If (2x + 3)(3x - 1) = 6x^2 + kx - 3,then find | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) To find the value of $k$,we expand the left side of the equation:$(2x + 3)(3x - 1) = 2x(3x) + 2x(-1) + 3(3x) + 3(-1)$$= 6x^2 - 2x + 9x - 3$$= 6x^2

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) To find the value of $k$,we expand the left side of the equation:$(2x + 3)(3x - 1) = 2x(3x) + 2x(-1) + 3(3x) + 3(-1)$$= 6x^2 - 2x + 9x - 3$$= 6x^2 + 7x - 3$Comparing this with the given equation $6x^2 + kx - 3$,we can see that the coefficient of $x$ is $k$ on the right side and $7$ on the expanded left side.Therefore,$k = 7$.

If $(2x + 3)(3x - 1) = 6x^2 + kx - 3$,then find the value of $k$.

Similar Questions

Without actually calculating the cubes,find the value of each of the following:
$(31)^{3} - (16)^{3} - (15)^{3}$

Factorise the following using appropriate identities: $x^{2} - \frac{y^{2}}{100}$

Find the following product:
$(x^{2}-1)(x^{4}+x^{2}+1)$

Expand $(2a + 3b)(2a - 5b)$.

Expand $(3x - 2)(3x - 6)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module