15 x 2 − 1 = 2 x 15 x 2 − 2 x − 1 = 0 a = 15 , b = − 2 , c = − 1 Δ = b 2 − 4 a c = ( − 2 ) 2 − 4 ⋅ 15 ⋅ ( − 1 ) = 4 + 60 = 64 x 1 = 2 a − b − Δ = 2 ⋅ 15 − ( − 2 ) − 64 = 30 2 − 8 = 30 − 6 = − 5 1 x 2 = 2 a − b + Δ = 2 ⋅ 15 − ( − 2 ) + 64 = 30 2 + 8 = 30 10 = 3 1
15 x 2 − 1 = 2 x 15 x 2 − 2 x − 1 = 0 a = 15 ; b = − 2 ; c = − 1 Δ = b 2 − 4 a c → Δ = ( − 2 ) 2 − 4 ⋅ 15 ⋅ ( − 1 ) = 4 + 60 = 64 x 1 = 2 − b − Δ a ; x 2 = 2 a − b + Δ Δ = 64 = 8 x 1 = 2 ⋅ 15 2 − 8 = 30 − 6 = − 5 1 ; x 2 = 2 ⋅ 15 2 + 8 = 30 10 = 3 1
To solve the equation 15 x 2 − 1 = 2 x , we rearranged it to standard form and used the quadratic formula. The solutions found are x = − 5 1 and x = 3 1 . These solutions indicate the points where the parabola intersects the x-axis.
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