\[ F(x) = 4x^2 - 3x + 2kx + 1 \]
What is the value of \( k \) for which the function has two zeros?
Can someone show me step by step?
Answer (3)
0\\\\\Delta=(2k-3)^2-4\cdot4\cdot1=4k^2-12k+9-16=4k^2-12k-7 > 0\\\\a_k=4;\ b_k=-12;\ c_k=-7\\\\\Delta_k=(-12)^2-4\cdot4\cdot(-7)=144+112=256\\\\k_1=\frac{-b_k-\sqrt{\Delta_k}}{2a_k};\ k_2=\frac{-b_k+\sqrt{\Delta_k}}{2a_k}"> f ( x ) = 4 x 2 − 3 x + 2 k x + 1 = 4 x 2 + ( 2 k − 3 ) x + 1 a = 4 ; b = 2 k − 3 ; c = 1 f u n c t i o n ha s tw o zeros w h e n Δ = b 2 − 4 a c > 0 Δ = ( 2 k − 3 ) 2 − 4 ⋅ 4 ⋅ 1 = 4 k 2 − 12 k + 9 − 16 = 4 k 2 − 12 k − 7 > 0 a k = 4 ; b k = − 12 ; c k = − 7 Δ k = ( − 12 ) 2 − 4 ⋅ 4 ⋅ ( − 7 ) = 144 + 112 = 256 k 1 = 2 a k − b k − Δ k ; k 2 = 2 a k − b k + Δ k
0\ (up\ parabola\ arms-see\ the\ picture)\\\\Answer:k\in(-\infty;-\frac{1}{2})\ \cup\ (\frac{7}{2};\ \infty)"> Δ k = 256 = 16 k 1 = 2 ⋅ 4 12 − 16 = 8 − 4 = − 2 1 ; k 2 = 2 ⋅ 4 12 + 16 = 8 28 = 2 7 a k = 4 > 0 ( u p p a r ab o l a a r m s − see t h e p i c t u re ) A n s w er : k ∈ ( − ∞ ; − 2 1 ) ∪ ( 2 7 ; ∞ )
The value of 'k' for which the given quadratic function has two zeros can be found by setting the determinant (b² - 4ac) of the quadratic equation greater than zero and solving for 'k'. ;
The value of k for the function to have two zeros is found by ensuring the discriminant is greater than zero. The quadratic inequality leads to the solution k ∈ ( − ∞ , − 2 1 ) ∪ ( 2 7 , ∞ ) . This means k must be either less than -0.5 or greater than 3.5.
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