21. If not root of the quadratic equation 2x + x – 6 = 0 is 2, find the value of k. Also, find the other root​

2 thoughts on “21. If not root of the quadratic equation 2x + x – 6 = 0 is 2, find the value of k. Also, find the other root​”

1. Correct Question:

   If one root of the quadratic equation 2x² + kx – 6 = 0 is 2, find the value of k. Also find the other root.

   Solution:

   Let us assume that the other root of the given quadratic equation is m

   then two roots of the equation are: 2 and m

   Comparing 2x² + x – 6 = 0 with ax² + bx + c = 0

   we will get, a = 2, b = k, c = –6

   Now we know the relation between roots and coefficient of the quadratic equation, so

   Sum of roots = 2 + m = -b/a = -k/2

   so, 4 + 2m = –k ___equation(1)

   also,

   product of roots = 2m = c/a = -6/2 = -3

   so, 2m = –3 ___equation(2)

   putting value of 2m from equation(2) into equation(1)

   ➡ 4 + 2m = -k

   ➡ 4 – 3 = -k

   ➡ k = –1

   and putting value of k in equation(1)

   ➡ 4 + 2m = -k

   ➡ 4 + 2m = -(-1)

   ➡ 4 + 2m = 1

   ➡ 2m = -3

   ➡ m = –3/2

   so, value of k is -1 and other root is -3/2.

   Reply

2. [tex]\underline{\bf{Correct \: Question \: :-}} \\ [/tex]

   If root of the quadratic equation 2x² + kx – 6 = 0 is 2, find the value of k. Also, find the other root.

   Methods for doing such questions :

   [tex]\bullet[/tex]By Using Splitting Middle Term Method

   [tex]\bullet[/tex]By Using Quadratic Formula

   Now, Firstly

   Steps for Splitting Middle Term Method :

   ★ ➤ We have to split the middle term of equation, so that their sum is equal to the middle term and their product is equal to the product of coefficient of x² and constant term.

   Let’s solve it :

   [tex]\dashrightarrow\:\:\sf 2 \times ( 2 )^{2} + k × 2 – 6 = 0 \\ [/tex]

   [tex]\dashrightarrow\:\:\sf 8 + 2k – 6 = 0 \\ [/tex]

   [tex]\dashrightarrow\:\:\sf 2 + 2k = 0 \\ [/tex]

   [tex]\dashrightarrow\:\:\sf k = \dfrac{-2}{2} \\ [/tex]

   [tex]\dashrightarrow\:\:\sf k = – 1 \\ [/tex]

   [tex]\boxed{ \bf{k = – 1}}\\ [/tex]

   Now , By using Splitting method :

   [tex]\rightarrow\:\:\sf 2x^{2} – x – 6 = 0 \\ [/tex]

   [tex]\rightarrow\:\:\sf 2x^{2} – ( 4 – 3 )x – 6 = 0 \\ [/tex]

   [tex]\rightarrow\:\:\sf 2x^{2} – 4x + 3x – 6 = 0 \\ [/tex]

   [tex]\rightarrow\:\:\sf 2x ( x – 2 ) + 3 ( x – 2 ) = 0 \\ [/tex]

   [tex]\rightarrow\:\:\sf x – 2 = 0 \\ [/tex]

   [tex]\rightarrow\:\:\sf x = 2 \\ [/tex]

   or

   [tex]\rightarrow\:\:\sf 2x + 3 = 0 \\ [/tex]

   [tex]\rightarrow\:\:\sf x = \dfrac{-3}{2} \\ [/tex]

   Hence, The other root is :

   [tex]\boxed{\therefore \sf{x = \dfrac{-3}{2} }}\\ [/tex]

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   Alternate Method :

   [tex]\large\mathtt{|| \: Quadratic \; \; Formula \: ||} \\ [/tex]

   [tex]\;\tt{\rightarrow\;\; \dfrac{-b\:\pm\:\sqrt{b^{2}\:-\:4ac}}{2a}}[/tex]

   Here,

   • 2x² – x – 6 = 0

   By comparing it with the general form of quadratic equation which is ,

   • [tex]\boxed{\boxed{\sf{ax^{2} + bx + c = 0}}} \\ [/tex]

   ⠀⠀[tex]\dag[/tex] a = 2

   ⠀⠀[tex]\dag[/tex] b = (-1)

   ⠀⠀[tex]\dag[/tex] c = (-6)

   Finding the Discriminant first, which is nothing but :

   [tex]\bigstar {\sf { D = b^{2} – 4 ac}} \\ \\ [/tex]

   Substituting the respective values here, we get :-

   [tex]\implies\sf D = b^{2} – 4 ac \\ \\ [/tex]

   [tex]\implies\sf D = (-1)^{2} – 4 \times 2 \times (-6) \\ \\ [/tex]

   [tex]\implies\sf D = 1 – 8 \times (-6) \\ \\ [/tex]

   [tex]\implies\sf D = 1 + 48 \\ \\ [/tex]

   [tex]\rightarrow [/tex] D = 49

   Plugging the values now ,

   [tex]\;\sf{\rightarrow\;\; \dfrac{-b\:\pm\:\sqrt{b^{2}\:-\:4ac}}{2a}} \\ \\ [/tex]

   [tex]\;\sf{\rightarrow\;\; \dfrac{- (-1) \:\pm\:\sqrt{49}}{2(2)}} \\ \\ [/tex]

   [tex]\;\sf{\rightarrow\;\; \dfrac{1 \:\pm\:\sqrt{49}}{4}} \\ \\ [/tex]

   [tex]\;\sf{\rightarrow\;\; \dfrac{1 \:\pm\:{7}}{4}} \\ \\ [/tex]

   Now,

   [tex]\;\sf{\rightarrow\;\; \dfrac{1 \: + \:{7}}{4}} \\ \\ [/tex]

   [tex]\;\sf{\rightarrow\;\; \dfrac{{8}}{4}} \\ \\ [/tex]

   [tex]\;\sf{\rightarrow\;\; 2 } \\ \\ [/tex]

   Then,

   [tex]\;\sf{\rightarrow\;\; \dfrac{1 \: – \:{7}}{4}} \\ \\ [/tex]

   [tex]\;\sf{\rightarrow\;\; \dfrac{{-6}}{4}} \\ \\ [/tex]

   [tex]\;\sf{\rightarrow\;\; \dfrac{{-3}}{2}} \\ \\ [/tex]

   Hence, The other root is :

   [tex]\boxed{\therefore \sf{x = \dfrac{-3}{2} }}\\ [/tex]

   _________________________

   [tex]\underline{\underline{\maltese\:\: \textbf{\textsf{Quadratic \: Equation}}}}[/tex]

   [tex]\bigstar[/tex] If p(x) is a quadratic polynomial , then p(x) = 0 is called a quadratic equation .

   [tex]\underline{\underline{\maltese\:\: \textbf{\textsf{Roots \: of \: Quadratic \: Equation}}}}[/tex]

   [tex]\bigstar[/tex]Let p(x) = 0 be a quadratic equation , then the zeros of the polynomial p(x) are called the roots of the equation p(x) = 0.

   ⠀⠀[tex]\underline{\underline{\maltese\:\: \textbf{\textsf{General \: form \: of \: Quadratic \: Equation}}}}[/tex]

   ⠀⠀⠀⠀⠀⠀[tex]\underline{\bf{ax^{2} + bx + c = 0}}[/tex]

   Where,

   [tex]\bullet[/tex]a , b and c are real numbers respectively.

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