will be used.

Let's first get a, b and c from the equation.

$\sf x^2\ -\ 4x\ +\ k\ =\ 0\\\\From\ here,\\\\\a\ =\ 1,\ b\ =\ -\ 4\ and\ c\ =\ k.\\\\\\\\D\ =\ 0\\\\b^2\ -\ 4ac\ =\ 0\\\\(-\ 4)^2\ -\ 4\ \times\ 1\ \times\ k\ =\ 0\\\\16\ -\ 4k\ =\ 0\\\\16\ =\ 4k\\\\\\\dfrac{16}{4}\ =\ k\\\\\\\sf \bf k\ =\ 4$