Solve the following quadratic equation using | vedclass

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

Question

(B) To solve the quadratic equation $2x^2 + ax - a^2 = 0$ by factorization,we need to split the middle term $ax$ into two parts such that their produc

Vedclass · Accepted Answer

$-a, \frac{a}{2}$

Vedclass · Answer

(B) To solve the quadratic equation $2x^2 + ax - a^2 = 0$ by factorization,we need to split the middle term $ax$ into two parts such that their product is $(2x^2) 	imes (-a^2) = -2a^2x^2$ and their sum is $ax$.We can write $ax$ as $2ax - ax$.Substituting this into the equation: $2x^2 + 2ax - ax - a^2 = 0$.Grouping the terms: $2x(x + a) - a(x + a) = 0$.Factoring out the common term $(x + a)$: $(2x - a)(x + a) = 0$.Setting each factor to zero:$1$) $2x - a = 0 \implies 2x = a \implies x = \frac{a}{2}$.$2$) $x + a = 0 \implies x = -a$.Therefore,the solutions are $x = -a$ and $x = \frac{a}{2}$.

Solve the following quadratic equation using the method of factorization: $2x^2 + ax - a^2 = 0$.

Similar Questions

State whether the following statement is 'True' or 'False'. Justify your answer: If in a quadratic equation the coefficient of $x$ is zero,then the quadratic equation has no real roots.

Find the roots of the following quadratic equation by the method of completing the square: $4x^{2} + 20x + 23 = 0$.

Find the discriminant of the following quadratic equation and hence determine the nature of the roots of the equation: $5x^{2} - 4\sqrt{5}x + 4 = 0$.

If the following quadratic equation has two equal and real roots,then find the value of $k$: $(k+1)x^2 - 2(k-1)x + 1 = 0$.

Find the discriminant of the following quadratic equation: $x^{2}+5x+1=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module