Examine whether the following equation is qua | vedclass

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

Question

(A) To determine if the equation $16x^2 - 3 = (2x + 5)(5x - 3)$ is quadratic,we first expand the right side:$(2x + 5)(5x - 3) = 2x(5x) + 2x(-3) + 5(5x

Vedclass · Accepted Answer

Yes,it is a quadratic equation.

Vedclass · Answer

(A) To determine if the equation $16x^2 - 3 = (2x + 5)(5x - 3)$ is quadratic,we first expand the right side:$(2x + 5)(5x - 3) = 2x(5x) + 2x(-3) + 5(5x) + 5(-3)$$= 10x^2 - 6x + 25x - 15$$= 10x^2 + 19x - 15$Now,substitute this back into the original equation:$16x^2 - 3 = 10x^2 + 19x - 15$Rearrange all terms to one side to set the equation to $0$:$16x^2 - 10x^2 - 19x - 3 + 15 = 0$$6x^2 - 19x + 12 = 0$Since this equation is in the form $ax^2 + bx + c = 0$ where $a 
eq 0$,it is a quadratic equation.

Part $I$	Part $II$
$1.$ The discriminant of $x^{2}+5x+6=0$	$a. 1$
$2.$ The discriminant of $x^{2}+5x+4=0$	$b. 9$
$3.$ The discriminant of $x^{2}+4x+3=0$	$c. 4$
$4.$ The discriminant of $x^{2}+6x+5=0$	$d. 16$

Examine whether the following equation is quadratic or not: $16x^2 - 3 = (2x + 5)(5x - 3)$

Similar Questions

Find the roots of $6x^{2} - \sqrt{2}x - 2 = 0$ by the factorisation of the corresponding quadratic polynomial.

Solve the following equation using the method of factorization and write its solution set: $\frac{x+1}{x-3}+\frac{x-3}{x+1}=\frac{5}{2}$

If $1$ is the solution of a quadratic equation $3x^{2}-kx+2=0$,then $k=\ldots$

Examine whether the following equation is quadratic or not: $x^{4}-5x^{2}+3x-1=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module