Question

Find the value of I got which the quadratic equation (k-12)x^2(k-12)x+2=0

Answer 1

Question:

Find the value of k for which the quadratic equation (k-12)x² + (k-12)x + 2 = 0 has equal roots.

Answer:

k = 12 , 20

Note:

• An equation of degree 2 is know as quadratic equation .

• Roots of an equation is defined as the possible values of the unknown (variable) for which the equation is satisfied.

• The maximum number of roots of an equation will be equal to its degree.

• A quadratic equation has atmost two roots.

• The general form of a quadratic equation is given as , ax² + bx + c = 0 .

• The discriminant of the quadratic equation is given as , D = b² - 4ac .

• If D = 0 , then the quadratic equation would have real and equal roots .

• If D > 0 , then the quadratic equation would have real and distinct roots .

• If D < 0 , then the quadratic equation would have imaginary roots .

Solution:

The given quadratic equation is ;

(k-12)x² + (k-12)x + 2 = 0

Clearly , we have ;

a = (k-12)

b = (k-12)

c = 2

We know that ,

The quadratic equation will have equal roots if its discriminant is equal to zero .

=> D = 0

=> (k-12)² - 4•(k-12)•2 = 0

=> (k-12)² - 8(k-12) = 0

=> (k-12)(k-12-8) = 0

=> (k-12)(k-20) = 0

=> k = 12 , 20

Hence,

The required values of k are 12 and 20 .

Answer 2

CORRECT QUESTION :

Find the value of k , which the quadratic equation (k-12)x^2+(k-12)x+2=0.

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Value\:of\:k=12\:and\:20}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given : }} \\ \tt{ : \implies (k - 12)x^{2} +( k - 12)x + 2 = 0 }\\ \\ \red{ \underline \bold{To \: Find : }} \\ \tt{: \implies value \: of \: k = ?}$

• According to given question :

$\tt{ : \implies ( k - 12)x^{2} +( k - 12)x + 2 = 0} \\ \\ \tt{\circ \: a = (k - 12)} \\ \tt{\circ \: b = (k - 12)}\\ \tt{\circ \:c = 2}\\ \bold{Discriminant \: = 0} \\ \\ \tt{: \rightarrow \: D \implies {b}^{2} - 4ac = 0 } \\ \\ \tt{: \implies {b}^{2} - 4ac = 0} \\ \\ \text{Putting \: the \: given \: values} \\ \tt{: \implies (k - 12)^{2} - 4 \times( k - 12) \times 2 = 0 } \\ \\ \tt{: \implies \: {k}^{2} + 144-24k - 8k + 96 = 0 } \\ \\ \tt{ : \implies \: {k}^{2} - 32k + 240 = 0 } \\ \\ \text{Solving \: quadratic \: by \: middle \: term \: spliting \: method} \\ \tt{ : \implies {k}^{2} - 12k - 20k + 240 = 0} \\ \\ \tt{: \implies k(k - 12) - 20(k - 12) = 0} \\ \\ \tt{: \implies (k - 12)(k - 20) = 0} \\ \\ \green{\tt{: \implies k = 12 \: and \: 20}}$