The real number k for which the equation 2x^2 | vedclass

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(D) Let $f(x) = 2x^2 + 3x + k$.For the quadratic equation $f(x) = 0$ to have two distinct real roots,the discriminant $D$ must be greater than $0$.$D

Vedclass · Accepted Answer

does not exist

Vedclass · Answer

(D) Let $f(x) = 2x^2 + 3x + k$.For the quadratic equation $f(x) = 0$ to have two distinct real roots,the discriminant $D$ must be greater than $0$.$D = b^2 - 4ac = 3^2 - 4(2)(k) = 9 - 8k > 0 \implies k For the roots to lie in the interval $[0, 1]$,the vertex of the parabola $x = -\frac{b}{2a} = -\frac{3}{4}$ must lie within $[0, 1]$.Since $-\frac{3}{4}$ is not in $[0, 1]$,it is impossible for both roots to lie in the interval $[0, 1]$.Thus,no such real number $k$ exists.

The real number $k$ for which the equation $2x^2 + 3x + k = 0$ has two distinct real roots in the interval $[0, 1]$.

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