If the sum of the two roots of the equation 4 | vedclass

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

Question

(D) Given the cubic equation $4x^3 + 16x^2 - 9x - 36 = 0$. We can factorize the expression by grouping terms: $4x^2(x + 4) - 9(x + 4) = 0$. Taking $(x

Vedclass · Accepted Answer

$-4, \frac{3}{2}, -\frac{3}{2}$

Vedclass · Answer

(D) Given the cubic equation $4x^3 + 16x^2 - 9x - 36 = 0$. We can factorize the expression by grouping terms: $4x^2(x + 4) - 9(x + 4) = 0$. Taking $(x + 4)$ as a common factor: $(x + 4)(4x^2 - 9) = 0$. This gives the roots $x + 4 = 0$ or $4x^2 - 9 = 0$. For $x + 4 = 0$,we get $x = -4$. For $4x^2 - 9 = 0$,we get $x^2 = \frac{9}{4}$,which implies $x = \pm \frac{3}{2}$. The roots are $-4, \frac{3}{2}, -\frac{3}{2}$. Note that the sum of the two roots $\frac{3}{2}$ and $-\frac{3}{2}$ is zero,which satisfies the given condition. Thus,the roots are $-4, \frac{3}{2}, -\frac{3}{2}$.

If the sum of the two roots of the equation $4x^3 + 16x^2 - 9x - 36 = 0$ is zero,then the roots are

Similar Questions

Consider the equation $(1+a+b)^2=3(1+a^2+b^2)$ where $a, b$ are real numbers. Then,

The value of $\sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}$ is equal to:

If $x$ is real,then the minimum value of ${x^2} - 8x + 17$ is

The product of the roots of the equation $9x^{2}-18|x|+5=0$ is

Let $x_1, x_2, x_3 \in \mathbb{R} \setminus \{0\}$,$x_1 + x_2 + x_3 \neq 0$ and $\frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = \frac{1}{x_1 + x_2 + x_3}$. Then $\frac{1}{x_1^n + x_2^n + x_3^n} = \frac{1}{x_1^n} + \frac{1}{x_2^n} + \frac{1}{x_3^n}$ holds good for:

Vedclass Test Series

Exam Paper Generator

Online Exam Module