If the equation 2x^2 - kx + x + 8 = 0 has equ | vedclass

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

Question

(D) The given equation is $2x^2 - (k - 1)x + 8 = 0$.For equal and real roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.Here,$a = 2$,$b = -

Vedclass · Accepted Answer

$9$ and $-7$

Vedclass · Answer

(D) The given equation is $2x^2 - (k - 1)x + 8 = 0$.For equal and real roots,the discriminant $D = b^2 - 4ac$ must be equal to $0$.Here,$a = 2$,$b = -(k - 1)$,and $c = 8$.Substituting these values into $D = 0$:$(-(k - 1))^2 - 4(2)(8) = 0$$(k - 1)^2 - 64 = 0$$(k - 1)^2 = 64$Taking the square root on both sides:$k - 1 = \pm 8$Case $1$: $k - 1 = 8 \Rightarrow k = 9$Case $2$: $k - 1 = -8 \Rightarrow k = -7$Therefore,the values of $k$ are $9$ and $-7$.

If the equation $2x^2 - kx + x + 8 = 0$ has equal and real roots,then the value of $k$ will be .....

Similar Questions

Let $a_n, a_{n-1}, \ldots, a_1, a_0 \in \mathbb{C}$ and $f(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$ be a polynomial. If the polynomial $f(x)$ is monic,then:

If $f(x) \in \mathbb{Q}[x]$ is a non-zero polynomial such that all its roots are irrational,then the degree of $f(x)$ is

The equation $\sqrt{x + 1} - \sqrt{x - 1} = \sqrt{4x - 1}$ for $x \in R$ has:

Let $A, G$ and $H$ be the arithmetic mean,geometric mean and harmonic mean,respectively,of two distinct positive real numbers. If $\alpha$ is the smallest of the two roots of the equation $A(G-H) x^2 + G(H-A) x + H(A-G) = 0$,then:

For the equation $|x^2| + |x| - 6 = 0$,the roots are

Vedclass Test Series

Exam Paper Generator

Online Exam Module