Question

4. If the sum of the zeroes of the quadratic polynomial 6x2 - kx + 3 is 2, then find the valueof k​

Answer 1

Answer:

The required value of k is 12.

Step-by-step explanation:

Given that,

The sum of the zeroes of the quadratic polynomial 6x² - kx + 3 is 2, and we need to find out the value of k.

So,

Let's consider that, α and β be the zeroes of the polynomial (6x² - kx + 3).

Now, this equation is in the form of ax² + bx + c, where,

• a = 6
• b = k
• c = 3.

Then we know that,

→ α + β = -b/a

But, α + β = 2 (given).

→ 2 = -(-k)/6

→ 2 = k/6

→ k = 6 * 2

→ k = 12.

Hence, the required value of k is 12.

Answer 2

${\pmb{\sf{\underline{Explanation...}}}}$

★ It is given that we have to find out the value of k if the sum of the zeroes of the quadratic polynomial is 6x² - Kx + 3 is equal to 2

• The value of k is 12.

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${\pmb{\sf{\underline{Knowledge...}}}}$

Some knowledge about Quadratic Equations -

★ Sum of zeros of any quadratic equation is given by ➝ α+β = -b/a

★ Product of zeros of any quadratic equation is given by ➝ αβ = c/a

★ Discriminant is given by b²-4ac

• Discriminant tell us about there are solution of a quadratic equation as no solution, one solution and two solutions.

★ A quadratic equation have 2 roots

★ ax² + bx + c = 0 is the general form of quadratic equation

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$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c}\\ \bf \bigstar \: Required \; Solution \\ \\ \star \: \sf{As \: given \: = 6x^2 - kx + 3 = 2} \\ \\ \star\sf \: {\alpha + \beta \: = -b/a} \\ \\ \star \sf \: {General \: form \: of \: Quadratic \: Equation \: is \: ax^2 + bx + c = 0} \end{array}}\end{gathered}\end{gathered}$

~ Now let's see what to do! Firstly, by using the general form of quadratic equation we get the following,

$\begin{gathered} \: \: \: \: \: \: \: \: \: \sf Here, \begin{cases} & \sf{a \: is \: \bf{6}} \\ \\ & \sf{b \: is \: \bf{k}} \\ \\ & \sf{c \: is \: \bf{3}} \end{cases}\\ \\\end{gathered}$

~ Now as we know that which formula we have to imply here, we have to use here that sum of zeros of any quadratic equation is given by ➝ α+β = -b/a. Butvas it's given that the sum of the zeroes of the quadratic polynomial is 6x² - Kx + 3 is 2. Henceforth, according to this statement:

${\small{\underline{\boxed{\sf{\alpha + \beta \: = -b/a}}}}} \\ \\ :\implies \sf \alpha + \beta \: = -b/a \\ \\ \dashrightarrow \sf According \: to \: the \: question, \\ \\ :\implies \sf 2 \: = -(-k)/6 \\ \\ :\implies \sf 2 \: = k/6 \\ \\ :\implies \sf 2 \times 6 = \: k \\ \\ :\implies \sf 12 \: = k \\ \\ :\implies \sf k \: = 12 \\ \\ \dashrightarrow \sf \therefore \: 12 \: is \: the \: value \: ok \: k$